contents 69 77 84

MATHEMATICS TODAY | NOVEMBER‘16 7

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Vol. XXXIV No. 11 November 2016Corporate Office:Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200e-mail : [email protected] website : www.mtg.inRegd. Office:406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor : Mahabir SinghEditor : Anil Ahlawat

CONTENTS

8

43

22

8170

69

77 84

41 66

315562

Clas

s XI

ICl

ass

XICo

mpe

titio

n Ed

ge

8 Concept Boosters

22 Brain@Work

31 Ace Your Way (Series 7)

41 MPP-5

43 Concept Boosters

55 Ace Your Way (Series 7)

62 Challenging Problems

66 MPP-5

69 Maths Musing Problem Set - 167

70 Mock Test Paper - WB JEE

77 JEE Work Outs

81 Math Archives

83 Maths Musing Solutions

84 You Ask We Answer

MATHEMATICS TODAY | NOVEMBER‘168

MEASURE OF ANGLESThere are three system for measuring angles1. Sexagesimal or English system : 1 right angle = 90 degree(= 90°) 1° = 60 minutes (= 60′) 1′ = 60 seconds (= 60′′)2. Centesimal or French system : 1 right angle = 100 grades (= 100g) 1 grade = 100 minutes (= 100′) 1 minute = 100 seconds (= 100′′)3. Circular system : The measure of an angle

subtended at the centre of a circle by an arc of length equal to the radius of the circle. 1 radian = 57°17 ′44.8′′ ≈ 57°17′45′′.If s is the length of an arc of a circle of radius r, then the angle θ (in radians) subtended by this arc at the centre of the circle is given by

θ θ= =sr

s ror

i.e., Length of arc = radius × angle in radians RELATIONSHIP BETWEEN DEGREE, GRADE AND RADIAN180° = 200g = π radian = 2 Right angles.

i.e. 1° = 109 180

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

g cπ and 1c = 57° 17′44.8′′ = 63g66′20′′

DOMAIN AND RANGE OF A TRIGONOMETRICAL FUNCTIONTrigonom-etrical Function

Domain Range

sinx R [–1, 1]cosx R [–1, 1]tanx

R n n I− + ∈⎧⎨⎩

⎫⎬⎭

( ) ,2 12π R

cosecx R – {nπ, n ∈ I} R – {x : –1 < x < 1}secx

R n n I− + ∈⎧⎨⎩

⎫⎬⎭

( ) ,2 12π R – {x : –1 < x < 1}

cotx R – {nπ, n ∈ I} R

TRIGONOMETRICAL RATIOS OF ALLIED ANGLESTwo angles are said to be allied when their sum or difference is either zero or a multiple of 90°.

Allied angles

Trigo. Ratio

sinθ cosθ tanθ

(–θ) –sinθ cosθ –tanθπ θ2−⎛

⎝⎜⎞⎠⎟

cosθ sinθ cotθ

This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.

*ALOK KUMAR, B.Tech, IIT Kanpur

* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91).He trains IIT and Olympiad aspirants.


π θ2

+⎛⎝⎜

⎞⎠⎟

cosθ –sinθ –cotθ

(π – θ) sinθ –cosθ –tanθ(π + θ) –sinθ –cosθ tanθ32π θ−⎛

⎝⎜⎞⎠⎟

–cosθ –sinθ cotθ

32π θ+⎛

⎝⎜⎞⎠⎟

–cosθ sinθ –cotθ

(2π – θ) –sinθ cosθ –tanθ(2π + θ) sinθ cosθ tanθ

TRIGONOMETRICAL RATIOS FOR VARIOUS ANGLES

θ 0° π/6 π/4 π/3 π/2 π 3π/2 2π

sinθ 0 1/2 1/ 2 3/2 1 0 –1 0

cosθ 1 3/2 1/ 2 1/2 0 –1 0 1

tanθ 0 1/ 3 1 3 ∞ 0 ∞ 0

SUM AND DIFFERENCES OF TWO ANGLESA + B) = sinAcosB + cosAsinBA – B) = sinAcosB – cosAsinBA + B) = cosAcosB – sinAsinBA – B) = cosAcosB + sinAsinB

tan( ) tan tantan tan

A B A BA B

+ = +−1

tan( ) tan tantan tan

A B A BA B

− = −+1

cot( ) cot cotcot cot

A B A BA B

+ = −+

1

cot( ) cot cotcot cot

A B A BB A

− = +−

1

A + B)sin(A – B) = sin2A – sin2B = cos2B – cos2AA + B)cos(A – B) = cos2A – sin2B = cos2B – sin2A

SUM AND DIFFERENCES OF THREE ANGLESA + B + C) = sinAcosBcosC + cosAsinBcosC

+ cosAcosB sinC –sinAsinBsinC = cosAcosBcosC(tanA + tanB + tanC – tanA· tanB · tanC)

A + B + C) = cosAcosBcosC – sinAsinBcosC – sinAcosBsinC – cosAsinBsinC = cosAcosBcosC(1 – tanAtanB – tanB tanC – tanCtanA)

A + B + C)

= + + −

− − −tan tan tan tan tan tan

tan tan tan tan tan tanA B C A B C

A B B C C A1A + B + C)

= − − −

+ + −cot cot cot cot cot cotcot cot cot cot cot cot

A B C A B CA B B C C A 1

In general : A1 + A2 + ... + An)

= cosA1cosA2 ... cosAn(S1 – S3 + S5 – S7 + ...) A1 + A2 + ... + An) = cosA1cosA2 ... cosAn(1 – S2

+ S4 – S6 + ...)

tan( .... )......

A A AS S S S

S S Sn1 21 3 5 7

2 4 61+ + + =

− + − +− + − +

where,S1 = tanA1 + tanA2 + .... + tanAn = The sum of the tangents of the separate angles.S2 = tanA1tanA2 + tanA1tanA3 + ... + tanAn–1tanAn = The sum of the tangent angles taken two at a time.S3 = tanA1tanA2tanA3 + tanA2tanA3tanA4 + ... = Sum of tangent angles taken three at a time, and so on.If A1 = A2 = ... = An = A, then S1 = ntanA, S2 = nC2tan2A, S3 = nC3tan3A,...

nA = cosnA(nC1tanA – nC3tan3A + nC5tan5A – ...)nA = cosnA(1 – nC2tan2A + nC4tan4A – ...)

tantan tan tan ....

tan tannA

C A C A C A

C A C A C

n n n

n n n=− + −

− + −1 3

35

5

22

441 66

6tan ...A +α) + sin(α + β) + sin(α + 2β) + .... +

sin(α + (n – 1)β) =+ − ⋅sin{ ( )( / )} sin( / )

sin( / )α β β

βn n1 2 2

2 cos(α) + cos(α + β) + cos(α + 2β) + ... +

cos(α + (n – 1)β) =+ − ⎛

⎝⎜⎞⎠⎟

⎧⎨⎩

⎫⎬⎭⋅ ⎛

⎝⎜⎞⎠⎟

⎧⎨⎩

⎫⎬⎭

⎛⎝⎜

⎞⎠⎟

cos ( ) sin

sin

α β β

β

n n12 2

2PRODUCT INTO SUM OR DIFFERENCE

AcosB= sin(A + B) + sin(A – B)AsinB = sin(A + B) – sin(A – B)AcosB = cos(A + B) + cos(A – B)AsinB = cos(A – B) – cos(A + B)

SUM AND DIFFERENCE INTO PRODUCTS

sin sin sin cosC D C D C D+ = + −22 2

sin sin cos sinC D C D C D− = + −22 2


cos cos cos cosC D C D C D+ = + −22 2

cos cos sin sinC D C D D C− = + −22 2

.

= − + −2

2 2sin sinC D C D

tan tan sin( )cos .cos

, ,A B A BA B

A n B m± = ± ≠ + ≠⎛⎝⎜

⎞⎠⎟

π π π2

cot cot sin( )sin sin

, ,A B B AA B

A n B m± = ± ≠ ≠ +⎛⎝⎜

⎞⎠⎟

π π π2

TRIGONOMETRIC RATIO OF MULTIPLE OF AN ANGLE

sin sin cos tantan

2 2 21 2A A A A

A= =

+A = 2cos2A – 1 = 1 – 2 sin2A

= − = −+

cos sin tantan

;2 22

211

A A AA

where A n≠ +( )2 14π

tan tantan

2 21 2A A

A=

−A = 3sinA – 4sin3A A = 4cos3A – 3cosA

tan tan tantan

3 31 3

3

2A A AA

= −−

A = 4sinA · cos3A – 4cosAsin3AA = 8cos4A – 8cos2A + 1

tan tan tantan tan

4 4 41 6

3

2 4A A AA A

= −− +

A = 16sin5A – 20sin3A + 5sinAA = 16cos5A – 20cos3A + 5cosA

M A X I M U M A N D M I N I M U M VA LU E O F a sinθ + b cosθLet a = rcosα ...(i) and b = rsinα ...(ii)Squaring and adding (i) and (ii), then a2 + b2 = r2 or,

r a b= +2 2

∴ asinθ + bcosθ = r(s inθcosα + cosθ sinα) = rsin(θ + α) Since, –1 ≤ sin(θ + α) ≤ 1 Then –r ≤ rsin(θ + α) ≤ r

Hence, sin cos− + ≤ + ≤ +a b a b a b2 2 2 2θ θ

Then the greatest and least values of asinθ + bcosθ are

a b a b2 2 2 2+ − +and respectively.

C O N D I T I O N A L T R I G O N O M E T R I C A L IDENTITIES1. If A + B + C = π, then

A + sin2B + sin2C = 4sinA sinB sinCA + sin2B – sin2C = 4cosA cosB sinCB + C – A) + sin(C + A – B)

+ sin(A + B – C) = 4sinA sinB sinCA + cos2B + cos2C = –1 – 4cosA cosB cosCA + cos2B – cos2C = –1 – 4sinA sinB cosC

sin sin sin cos cos cosA B C A B C+ + = 42 2 2

sin sin sin sin sin cosA B C A B C+ − = 42 2 2

cos cos cos sin sin sinA B C A B C+ + = +1 42 2 2

cos cos cos cos cos sinA B C A B C+ − = − +1 42 2 2

cossin sin

cossin sin

cossin sin

AB C

BC A

CA B

+ + = 2

2A + sin2B – sin2C = 2sinA sinB cosC2A + cos2B + cos2C = 1 – 2cosA cosB cosC2A + sin2B + sin2C = 1 – 2sinA sinB cosC

sin sin sin sin sin sin2 2 22 2 2

1 22 2 2

A B C A B C+ + = −

cos cos cos sin sin sin2 2 22 2 2

2 22 2 2

A B C A B C+ + = +

sin sin sin cos cos sin2 2 22 2 2

1 22 2 2

A B C A B C+ − = −

cos cos cos cos cos sin2 2 22 2 2

22 2 2

A B C A B C+ − =

A + tanB + tanC = tanA tanB tanCB cotC + cotC cotA + cotA cotB = 1

tan tan tan tan tan tanB C C A A B2 2 2 2 2 2

1+ + =

cot cot cot cot cot cotA B C A B C2 2 2 2 2 2

+ + =

2. If thenA B C+ + = π2

,

2A + sin2B + sin2C = 1 – 2sinA sinB sinC2A + cos2B + cos2C = 2 + 2sinA sinB sinCA + sin2B + sin2C = 4cosA cosB cosC


VALUES OF TRIGONOMETRICAL RATIOS OF SOME IMPORTANT ANGLES

sin15 3 12 2

° = −

=+3 1

2 2

tan cot15 2 3 75° = − = °

3 = tan75°

sin cos18 5 14

72° = − = °

cos sin18 10 2 54

72° = + = °

sin cos36 10 2 54

54° = − = °

cos sin36 5 14

54° = + = °

NAPOLIAN OR NA PIER’ S ANALO GY OR TANGENTS LAW

A B−⎛⎝⎜

⎞⎠⎟2

=a ba b

C−+

cot2

B C−⎛⎝⎜

⎞⎠⎟2

=b cb c

A−+

cot2

C A−⎛⎝⎜

⎞⎠⎟2

=c ac a

B−+

cot2

PROBLEMSSingle Correct Answer Type

1. Which of the following relation is correct ?(a) sin1 < sin1° (b) sin1 > sin1°

(c) sin1 = sin1° (d) π180

sin1 = sin1°

2. If sinθ + cosecθ = 2, the value of sin10θ + cosec10 is(a) 10 (b) 210 (c) 29 (d) 2

3. If sinθ = −45

and θ lies in the third quadrant, then

cos θ2

=

(a) 15

(b) − 15

(c) 25

(d) − 25

4. (m + 2)sinθ + (2m – 1)cosθ = 2m + 1,if

(a) tanθ = 34

(b) tanθ = 43

(c) tanθ =+

212

mm

(d) None of these

5. If sinx + siny = 3(cosy – cosx), then the value of sinsin

33

xy

is

(a) 1 (b) –1(c) 0 (d) None of these6. If tanθ + secθ = ex, then cosθ equals

(a) ( )e ex x+ −

2 (b)

2( )e ex x+ −

(c) ( )e ex x− −

2 (d) ( )

( )e ee e

x x

x x−+

−

−

7. If xx

+ =1 2cos ,α then xx

nn+ =1

(a) 2ncosα (b) 2ncosnα(c) 2isinnα (d) 2cosnα

8. 11

+ −+ +

=sin cossin cos

A AA A

(a) sin A2

(b) cos A2

(c) tan A2

(d) cot A2

9. If 2ycosθ = xsinθ and 2xsecθ – y cosecθ = 3, then x2 + 4y2 =(a) 4 (b) –4(c) 0 (d) None of these10. If x = secφ – tanφ, y = cosecφ + cotφ, then

(a) x yy

=+−

11

(b) x yy

=−+

11

(c) y xx

= −+

11

(d) None of these

11. If tansin

cosθ

φφ

=−x

x1and tan

sincos

,φθθ

=−y

y1 then

xy

=

(a) sinsin

φθ

(b) sinsin

θφ

(c) sincosφθ1−

(d) sin

cosθφ1−

12. If tanθ + sinθ = m and tanθ – sinθ = n, then (a) m2 – n2 = 4mn (b) m2 + n2 = 4mn

(c) m2 – n2 = m2 + n2 (d) m n mn2 2 4− =13. If cotθ + tanθ = m and secθ – cosθ = n, then which of the following is correct(a) m(mn2)1/3 – n(nm2)1/3 = 1(b) m(m2n)1/3 – n(mn2)1/3 = 1 (c) n(mn2)1/3 – m(nm2)1/3 = 1(d) n(m2n)1/3 – m(mn2)1/3 = 1


14. If sinx + sin2x = 1, then the value of cos12x + 3cos10x + 3cos8x + cos6x – 2 is equal to(a) 0 (b) 1 (c) – 1 (d) 215. If xsin3α + ycos3α = sinαcosα and xsinα – ycosα = 0, then x2 + y2 = (a) – 1 (b) ±1(c) 1 (d) None of these

16. If sin ,22 2 1

2θ =

+ +x yx

then x must be

(a) – 3 (b) – 2(c) 1 (d) None of these17. If tanθ – cotθ = a and sinθ + cosθ = b, then (b2 – 1)2(a2 + 4) is equal to (a) 2 (b) – 4 (c) 3 (d) 418. If tan2α tan2β + tan2β tan2γ + tan2γ tan2α + 2tan2α tan2β tan2γ = 1 then the value of sin2α + sin2β + sin2γ is(a) 0 (b) – 1(c) 1 (d) None of these19. cos1°+ cos2° + cos3°+.…+ cos180° equals(a) 1 (b) –1(c) 0 (d) None of the above20. If angle θ be divided into two parts such that the tangent of one part is k times the tangent of the other and φ is their difference, then sinθ =

(a) kk

+−

11

sinφ (b) kk−+

11

sinφ

(c) 2 12 1

kk−+

sinφ (d) None of these

21. Given that π α π< < 32

, then the expression

( sin sin ) cos4 2 44 2

4 2 2α α π α+ + −⎛⎝⎜

⎞⎠⎟

is equal to

(a) 2 (b) 2 + 4sinα(c) 2 – 4sinα (d) None of these22. tan100° + tan125° + tan100° tan125° =

(a) 0 (b) 12

(c) – 1 (d) 1

23. If (1 + sinA)(1 + sinB)(1 + sinC) = (1 – sinA) (1 – sinB)(1 – sinC), then each side is equal to(a) ±sinAsinBsinC (b) ±cosAcosBcosC(c) ±sinAcosBcosC (d) ±cosAsinBsinC24. If tanα = (1 + 2–x)–1, tanβ = (1 + 2x + 1)–1, then α + β equals(a) π/6 (b) π/4 (c) π/3 (d) π/2

25. The sum S = sinθ + sin2θ + ... + sinnθ, equals

(a) sin ( ) sin / sin12

1 12 2

n n+ ⋅ θ θ θ

(b) cos ( ) sin / sin12

1 12 2

n n+ ⋅ θ θ θ

(c) sin ( ) cos / sin12

1 12 2

n n+ ⋅θ θ θ

(d) cos ( ) cos / sin12

1 12 2

n n+ ⋅θ θ θ

26. The value of cot70° + 4cos70° is

(a) 13

(b) 3 (c) 2 3 (d) 12

27. tan20°tan40°tan60°tan80° = (a) 1 (b) 2 (c) 3 (d) 3 2/28. sin36°sin72°sin108°sin144° =(a) 1/4 (b) 1/16 (c) 3/4 (d) 5/16 29. If x = cos10°cos20°cos40°, then the value of x is

(a) 14

10tan ° (b) 18

10cot °

(c) 18

10cosec ° (d) 18

10sec °

30. tan9° – tan27° – tan63° + tan81° = (a) 1/2 (b) 2 (c) 4 (d) 8

31. sin sin sin sincos cos cos cos

3 5 7 93 5 7 9θ θ θ θθ θ θ θ

+ + ++ + +

=

(a) tan3θ (b) cot3θ (c) tan6θ (d) cot6θ

32. sin( ) cos( )sin( ) cos( )

B A B AB A B A

+ + −− + +

=

(a) cos sincos sin

B BB B

+−

(b) cos sincos sin

A AA A

+−

(c) cos sincos sin

A AA A−+

(d) None of these

33. If mtan(θ – 30°) = ntan(θ + 120°) then m nm n

+−

=

(a) 2cos2θ (b) cos2θ (c) 2sin2θ (d) sin2θ 34. The value of tan20° + 2tan50° – tan70° is equal to(a) 1 (b) 0(c) tan50° (d) None of these

35. If cos cos( )cos( )

,2B A CA C

= +−

then tanA, tanB, tanC are in

(a) A.P. (b) G.P.(c) H.P. (d) None of these


Multiple Correct Answer Type

36. If sinsin

tantan

θφ

θφ

⎛⎝⎜

⎞⎠⎟

= =2

3 then

(a) tanφ = 13

(b) tanφ = − 13

(c) tanθ = 3 (d) tanθ = − 3

37. For α π=7

which of the following hold(s) good?

(a) tanαtan2αtan3α = tan3α – tan2α – tanα(b) cosecα = cosec2α + cosec4α

(c) cos cos cosα α α− + =2 3 12

(d) 8cosαcos2αcos4α = 138. Which of the following quantities are rational?

(a) sin sin1112

512

π π⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

(b) cosec 910

45

π π⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

sec

(c) sin cos4 48 8π π⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

+

(d) 1 29

1 49

1 89

+⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

cos cos cosπ π π

39. For 0 ≤ x ≤ 2π then 22

1 22 2

cosec x y y− + ≤ is

(a) satisfied by exactly one value of y (b) satisfied by exactly two values of x (c) satisfied by x for which cosx = 0 (d) satisfied by x for which sinx = 040. If xcosα + ysinα = xcosβ + ysinβ = 2a and

22 2

1sin sinα β⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= then

(a) cosα + cosβ = cosαcosβ

(b) cos cosα β⋅ =++

4 2 2

2 2a y

x y

(c) cos cosα β+ =+

42 2

axx y

(d) none of these41. Which of the following is true for a ΔABC

(a) R abca b c

2 ≥+ +

⎛⎝⎜

⎞⎠⎟

(b) r + 2R = s if C = 90°

(c) sin sin sin2 2 2 3 32

A B C+ + ≤

(d) 2 1 1 1

1 2 3R r r r≤ + +

Comprehension TypeParagraph for Q. No. 42 to 44

Let cos ,cos ,cosπ π π7

37

57

are the roots of equation

8x3 – 4x2 – 4x + 1 = 0

42. The value of sec sec secπ π π7

37

57

⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟ is

(a) 2 (b) 4(c) 8 (d) None of these

43. The value of sin sin sinπ π π14

314

514

is

(a) 14

(b) 18

(c) 74

(d) 78

44. The value of cos cos cosπ π π14

314

514

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

is

(a) 14

(b) 18

(c) 74

(d) 78

Paragraph for Q. No. 45 to 47

sin sin cos cosα β α β+ = + =14

13

and

45. The value of sin(α + β) is

(a) 2425

(b) 1325

(c) 1213

(d) None of these

46. The value of cos(α + β) is

(a) 1225

(b) 725

(c) 1213

(d) None of these

47. The value of tan(α + β) is

(a) 257

(b) 2512

(c) 2513

(d) 247

Matrix-Match Type48. Match the following trigonometric ratios with the equations whose one of the roots is given

Column I Column IIA. cos20° P. x3 – 3x2 – 3x + 1 = 0B. sin10° Q. 32x5 – 40x3 + 10x – 1 = 0C. tan15° R. 8x3 – 6x – 1 = 0D. sin6° S. 8x3 – 6x + 1 = 0


49. Match the following :

Column I Column IIA. T h e m a x i m u m v a l u e o f

cos(2A + θ) + cos(2B + θ) (θ ∈ R and A, B are constants)

P. 2sin(A + B)

B. Maximum value ofcos2A + cos2B

( , , ,A B A B∈⎛⎝⎜

⎞⎠⎟

+02π is

constant)

Q. 2sec(A + B)

C. Minimum value ofsec2A + sec2B

( , , ,A B A B∈⎛⎝⎜

⎞⎠⎟

+04π

is constant)

R. 2cos(A + B)

D. Minimum value of tanθ θ+ − +cot cos( )2 2 2A B

(θ ∈ R, A, B are constants)

S. 2cos(A – B)


Column I Column IIA. If Δ = a2 – (b – c)2, where Δ is

the area of the triangle ABC, then tanA is equal to

P. 12

B. I n a Δ A B C , g i v e n t h a t tanA : tan B : tan C = 3 : 4 : 5, then the value of sinA sinB sinC is

Q. 815

C. Let f ( ) sin coscos

θ θ θθ

= − +1 2 22 2

and

α β π+ = 54

, then the value of

f(α) f(β) is

R. π4

S. 2 57

Integer Answer Type

51. If tan tan tanx y z2 3 5

= = and x + y + z = π,

tan2x + tan2y + tan2z = 38K

then K =

52. If tanα is an integral solution of 4x2 – 16x + 15 < 0 and cosβ is the slope of the bisector of the angle in the first quadrant between the x and y axis. Then sin(α + β) : sin(α – β) = 53. If sinθ + sin2θ + sin3θ = 1, then the value of cos6θ – 4cos4θ + 8cos2θ must be

54. In a ΔABC, a = 5, b = 4 and cos ,A B−( ) = 3132

then c must be55. In a triangle ABC, if r1, r2, r3 are the ex-radius thenbcr

acr

abr

k abc sa

sb

sc1 2 3 2

3+ + = + + −⎡⎣⎢

⎤⎦⎥Δ

then k is equal to

SOLUTIONS

1. (b) : The true relation is sin1 > sin1°Since, value of sinθ is increasing2. (d) : We have, sinθ + cosecθ = 2⇒ sin2θ + 1 = 2sinθ ⇒ sin2θ – 2sinθ + 1 = 0 ⇒ (sinθ – 1)2 = 0 ⇒ sinθ = 1

⇒ sin10θ + cosec10θ = + =( )( )

1 11

21010

3. (b) : Given that sinθ = − 45

and θ lies in the III quadrant.

⇒ = − = −cosθ 1 1625

35

cos cos /θ θ

21

21 3 5

215

= ± + = ± − = ±

Since θ/2 will be in III quadrant.

Hence, cos θ2

15

= − .

4. (b) : Squaring the given relation and putting tanθ = t( ) ( )( ) ( ) ( ) ( )m t m m t m m t+ + + − + − = + +2 2 2 2 1 2 1 2 1 12 2 2 2 2

⇒ − + + − − =3 1 4 6 4 8 02 2 2( ) ( )m t m m t m ⇒ − − + =( )[( ) ]3 4 1 2 02t m t m ,

which is true if t = =tanθ 43

or tanθ =−

212

mm

5. (b) : We have sinx + siny = 3(cosy – cosx)⇒ sinx + 3cosx = 3cosy – siny ...(i)⇒ rcos(x – α) = rcos(y + α),

where r = =10 13

, tanα

⇒ x – α = ±(y + α) ⇒ x = –y or x – y = 2αClearly, x = –y satisfies (i)

∴ =−

= −sinsin

sinsin

33

33

1xy

yy

6. (b) : We have, tanθ + secθ = ex ...(i)∴ secθ – tanθ = e–x ...(ii)From (i) and (ii), we get

2 2sec cos .θ θ= + ⇒ =+

−−e e

e ex x

x x


7. (d) : We have, xx

+ =1 2cosα

⇒ xx

22

21 4 2+ = −cos α

⇒ xx

22

21 2 2 1 2 2+ = − =( cos ) cosα α

Similarly, xx

nnn+ =1 2cos α

8. (c) : 11

+ −+ +

sin cossin cos

A AA A

=+

+

22

22 2

22

22 2

2

2

sin sin cos

cos sin cos

A A A

A A A

=+⎛

⎝⎜⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

=2

2 2 2

22 2 2

2

sin sin cos

cos cos sintan

A A A

A A AA

9. (a) : Given that 2ycosθ = xsinθ ...(i)and 2xsecθ – ycosecθ = 3 ...(ii)

⇒ − =2 3x ycos sinθ θ

⇒ 2xsinθ – ycosθ – 3sinθcosθ = 0 ...(iii)Solving (i) and (iii), we get y = sinθ and x = 2cosθNow, x2 + 4y2 = 4cos2θ + 4sin2θ = 410. (b) : We have, xy = (secφ – tanφ)(cosecφ + cotφ)

=

− +1 1sincos

.cos

sinφφ

φφ

⇒ + =− + − +

xy 11 sin cos sin cos sin cos

cos sinφ φ φ φ φ φ

φ φ

= − +1 sin coscos sin

φ φφ φ

...(i)

Also, x – y = (secφ – tanφ) – (cosecφ + cotφ)

= − − + = − − −1 1 2 2sincos

cossin

sin sin cos coscos sin

φφ

φφ

φ φ φ φφ φ

=− −sin cos

cos sinφ φ

φ φ1 ...(ii)

Adding (i) and (ii) we get, xy + 1 + (x – y) = 0

⇒ = −+

x yy

11

11. (b) : We have tansin

cosθ

φφ

=−x

x1⇒ xsinφ = tanθ – xcosφ tanθ

⇒ =+

xtan

sin cos tanθ

φ φ θ

=

+=

+sin

cos sin cos sinsin

sin( )θ

θ φ φ θθ

θ φ

Similarly, y xy

=+

∴ =sin

sin( ); sin

sin.

φθ φ

θφ

12. (d) : (m + n) = 2tanθ, (m – n) = 2sinθ∴ m2 – n2 = 4tanθ · sinθ ...(i)and 4 4 42 2mn = − = ⋅tan sin sin tanθ θ θ θ ...(ii)

From (i) and (ii), m n mn2 2 4− =13. (a) : As given,

1 1 2

tantan tan tan

θθ θ θ+ = ⇒ + =m m

⇒ sec2θ = mtanθ ...(i)and secθ – cosθ = n ⇒ sec2θ – 1 = nsecθ⇒ tan2θ = nsecθ⇒ tan4θ = n2sec2θ = n2 m tanθ {by (i)}⇒ tan3θ = n2m [∵ tanθ ≠ 0]⇒ tanθ = (n2m)1/3 ...(ii)Also, sec2θ = mtanθ = m(n2m)1/3 {by (i) and (ii)}⇒ m(mn2)1/3 – (n2m)2/3 = 1⇒ m(mn2)1/3 – n(nm2)1/3 = 114. (c) : We have, sinx + sin2x = 1or sinx = 1 – sin2x or sinx = cos2x∴ cos12x + 3cos10x + 3cos8x + cos6x – 2 = sin6x + 3sin5x + 3sin4x + sin3x – 2 = (sin2x)3 + 3(sin2x)2 sinx + 3(sin2x)(sinx)2

+ (sinx)3 – 2 = (sin2x + sinx)3 – 2 = (1)3 – 2 = – 1.15. (c) : We have, xsin3α + ycos3α = sinαcosα ... (i)and xsinα – ycosα = 0 ... (ii)From (i) and (ii), we get⇒ ycosα sin2α + ycos3α = sinα cosα⇒ ycosα{sin2α + cos2α} = sinα cosα⇒ ycosα = sinα cosα ⇒ y = sinα and x = cosαHence, x2 + y2 = sin2α + cos2α = 116. (d) : Since, sin2θ ≤ 1

∴ + + ≤x yx

2 2 12

1

x2 + y2 – 2x + 1 ≤ 0 (x – 1)2 + y2 ≤ 0 It is possible, iff x = 1 and y = 0, i.e., It also depends on value of y.17. (d) : Given that tanθ – cotθ = a ...(i)and sinθ + cosθ = b ...(ii)Now, (b2 – 1)2 (a2 + 4)= {(sinθ + cosθ)2 – 1}2 {(tanθ – cotθ)2 + 4}= [1 + sin2θ – 1]2 [tan2θ + cot2θ – 2 + 4]= sin22θ(cosec2θ + sec2θ)

= +⎡⎣⎢

⎤⎦⎥

=4 1 1 42 22 2sin cos

sin cosθ θ

θ θ


18. (c) : sin2α + sin2β + sin2γ

=+

++

++

tantan

tantan

tantan

2

2

2

2

2

21 1 1αα

ββ

γγ

=+

++

++

= = =

xx

yy

zz

x y z1 1 1

2 2 2( tan , tan , tan )Here, α β γ

= + + + + + + + + + ++ + +

( ) ( )( )( )( )

x y z xy yz zx xyz xy yz zx xyzx y z

21 1 1

= + + + + + + ++ + +

=11 1 1

1x y z xy yz zx xyzx y z( )( )( )

(∵ xy + yz + zx + 2xyz = 1)19. (b) : (cos1° + cos179°) + (cos2° + cos178°) + ... + (cos89° + cos91°) + cos90° + cos180° = –120. (a) : Let A + B = θ and A – B = φ.

Then tanA = ktanB or k AB

A BA B1

= =tantan

sin coscos sin

Applying componendo and dividendo

⇒ +−

= +−

= +−

kk

A B A BA B A B

A BA B

11

sin cos cos sinsin cos cos sin

sin( )sin( ))

sinsin

= θφ

⇒ = +−

sin sinθ φkk

11

21. (c) : Given that π α π< < 32

i.e., α is in third quadrant.

Now, ( sin sin ) cos4 2 44 2

4 2 2α α π α+ + −⎛⎝⎜

⎞⎠⎟

= + + ⋅ −⎛⎝⎜

⎞⎠⎟

( sin sin cos ) cos4 4 2 24 2

4 2 2 2α α α π α

= + + + −⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

4 2 12

2 2 2sin (sin cos ) cosα α α π α

= ± 2sinα + 2 + 2sinαOn taking –ve, answer is 2 and on taking +ve, answer

is 2 + 4sinα.But π α π< < 3

2, Hence answer is 2 – 4sinα because

sinα is –ve in third quadrant.

22. (d) : tan( )tan tan

tan tan100 125

100 1251 100 125

° + ° =° + °

− ° °

∴ ° =° + °

− ° °tan

tan tantan tan

225100 125

1 100 125

i e. .,tan tan

tan tan1

100 1251 100 125

=° + °

− ° °

i.e., tan100° + tan125° + tan100° tan125° = 1

23. (b) : Multiplying both sides by (1 – sinA)(1 – sinB)(1 – sinC), we have,(1 – sin2A)(1 – sin2B)(1 – sin2C) = (1 – sinA)2 (1 – sinB)2 (1 – sinC)2

⇒ (1 – sinA)(1 – sinB)(1 – sinC) = ±cosA cosB cosCSimilarly, (1 + sinA)(1 + sinB)(1 + sinC) = ±cosA cosB cosC

24. (b) : tan( ) tan tan

tan tanα β α β

α β+ = +

−1

⇒ + =+

++

−+

×+

+

+

tan( )

/

α β

1

1 12

11 2

1 11 1 2

11 2

1

1

xx

x x

⇒ + = + ⋅ + ++ + ⋅ + ⋅ −

+

+tan( )α β 2 2 2 2 11 2 2 2 2 2 2

x x x x

x x x x x

⇒ + = ⇒ + =tan( )α β α β π14

25. (a) : S = sinθ + sin2θ + sin3θ + ... + sinnθWe know, sinθ + sin(θ + β) + sin(θ + 2β) + ... n terms

= + −⎡⎣⎢

⎤⎦⎥

sin

sinsin ( )

n

n

β

β θ β2

2

12

Put β = θ, then S

n n

=

+sin .sin ( )

sin

θ θ

θ2

12

226. (b) : Now,

cot cos cos sin cossin

70 4 70 70 4 70 7070

° + ° = ° + ° °°

= ° + °°

= ° + °− °°

cos sinsin

cos sin( )sin

70 2 14070

70 2 180 4070

= ° + ° + °°

= ° ° + °°

sin sin sinsin

sin cos sinsin

20 40 4070

2 30 10 4070

= ° + °°

= ° °°

=sin sinsin

sin cossin

80 4070

2 60 2070

3

27. (c) : tan20° tan40° tan60° tan80°

=° ° ° °

° ° °sin sin sin tan

cos cos cos20 40 80 60

20 40 80Let Nr = (sin20° sin40° sin80°)

=

°° °

sin( sin sin )

202

2 40 80

= ° °− °sin (cos cos )20

240 120


= ° − ° +⎛

⎝⎜⎞⎠⎟

12

20 1 2 20 12

2sin sin

= ° − °⎛

⎝⎜⎞⎠⎟

=°

=12

20 32

2 20604

38

2sin sinsin

Now, we take Dr = cos20° cos40° cos80°

=

°

°=

°°

= °°

=sin

sinsin

sinsinsin

2 202 20

1608 20

208 20

18

3

3

∴ Hence tan tan tan //

20 40 80 3 81 8

° ° ° =

Therefore tan20° tan40° tan60° tan80° = 3 3 3⋅ =28. (d) : sin36° sin72° sin108° sin144°

= ° ° = ° °sin sin {( sin ) ( sin )}2 2 2 236 72 14

2 36 2 72

= − ° − °1

41 72 1 144{( cos ) ( cos )}

= − ° + °1

41 18 1 36{( sin ) ( cos )}

= − −⎛

⎝⎜⎞

⎠⎟+ +⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

= × =14

1 5 14

1 5 14

2016

14

516

29. (b) : x = cos10° cos20° cos40°

=

°° ° ° °1

2 102 10 10 20 40

sin[ sin cos cos cos ]

=

⋅ °° ° °1

2 2 102 20 20 40

sin[ sin cos cos ]

=

⋅ °° ° =

°°1

2 4 102 40 40 1

8 1080

sin[ sin cos ]

sin(sin )

=

°° = °1

8 1010 1

810

sincos cot

30. (c) : tan9° – tan27° – tan63° + tan81° = tan9° – tan27° – cot27° + cot9° = (tan9° + cot9°) – (tan27° + cot27°)

=

°−

°218

254sin sin

= °− °

° °⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

⋅ ° ⋅ °2 54 18

18 542

2 36 18sin sinsin sin

.cos sinsin118 54

4° °

=sin

31. (c) : sin sin sin sincos cos cos cos

3 5 7 93 5 7 9θ θ θ θθ θ θ θ

+ + ++ + +

=

+ + ++ + +

(sin sin ) (sin sin )(cos cos ) (cos cos )

3 9 5 73 9 5 7θ θ θ θθ θ θ θ

=

++

2 6 3 2 62 6 3 2 6

sin cos sin coscos cos cos cos

θ θ θ θθ θ θ θ

=

++

=2 6 32 6 3

6sin (cos cos )cos (cos cos )

tanθ θ θθ θ θ

θ

32. (b) : sin( ) cos( )sin( ) cos( )

B A B AB A B A

+ + −− + +

=

+ + °− −− + °− +

sin( ) sin( )sin( ) sin( )

B A B AB A A B

9090

=

+ ° °−°− °−

2 45 452 45 45

sin( )cos( )sin( )cos( )

A BA B

=

+ °°−

= +−

sin( )sin( )

cos sincos sin

AA

A AA A

4545

33. (a) : mn

=° +− °

tan( )tan( )

12030θ

θ

⇒ +−

=+ ° + − °+ ° − − °

m nm n

tan( ) tan( )tan( ) tan( )

θ θθ θ

120 30120 30

( B y c o m p o n e n d o a n d dividendo)

= + ° − ° + + ° − °+ °

sin( )cos( ) cos( )sin( )sin( )cos

θ θ θ θθ

120 30 120 30120 (( ) cos( )sin( )θ θ θ− ° − + ° − °30 120 30

=+ °

°= =

sin( )sin( )

cos/

cos2 90

1502

1 22 2

θ θθ

34. (b) : tan20° + 2tan50° – tan70°

= °

°− °

°+ °sin

cossincos

tan2020

7070

2 50

= ° °− ° °

° °+ °sin cos cos sin

cos costan20 70 20 70

20 702 50

= °− °

° + ° + °− °+ °sin( )

[cos( ) cos( )]tan20 70

12

70 20 70 202 50

= − °

° + °+ °2 50

90 502 50sin( )

cos costan

= − °

+ °+ °2 50

0 502 50sin

costan

= –2tan50° + 2tan50° = 035. (b) : We have,

cos cos( )cos( )

cos cos sin sincos cos sin sin

2B A CA C

A C A CA C A C

= +−

= −+

⇒ −+

= −+

11

11

2

2tantan

tan tantan tan

BB

A CA C

⇒ + − −1 2 2tan tan tan tan tan tanB A C A C B= − + −1 2 2tan tan tan tan tan tanB A C A C B⇒ = ⇒ =2 22 2tan tan tan tan tan tanB A C B A CHence, tan A, tan B and tan C will be in G.P.


36. (a, b, c, d) : Since, sinsin

tantan

θφ

θφ

⎛⎝⎜

⎞⎠⎟

=2

⇒ =sinsin

sinsin

sinsin

coscos

θφ

θφ

θφ

φθ

⇒ = ⇒ =sinsin

coscos

sin sinθφ

φθ

θ φ2 2

⇒+

=+

⇒+

=+

21

21

61 9

212 2 2 2

tantan

tantan

tantan

tantan

θθ

φφ

φφ

φφ

⇒ = = ±tan tan2 13

3φ θand

37. (a, b, c) :(a) 3α = 2α + α⇒ tan3α = tan(2α + α)⇒ tan3α – tan2α – tanα = tanα tan2α tan3α

(b) R.H.S = + =sin sinsin sin

sin cossin sin

4 22 4

2 32 4

α αα α

α αα α

= = =2 3

7 727

47

1sin cos

sin sin sin

π π

π π ααcosec

(c) cos cos cos cos cos cosα α α π π π− + = − +2 37

27

37

= 1/2(d) 8cosα cos2α cos4α = –138. (a, b, c, d) :

(a) sin sin sin cos1112

512 12 12

π π π π= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ⎛

⎝⎜⎞⎠⎟

= ∈12 6

14

sin π Q

(b) cosec cosec910

45 10 5

4π π π π⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= − ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= − ∈sec sec QQ

(c)

1 28 8

1 14

34

2 2− = − = ∈⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

sin cosπ π Q

(d) 29

2 29

2 49

18

2 2 2cos cos cosπ π π⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

= ∈Q

39. (a, b, c) : 2 1 1 22 2cosec x y( )− + ≤

⇒ cosec2x = 1 and y = 1 ⇒ = =x yπ π2

32

1, and

40. (a, c) : Since, α and β satisfy xcosθ + ysinθ = 2a⇒ (x2 + y2)cos2θ – 4axcosθ + (4a2 – y2) = 0

cos cos ,cos cosα β α β+ =+

⋅ = −+

4 42 2

2 2

2 2ax

x ya yx y

22 2

1 42 2

12 2sin sin sin sinα β α β⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= ⇒ =⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⇒ cosα + cosβ = cosα ⋅ cosβ

41. (a, b, c, d) : R r R RS

R abca b c

≥ ⇒ ≥ ⇒ ≥+ +

2 42

2 2Δ

If ∠C = 90° ⇒ a2 + b2 = c2, c = 2R

r + 2R = (s – c) tan C c2

+ = s – c + c = s

42. (b) : sec ,sec ,secπ π π7

37

57

are the roots of x3 – 4x2 – 4x + 8 = 0

sec sec secπ π π7

37

57

4⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

=

43. (b) : 8 4 4 1

87

37

57

3 2x x x

x x x

− − +

= −⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

cos cos cosπ π π ...(i)

Put x = 1 ⇒ ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=sin sin sinπ π π14

314

514

18

44. (d) : Put x = –1 in (i), we get

⇒ ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=cos cos cosπ π π14

314

514

78

(45 - 47) :45. (a) 46. (b) 47. (d) sinα + sinβ = 1/4 ...(i)

⇒ +⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

=22 2

1 4sin cos /α β α β

...(ii)

cosα + cosβ = 1/3 ...(iii)

⇒ +⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

=22 2

1 3cos cos /α β α β ...(iv)

Dividing (iv) by (ii), we have

tan α β+⎛⎝⎜

⎞⎠⎟

=2

34

⇒ + =

+⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

=×

+ ⎛⎝⎜⎞⎠⎟

=sin( )tan

tanα β

α β

α β

22

12

2 34

1 34

22 2

4425

And cos( )tan

tanα β

α β

α β+ =

− +⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

=− ⎛⎝⎜

⎞⎠⎟

+ ⎛

12

12

1 34

1 34

2

2

2

⎝⎝⎜⎞⎠⎟

=2725

48. A → R; B → S; C → P; D → Q(A) A A A= ° ⇒ = ° ⇒ =20 3 60 3 1

2cos

⇒ 8x3 – 6x – 1 = 0 where x = cos20°


(B) A = 10° ⇒ sin3A = 1/2⇒ 8x3 – 6x + 1 = 0 where x = sin10°(C) A = 15° ⇒ tan3A = 45°⇒ x3 – 3x2 – 3x + 1 = 0 where x = tan15°(D) A = 6° ⇒ sin5A = 1/2 ⇒ 32x5 – 40x3 + 10x – 1 = 0 where x = sin6°49. A → S; B → R; C → Q; D → P(A) cos(2A + θ) + cos(2B + θ) = 2cos(A + B + θ) cos(A – B) ≤ 2cos(A – B) (B) cos2A + cos2B = 2cos(A + B) cos(A – B) ≤ 2cos(A + B) (C) y = secx always concave up

∴ + ≥ +sec sec sec( )2 22

A B A B

(D) tan cot cos( )θ θ+ − +2 2 2A B

= + + +( tan cot ) sin ( )θ θ 2 24 A B ≥ 2sin(A + B)50. A → Q; B → S; C → P(A) We have, Δ = a2 – (b – c)2 ⇒ Δ = a2 – b2 – c2 + 2bc⇒ b2 + c2 – a2 = 2bc – Δ

⇒ 2 2 12

bc A bc bc Acos sin= − ⇒ 4cosA + sinA = 4

⇒ −⎛⎝⎜

⎞⎠⎟

+ =4 1 22

22 2

42sin sin cosA A A

⇒ = ⇒ =tan tanA A2

14

815

(B) Let tan A = 3k, tanB = 4k, tanC = 5k ∴ tanA + tanB + tanC = tanA tanB tanC

⇒ 12k = 60 k3 ⇒ =k 15

∴ = = =tan , tan , tanA B C35

45

5

∴ =sin sin sinA B C 2 57

(C) We have, f ( ) cos sin cos(cos sin ) tan

θ θ θ θθ θ θ

= −−

=+

2 22

11

2

2 2

∵ tan(α + β) = 1

⇒ f f( )tan

, ( )tan

αα

ββ

=+

=+

11

11

∴ =f f( ) ( )α β 12

51. (3) : Let tanx = 2t, tany = 3t, tanz = 5t

∑ = ⇒ =tan (tan tan tan )x x y z t2 13

tan2x + tan2y + tan2z = t2(4 + 9 + 25) = 38t2 ⇒ K = 3

52. (1) : 4x2 – 16x + 15 < 04x2 – 10x – 6x + 15 < 0 2x(2x – 5) – 3(2x – 5) < 0

⇒ 32

52

2< < ⇒ =x x

∴ tanα = 2, cosβ = 1

⇒ sin( )sin( )

tan tantan tan

α βα β

α βα β

+−

= +−

= +−

=2 02 0

1

53. (4) : We have, sinθ(1 + sin2θ) = 1 – sin2θ⇒ sinθ(2 – cos2θ) = cos2θSquaring both sides, we getsin2θ(2 – cos2θ)2 = cos4θ⇒ (1 – cos2θ)(4 – 4cos2θ + cos4θ) = cos4θ⇒ –cos6θ + 5cos4θ – 8cos2θ + 4 = cos4θ∴ cos6θ – 4cos4θ + 8cos2θ = 4

54. (6) : costan

tanA B

A B

A B−( ) =

− −⎛⎝⎜

⎞⎠⎟

+ −⎛⎝⎜

⎞⎠⎟

12

12

2

2

⇒ tan A B−⎛⎝⎜

⎞⎠⎟

=2

163

Use Napier’s analogy, we will get cot C2

963

=

Then cos ,C c= =18

6

55. (2) : rs a

rs b

rs c1 2 3=

−=

−=

−Δ Δ Δ, ,

Substitute this value in given equation and take abc common

L.H.S = − + − + −⎡⎣⎢

⎤⎦⎥

abc s aa

s bb

s ccΔ

= −⎡⎣⎢

⎤⎦⎥

∑abc saΔ

3

⇒ k = 2


DEFINITION OF A STRAIGHT LINEA straight line is a curve such that every point on

the line segment joining any two points on it lies on it.Every first degree equation in x, y represents a straight line.A first degree equation in x, y i.e., ax + by + c = 0 represents a line, it means that all points (x, y) satisfying ax + by + c = 0 lie along a line. Thus, a line is also defined as the locus of a point satisfying the conditions ax + by + c = 0, where a, b, c are constants.

ANGLE OF INCLINATION OF LINEThe angle of inclination of a line which crosses the x-axis is the smallest angle θ which the line makes with the positive direction of x-axis in anticlockwise direction. Note : (i) When two lines are parallel, they have the same

inclination.(ii) The inclination of a line which is parallel to x-axis

or coinciding with x-axis is 0°.(iii) If θ is the angle of inclination of the line, then

0 ≤ θ < π.SLOPE OF A LINEIf a line makes an angle θ (0 ≤ θ < π and θ ≠ π/2) with the positive direction of x-axis, then tan θ is called the slope or gradient of the line. Slope of a line is usually denoted by m.

Slope of the line ax + by + c = 0

is

coefficient ofcoefficient of

− = −ab

xy

Slope of the line joining the point A (x1, y1) and

B(x2, y2) is y yx x

yx

1 2

1 2

−−

=Difference of coordinatesDifference of coordinatess

ANGLE BETWEEN TWO LINESThe angle θ between the lines y = m1x + c1 and

y = m2x + c2 is given by tan ,θ =−

+m m

m m1 2

1 21

Two lines are parallel, if m1 = m2.Two lines are perpendicular, if m1m2 = –1.

INTERCEPT OF A LINE ON AXESIf a line cuts x-axis at (a, 0), then a is called the intercept of the line on x-axis.If a line cuts y-axis at (0, b), then b is called the intercept of the line on y-axis.

Note : (i) If a line is making equal intercept on the axes,

then its slope is –1.(ii) If a line is making equal length of intercept on the

axes, then its slope is ± 1.EQUATION OF LINES PARALLEL TO AXES(a) Equation of x-axis is y = 0 Equation of y-axis is x = 0(b) E q u at i on o f a l i n e p a r a l l e l t o x - a x i s i s

y = c (constant) Equation of the line parallel to x-axis and passing

through (α, β) is y = β.(c) Equation of a line parallel to y-axis is x = c. Equation of the line parallel to y-axis and passing

through (α, β) is x = α.

This article is a collection of shortcut methods, important formulas and MCQs along with their detailed solutions which provides an extra edge to the readers who are preparing for various competitive exams like JEE(Main & Advanced) and other PETs.

Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231


EQUATION OF STRAIGHT LINE IN DIFFERENT STANDARD FORMS(a) General Equation : Any equation of the form ax + by + c = 0, where a,

b, c are constants and a, b are not simultaneously zero (i.e., a2 + b2 ≠ 0), always represents a straight line.

(b) Slope-Intercept Form: If OL = c, ∠BAX = θ, Equation of line AB is y = mx + c,

where m = tanθ = slope of AB. If L lies on the negative side of OY then c is taken as negative.

Note : (i) The equation of a line

passing through the origin is y = mx, where m is the slope

of the line. (ii) If the line is parallel to x-axis, then m = 0.

(c) Point Slope Form: The equation of a line which passes through

the point (x1, y1) and has the slope ‘m ’ is y – y1 = m(x – x1)

(d) Two Point Form: Equation of the line joining the points A(x1, y1)

and B(x2, y2) is

(i) y y y yx x

x x y yy y

x xx x

− = −−

− −−

= −−1

1 2

1 21

1

1 2

1

1 2( ) or

(ii) y y y yx x

x x y yy y

x xx x

− = −−

− −−

= −−2

1 2

1 22

2

1 2

2

1 2( ) or

(e) Determinant Form: The equation of a line passing through two points

(x1, y1) and (x2, y2) can also be written in the

determinant form as x yx yx y

111

01 1

2 2

=

(f) Intercept Form: The equation of a line which cuts off intercepts

a and b from the x-axis and y-axis respectively, is xa

yb

ab+ = ≠1 0,( )

RELATION BETWEEN TWO LINES Two lines given by the equation a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are (where a1

2 + b12 ≠ 0 and

a22 + b2

2 ≠ 0)

(i) Intersecting, consistent and having unique solution

if aa

bb

1

2

1

2≠ (slopes are unequal)

(ii) Parallel and not coincident, inconsistent and having

no solution, if aa

bb

cc

1

2

1

2

1

2= ≠ (slopes are equal)

(iii) Perpendicular, consistent and having unique solution, if a1a2 + b1b2 = 0

(iv) Identical or coincident, consistent having infinitely

many solutions, if aa

bb

cc

1

2

1

2

1

2= = .

E Q UAT I O N O F L I N E S PA R A L L E L A N D PERPENDICULAR TO A GIVEN LINE

The general form of equation of lines parallel to line

ax + by + c = 0 is ax + by + k = 0 Equation of the line parallel to the line ax + by + c = 0 and passing through (α, β) is ax + by – (aα + bβ) = 0The general form for the equation of lines

perpendicular to ax + by + c = 0 is bx – ay + k = 0Equation of the line perpendicular to ax + by + c = 0 and passing through (α, β) is bx – ay – (bα – aβ) = 0

DISTANCE BETWEEN TWO PARALLEL LINESDistance between two parallel lines ax + by + c1 = 0

and ax + by + c2 = 0 is d c c

a b=

−

+

| |1 22 2

DISTANCE OF A POINTDistance ( d) of a point P(x1, y1) (not lying on the line) from the line L : ax + by + c = 0 is

d ax by c

a b=

+ +

+

| |1 12 2

Distance of line L : ax + by + c = 0 from the origin (0, 0) is

d c

a b=

+

| |2 2

EQUATIONS OF STRAIGHT LINES PASSING THROUGH A GIVEN POINT AND MAKING A GIVEN ANGLE WITH A GIVEN LINEThe equation of the straight lines which pass through a given point (x1, y1) and make a given angle α with the given straight line y = mx + c are

y – y1 = ± −mm

x xtantan

( )αα1 1∓


CONCURRENT LINESThe three lines aix + bi y + ci = 0, i = 1, 2, 3 are concurrent,

if a b ca b ca b c

1 1 1

2 2 2

3 3 3

0=

The equation of lines passing through the intersection of lines ax + by + c = 0, a′x + b′y + c′ = 0 must be of the form ax + by + c + λ (a′x + b′y + c′) = 0, where a, b, c, a′, b′, c′ are constant and λ is a parameter. POSITION OF A GIVEN POINT RELATED TO A GIVEN LINE Let the given line L(x, y) ≡ ax + by + c = 0 (where b ≠ 0 ) and the given point is P(x1, y1) (i) P l i e s a b o v e t h e l i n e a x + b y + c = 0 ,

if L x yb

( , )1 1 0>

(ii) P lies below the line ax + by + c = 0, if L x yb

( , )1 1 0<

(iii) P lies on the line ax + by + c = 0, if L(x1, y1) = 0POSITION OF TWO POINTS (x1, y1) AND (x2, y2) W.R.T. TO A GIVEN LINELet the given line be L ≡ ax + by + c = 0 and P (x1, y1) and Q(x2, y2) are two given points.(i) If ax1 + by1 + c and ax2 + by2 + c both are of the

same sign or L x yL x y

( , )( , )

1 1

2 20> or L(x1, y1) . L(x2, y2) > 0

Then the point P(x1, y1) and Q(x2, y2) lie on same side of line ax + by + c = 0.

(ii) If ax1 + by1 + c and ax2 + by2 + c both are of opposite

in sign or L x yL x y

( , )( , )

1 1

2 20<

or L(x1, y1) . L(x2, y2) < 0 Then the point P(x1, y1) and Q (x2, y2) lie on opposite

side of the line ax + by + c = 0.(iii) (a) The side of the line where origin lies is known

as origin side. (b) A point (α, β) will lie on origin side of the

line ax + by + c = 0, if aα + bβ + c and c have same sign.

(c) A point (α, β) will lie on non-origin side of the line ax + by + c = 0, if aα + bβ + c and c have opposite sign.

(iv) The point (x1,y1) lies in the angle between the lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 in

(a) The acute angle, if

a a b ba x b y c a x b y c

1 2 1 2

1 1 1 1 1 2 1 2 1 20+

+ +( ) + +( ) <

(b) The obtuse angle, if

a a b ba x b y c a x b y c

1 2 1 2

1 1 1 1 1 2 1 2 1 20+

+ +( ) + +( ) >

(v) The origin lies in the angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 in

(a) The acute angle, if a a b b

c c1 2 1 2

1 20+

<

(b) The obtuse angle, if a a b b

c c1 2 1 2

1 20+

>

ANGLE BISECTORConsider two linesL1 ≡ a1x + b1y + c1 = 0 and L2 ≡ a2x + b2 y + c2 = 0Let P(x, y) be any point on either of bisectors. ⊥r Distance of P from L1 = ⊥ r Distance of P from L2

∴+ +

+( )=

+ +

+( )| | | |a x b y c

a b

a x b y c

a b1 1 1

12

12

2 2 2

22

22

PROBLEMS

Single Correct Answer Type

1. ABC is a variable triangle with the fixed vertex C(1, 2) and A, B having the coordinates (cost, sint), (sint, –cost) respectively, where t is a parameter. The locus of the centroid of the ΔABC is(a) 3(x2 + y2) – 2x – 4y – 1 = 0(b) 3(x2 + y2) – 2x – 4y + 1 = 0(c) 3(x2 + y2) + 2x + 4y – 1 = 0(d) 3(x2 + y2) + 2x + 4y + 1 = 02. The lines x + 2y + 3 = 0, x + 2y – 7 = 0 and 2x – y – 4 = 0 are the sides of a square. Equation of the remaining side of the square can be (a) 2x – y – 14 = 0 (b) 2x – y + 8 = 0(c) 2x – y – 10 = 0 (d) 2x – y – 6 = 03. If in a ΔABC (whose circumcentre is at the origin), and a ≤ sinA, then for any point (x, y) inside the circumcircle of ΔABC


(a) | |xy < 18

(b) | |xy > 18

(c) 18

12

< <xy (d) none of these

4. If the vertices of a triangle are A(10, 4), B(–4, 9) and C(–2, –1), then the equation of its altitudes are(a) x – 5y + 10 = 0, 12x + 5y + 3 = 0 and 14x + 5y – 23 = 0(b) 14x – 5y + 10 = 0, x + 5y + 3 = 0 and 14x – 5y + 23 = 0(c) x – 5y + 1 = 0, 12x + 5y + 3 = 0 and x – 5y + 23 = 0(d) x – 5y + 10 = 0, 12x + 5y + 3 = 0 and 14x – 5y + 23 = 05. If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x secθ + y cosecθ = a and x cosθ – y sinθ = a cos2θ respectively, then the value of 4p1

2 + p22 is

(a) 4a2 (b) 2a2

(c) a2 (d) none of these6. If (1, 1) and (–3, 5) are vertices of a diagonal of a square, then the equations of its sides through (1, 1) are(a) 2x – y = 1, y – 1 = 0 (b) 3x + y = 4, x – 1 = 0(c) x = 1, y = 1 (d) none of these7. A family of lines is given by (1 + 2λ)x + (1 – λ)y + λ = 0, λ being the parameter. The line belonging to this family at the maximum distance from the point (1, 4) is (a) 4x – y + 1 = 0 (b) 33x + 12y + 7 = 0(c) 12x + 33y = 7 (d) none of these

8. If A(sin , / )α 1 2 and B( / , cos )1 2 α , –π ≤ α ≤ π, are two points on the same side of the line x – y = 0 then α belongs to the interval

(a) −⎛⎝⎜

⎞⎠⎟∪⎛⎝⎜

⎞⎠⎟

π π π π4 4 4

34

, , (b) −⎛⎝⎜

⎞⎠⎟

π π4 4

,

(c) π π4

34

,⎛⎝⎜

⎞⎠⎟

(d) none of these

9. If the point (cosθ, sinθ) does not fall in that angle between the lines y = |x – 1| in which the origin lies, then θ belongs to

(a) π π2

32

,⎛⎝⎜

⎞⎠⎟ (b) −⎛

⎝⎜⎞⎠⎟

π π2 2

,

(c) (0, π) (d) none of these

10. If a ray travelling along the line x = 1 gets reflected from the line x + y = 1, then the equation of line along which the reflected ray travel is(a) y = 0 (b) x – y = 1(c) x = 0 (d) none of these

11. The point P(2, 1) is shifted by 3 2 parallel to the line x + y = 1, in the direction of increasing ordinate, to reach Q. The image of Q by the line x + y = 1 is (a) (5, –2) (b) (–1, –2)(c) (5, 4) (d) (–1, 4)12. In triangle ABC; A(1, 1), B(4, –2), C(5, 5). Equation of the internal angle bisector of ∠A is (a) y = 1 (b) x – y = 1(c) y = 5 (d) x + y = 5


13. A line which makes an acute angle θ with the positive direction of x-axis is drawn through the point P(3, 4) to meet the line x = 6 at R and y = 8 at S, then(a) PR = 3 secθ (b) PS = 4 cosecθ

(c) PR PS+ = +2 3 42

( sin cos )sinθ θ

θ

(d) 9 16 12 2( ) ( )PR PS

+ =

14. If the lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent (a + b + c ≠ 0), then(a) a3 + b3 + c3 – 3abc = 0(b) a = – b (c) a = b = c(d) a2 + b2 + c2 – bc – ca – ab = 015. If 6a2 – 3b2 – c2 + 7ab – ac + 4bc = 0, then the family of lines ax + by + c = 0 is concurrent at (a) (–2, –3) (b) (3, –1)(c) (2, 3) (d) (–3, 1)16. If (α, α2) lies inside the triangle formed by the lines 2x + 3y – 1 = 0, x + 2y – 3 = 0, 5x – 6y – 1 = 0, then(a) 2α + 3α2 – 1 > 0 (b) α + 3α2 – 3 < 0(c) α + 2α2 – 3 < 0 (d) 6α2 – 5α + 1 > 017. A(1, 2) and B(7, 10) are two points. If P(x, y) is a point such that the angle APB is 60° and the area of the triangle APB is maximum, then which of the following is/are true?(a) P lies on any line perpendicular to AB(b) P lies on the right bisector of AB(c) P lies on the straight line 3x + 4y = 36(d) P lies on the circle passing through the points

(1, 2) and (7, 10) and having a radius of 10 units


18. If the points aa

aa

bb

bb

3 2 3 2

13

1 13

1−−−

⎛

⎝⎜

⎞

⎠⎟ −

−−

⎛

⎝⎜

⎞

⎠⎟, , , and

cc

cc

3 2

13

1−−−

⎛

⎝⎜

⎞

⎠⎟, , where a ≠ b ≠ c ≠ 1, lies on the line

lx + my + n = 0 , then(a) a + b + c = –m/l(b) ab + bc + ca = n/l

(c) abc m nl

= +

(d) abc – (bc + ca + ab) + 3(a + b + c) = 0Comprehension Type

Paragraph for Q. No. 19 to 21Let A(0, β), B(–2, 0) and C(1, 1) be the vertices of a triangle then19. Angle A of the triangle ABC will be obtuse if β lies in

(a) (–1, 2) (b) 2 52

,⎛⎝⎜

⎞⎠⎟

(c) −⎛⎝⎜

⎞⎠⎟∪ ⎛⎝⎜

⎞⎠⎟

1 23

23

2, , (d) none of these

20. If I1 is the interval of values of β for which A is obtuse and I2 be the interval of values of β for which A is largest angle of ΔABC, then(a) I1 = I2 (b) I1 is a subset of I2(c) I2 is a subset of I1 (d) none of these21. All the value of β for which angle A of the triangle ABC is largest lie in interval

(a) (–2, 1) (b) −⎛⎝⎜

⎞⎠⎟∪⎛⎝⎜

⎞⎠⎟

2 23

23

1, ,

(c) −⎛⎝⎜

⎞⎠⎟∪ ⎛⎝⎜

⎞⎠⎟

2 23

23

6, , (d) none of these

Paragraph for Q. No. 22 to 24Given the equations of two sides of a square as 5x + 12y – 10 = 0, 5x + 12y + 29 = 0. Also given a point M(–3, 5) lying on one of its sides.22. The number of possible squares must be(a) one (b) two(c) four (d) none of these23. The area of the square must be(a) 9 sq. units (b) 6 sq. units(c) 5 sq. units (d) none of these24. If the possible equations of the remaining sides is 12x – 5y + λ = 0, then λ cannot be

(a) 61 (b) 22(c) 100 (d) 36

Paragraph for Q. No. 25 to 27The vertex A of triangle ABC is (3, –1). The equations of median BE and angular bisector CF are 6x + 10y – 59 = 0 and x – 4y + 10 = 0. Then

25. Slope of the side BC must be(a) 1/9 (b) –2/9 (c) 1/7 (d) none of these

26. The equation of AB must be(a) x + y = 2 (b) x + 4y = 0(c) 18x + 13y = 41 (d) 23x – y = 70

27. The length of the side AC must be

(a) 71 (b) 83(c) 85 (d) none of these

Integer Answer Type

28. The number of integral values of m for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is

29. The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then the value

of

1 1

1 12 2

2 2

a b

p q

+

+ is

30. If λ : 1 is the ratio in which the line joining the points (2, 3) and (4, 1) divides the segment joining the points (1, 2) and (4, 3), then the value of λ is

31. If the quadrilateral formed by the lines ax + by + c = 0, a′x + b′y + c = 0, ax + by + c′ = 0, a′x + b′y + c′ = 0 have

perpendicular diagonals, then the value of a b

a b

2 2

2 2+

′ + ′ is

32. The lines x + y = | a | and ax – y = 1 intersect each other in the first quadrant. Then the minimum sum of two different integral value of a is

33. A ray of light along x y+ =3 3 gets reflected upon reaching x-axis and the equation of the reflected ray is ax + by + c = 0 then the value of bc is


Matrix-Match Type

34. Match the following:

Column-1 Column-II(A) If P(1, 1), Q(4, 2) and R(x, 0) be

three points such that PR + RQ is minimum, then x is equal to

(P) 0

(B) The area bound by the curves max {|x|, |y|} = 1 is equal to

(Q) 1

(C) The number of circles that touch all the three lines 2x – y = 5, x + y = 3 and 4x – 2y = 7 is equal to

(R) 2

(D) If the point (a, a) lies between the lines |x + y| = 6 then [|a|] can be equal to, ([⋅] represents the integral part)

(S) 3

(T) 4

35. Match the following:Column-1 Column-II

(A) If the point (x1 + t(x2 – x1), y1 + t(y2 – y1)) divides the join of (x1, y1) and (x2, y2) internally, then

(P)− < <4 2

35 2

6t

(B) Set of values of ‘t ’ for which point P(t, t2 – 2) lies inside the triangle f o r m e d b y t h e l i n e s x + y = 1, y = x + 1 and y = –1 is

(Q)1 13 1

2< < −| |t

(C) If P t t( ( / ), ( / ))1 2 2 2+ + be any point on a line, then the value of t for which the point P lies between parallel lines x + 2y = 1 and 2x + 4y = 15, is

(R) 0 < t < 1

(D) If the point (1, t) always remains in the interior of the triangle formed by the lines y = x, y = 0 and x + y = 4, then

(S) 4 53

5 26

< <t

SOLUTIONS

1. (b) : Let G(α, β) be the centroid in any position.

Then ( , ) cos sin , sin cosα β = + + + −⎛⎝⎜

⎞⎠⎟

13

23

t t t t

∴ = + + = + −α β13

23

cos sin , sin cost t t t

or 3α – 1 = cost + sint .... (i)and 3β – 2 = sint – cost ... (ii)On squaring and adding (i) and (ii), we get (3α – 1)2 + (3β – 2)2 = 2∴ The equation of the locus of the centroid is (3x – 1)2 + (3y – 2)2 = 2⇒ 3(x2 + y2) – 2x – 4y + 1 = 02. (a) : Distance between x + 2y + 3 = 0 and

x + 2y – 7 = 0 is 105

Let the remaining side be 2x – y + λ = 0

We have, | |λ + =45

105

⇒ λ = 6, –14

Thus remaining side is 2x – y + 6 = 0 or 2x – y – 14 = 0

3. (a) : Since, a ≤ sinA ⇒ ≤aAsin

1

⇒ 2R ≤ 1 ⇒ R ≤ 1/2So, for any point (x, y) inside the circumcircle,

x y2 2 14

+ <

Using AM ≥ GM, x y xy2 2

2+ ≥

⎛

⎝⎜

⎞

⎠⎟| | ⇒ <| |xy 1

84. (d) : Let AD, BE and CF be three altitudes of ΔABC.Slope of BC = –5 ⇒ Slope of AD = 1/5Since AD passes through A(10, 4).∴ Equation of AD is x – 5y + 10 = 0 Similarly the equation of BE is 12x + 5y + 3 = 0and the equation of CF is 14x – 5y + 23 = 0

5. (c) : We have, p a1 2

= −

+

| |

sec θ θcosec2

⇒ =+

p a12

2

2sec θ θcosec2 = a2 2 2

1sin cosθ θ

⇒ 4p12 = a2sin22θ ...(i)

Also, p a a2 2 22 2= −

+= −| cos |

cos sin| cos |θ

θ θθ

⇒ p22 = a2cos22θ ...(ii)

On adding (i) and (ii), we get 4p12 + p2

2 = a2

6. (c) : Since each side of a square makes 45° angle with its diagonals, so equations of the sides through

(1, 1) are given by y x− = − °± − °

−1 1 451 1 45

1∓ tan( )tan

( )

[∵ Slope of diagonal, m = –1]

⇒ − = − −y x1 1 11 1

1∓∓

( ) ⇒ x – 1 = 0, y – 1 = 0.


7. (c) : Here, x + y + λ(2x – y + 1) = 0

(1, 4)

(–1/3, 1/3)

⇒ Each line passes through the point of intersection of the lines x + y = 0 and 2x – y + 1 = 0,

which is −⎛⎝⎜

⎞⎠⎟

13

13

, .

The required line passes through −⎛⎝⎜

⎞⎠⎟

13

13

, and is

perpendicular to the line joining (1, 4) and −⎛⎝⎜

⎞⎠⎟

13

13

,

∴ Slope of the required line = − 411

∴ The required line is y x− = − +⎛⎝⎜

⎞⎠⎟

13

411

13

or 12x + 33y = 7.

8. (a) : sinα− >12

0 and 12

0− >cosα ....(i)

or andsin cosα α− < − <12

0 12

0

.... (ii)

(i) ⇒ >sinα α12

12

and cos < ⇒ < <π α π4

34

(ii) ⇒ <sinα α12

12

and cos > ⇒ − < <π α π4 4

9. (b) : The lines are

= – 1

(cos , sin )

/4

/4= – + 1

y = x – 1 and y = –x + 1 and (cosθ, sinθ) is any point on the circle x2 + y2 = 1, where centre = (0, 0) and radius = 1.Clearly from the figure, θ can vary from

− π π2 2

to .

10. (a) :

Clearly from the figure, the reflected ray moves along the x-axis.11. (d) : Q = ± ±( cos , sin ),2 3 2 1 3 2θ θ where tanθ = –1

Q = ± × − ± ×⎛

⎝⎜⎞⎠⎟

= ±2 3 2 12

1 3 2 12

2 3 1 3, ( , )∓

From the question, the y-coordinate of Q should be more than 1( = y-coordinate of P).

12. (a) : mAB = − −−

= −2 14 1

1

and mAC = −−

=5 15 1

1

⇒ mAA1 = 0

Hence, equation of required bisector is, y = 1.13. (a, b, c, d) : Equation of any line through P(3, 4) making an angle θ with the positive direction of x-axis

is x y r− = − =3 4cos sinθ θ

...(i)

where r is the distance of any point on the line from P. Therefore, coordinates of any point on the line (i) are (3 + r cosθ, 4 + r sinθ) ...(ii)If (ii) represents R, then

3 + r cosθ = 6 ⇒ = =r PR3cosθIf (ii) represents S, then

4 + r sinθ = 8 ⇒ = =r PS4sinθ

Hence, PR = 3secθ, PS = 4cosecθ

⇒ + = + = +PR PS 3 4 2 3 42

sin cossin cos

( sin cos )sin

θ θθ θ

θ θθ

and 3 4 1

2 22 2

PR PS⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

= + =cos sinθ θ

⇒ + =9 16 12 2( ) ( )PR PS 14. (a, c, d)15. (a, b) : Given, 6a2 – 3b2 – c2 + 7ab – ac + 4bc = 0⇒ 6a2 + (7b – c)a – (3b2 – 4bc + c2) = 0

⇒ = − ± − + − +a c b b c b bc c7 7 24 3 412

2 2 2( ) ( )

⇒ 12a + 7b – c = ±(11b – 5c)⇒ 3a – b + c = 0 or 2a + 3b – c = 0⇒ (3, –1) or (–2, –3) lies on line ax + by + c = 016. (a, c, d) : O and the point (α, α2) lie to the opposite sides w.r.t. 2x + 3y – 1 = 0⇒ 2x + 3y – 1 = –1 < 0 ⇒ 2α + 3α2 – 1 > 0O and the point (α , α2) lie to the same side w.r.t x + 2y – 3 = 0

(,

)2

2 + 3 – 1 = 05 – 6 – 1 = 0

+ 2 – 3 = 0


x + 2y – 3 = –3 < 0 ⇒ α + 2α2 – 3 < 0Again O and the point (α, α2) lie to the same side w.r.t 5x – 6y – 1 = 0 ⇒ 5x – 6y – 1 = –1 < 0⇒ 5α – 6α2 – 1 < 0 ⇒ 6α2 – 5α + 1 > 017. (b, c) : For the area tobe maximum, P should lie on the right bisector of the side AB.Coordinates of mid-point of AB are (4, 6)

Slope of AB = 86

43

=

∴ Slope of perpendicular

bisector = − 34

Thus equation of line on which P lies is 3x + 4y = 3618. (a, b, d) : Since the given points lie on the line lx + my + n = 0 and a, b, c are the roots of the equation

l t

tm t

tn

3 2

13

10

−

⎛

⎝⎜

⎞

⎠⎟ + −

−

⎛

⎝⎜

⎞

⎠⎟ + =

or l t3 + mt2 + nt – (3m + n) = 0 ... (i)⇒ a + b + c = –m/l ; ab + bc + ca = n/l ... (ii)

and abc m nl

= +3 ... (iii)

So, that from (i), (ii) and (iii) we get abc – (bc + ca + ab) + 3(a + b + c) = 019. (c) : Angle A will be obtuse if BC2 > AB2 + AC2 and A, B, C are non-collinear.⇒ 10 > β2 + 4 + 1 + (β – 1)2

⇒ (β – 2)(β + 1) < 0 ⇒ β ∈ (–1, 2)But A, B, C are collinear for β = 2/3

⇒ Correct interval for β is −⎛⎝⎜

⎞⎠⎟∪⎛⎝⎜

⎞⎠⎟

1 23

23

2, ,

20. (b) : Whenever A is obtuse, it is the largest angle also. But there may be more values of β for which A is largest but is not obtuse. (i.e., right angle at A)⇒ Choice (b) is correct.21. (c) : Angle A will be largest if a > b, a > c and A, B, C are non-collinear⇒ a2 > b2, a2 > c2, β ≠ 2/3 ⇒ 10 > 1 + (β – 1)2, 10 > 4 + β2

⇒ β2 – 2β – 8 < 0, β2 < 6⇒ –2 < β < 4, − < <6 6β On taking intersection and rejecting β = 2/3, we get

β∈ −⎛

⎝⎜⎞⎠⎟∪ ⎛⎝⎜

⎞⎠⎟

2 23

23

6, ,

(22-24) :

22. (b) : It is evident that M(–3, 5) lies on some other side and possible squares are obviously two.23. (a) : The perpendicular distance between AB and CD

is 29 10 3913

32 2− −

+= =( )

a b ⇒ Area = (3)2 = 9 sq. units

24. (d) : The equations of remaining sides can easily be determined as 12x – 5y + 61 = 0, 12x – 5y + 22 = 0, 12x – 5y + 100 = 0⇒ (d) is correct.Equation of BC is 12x – 5y + 61 = 0⇒ Other possible sides are 12x – 5y + c = 0, where

| |c − =6113

3 ⇒ c = 100, c = 22

(25 - 27) :

(3, 1)–

( , )

6+

10– 59 =

0

4+ 10 = 0

–

25. (b) : Let coordinates of C be (h, k), then h – 4k + 10 = 0 ....(i)

and E h k+ −⎛⎝⎜

⎞⎠⎟

32

12

, lies on 6x + 10y – 59 = 0

⇒ +⎛⎝⎜

⎞⎠⎟

+ −⎛⎝⎜

⎞⎠⎟− =6 3

210 1

259 0h k

⇒ 3h + 5k = 55 ....(ii)From (i) and (ii), we get h = 10, k = 5 ⇒ Coordinates of C are (10, 5)

⇒ Slope of AC = 67

⇒ Slope of BC must be given by

m

m

−

+=

−

+ ×

14

14

67

14

1 67

14

⇒ = −m 67

29

,


Since 6/7 represents slope, we have slope BC = − 29

26. (c) : Equation of BC is y x− = − −5 29

10( )or 2x + 9y – 65 = 0Now the point B can be found by solving BC and BE whence equation of AB can be determined as 18x + 13y = 4127. (c) 28. (2) : 3x + 4(mx + 1) = 9

⇒ + = ⇒ =+

( )3 4 5 53 4

m x xm

For x to be an integer, 3 + 4m should be a divisor of 5i.e., 1, –1, 5 or –5. 3 + 4m = 1 ⇒ m = – 1/2 (not an integer) 3 + 4m = –1 ⇒ m = –1 (integer) 3 + 4m = 5 ⇒ m = 1/2 (not an integer) 3 + 4m = –5 ⇒ m = –2 (integer)Hence, there are two integral values of m.29. (1) : The equation of the line L in the two coordinate

systems are xa

yb

Xp

Yq

+ = + =1 1 and where (X, Y) are

the new coordinates of a point (x, y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed,

11 1

11 1

1 1 1 1

2 2 2 2

2 2 2 2

a b p qa b p q+

=+

⇒ + = +

⇒+

+=

1 1

1 1 12 2

2 2

a b

p q30. (1) : The equation of the line joining the points (2, 3) and (4, 1) is

y x− = −−

−3 1 34 2

2( ) ⇒ x + y – 5 = 0 ….(i)

Suppose the line joining (2, 3) and (4, 1) divides the segment joining (1, 2) and (4, 3) at the point P in the ratio of λ : 1.Then the coordinates of P are 4 1

13 2

1λλ

λλ

++

++

⎛⎝⎜

⎞⎠⎟,

Clearly, P lies on (i),

⇒ ++

+ ++

− = ⇒ =4 11

3 21

5 0 1λλ

λλ

λ

31. (1) : Since the diagonals are perpendicular, so the given quadrilateral is a rhombus.∴ Distance between two pairs of parallel sides are equal.

⇒ ′ −

+= ′ −

′ + ′

c c

a b

c c

a b2 2 2 2 ⇒ a2 + b2 = a′ 2 + b′ 2

32. (3) : Given lines intersect at the point

P a

aa a

a11

11

++

−+

⎛⎝⎜

⎞⎠⎟

| | , | |

This point will lie in first quadrant iff 1 + a > 0 and a | a | – 1 ≥ 0⇒ a > – 1 and a | a | – 1 ≥ 0⇒ a2 – 1 ≥ 0 if a ≥ 0 or – a2 – 1 > 0 if 0 < a < – 1 ⇒ a ∈ [1, ∞)33. (3) : The line x + 3 3y = cuts y-axis at P(0, 1). Clearly its image Q(0, –1) lies on the reflected ray AR produced backward.

So, the equation of the reflected ray AR is

y + 1 = 0 13 0+−

(x – 0) or 3 3y x= −

34. (A) – (R); (B) – (T) ; (C) – (R) ; (D) – (P,Q,R)(A) The image of P in the x-axis

is P′(1, –1). Equation of P ′Q is y + 1 = 3

3(x – 1), which

intersects x–axis at (2, 0).(B) The region is a square of side length 2, so the

desired area is 4.

(C) As two of the lines are parallel, so two such circles are possible.

(D) The line x = y cuts the linesx + y = ± 6 at (–3, –3) and (3, 3) ⇒ – 3 < a < 3.∴ [| a |] = 0, 1, 2.35. (A) – (R) ; (B) – (Q) ; (C) – (P) ; (D) – (R)


CLASS XI Series 7

DEFINITIONA path traced by a point in a constant direction and endlessly in its opposite direction is called straight line.

Distance FormulaDistance between A(x1, y1) and B(x2, y2) is

AB x x y y= − + −( ) ( )2 12

2 12

Section Formula

InternalDivision

Coordinates of M, if M divides the join of A(x1, y1) and B(x2, y2) internally

in the ratio m : n is mx nxm n

my nym n

2 1 2 1++

++

⎛⎝⎜

⎞⎠⎟,

ExternalDivision

Coordinates of M, if M divides the join the of A(x1, y1) and B(x2, y2)

externally in the ratio m : n is mx nxm n

my nym n

2 1 2 1−−

−−

⎛⎝⎜

⎞⎠⎟,

Mid Point FormulaCoordinates of M, if M bisects the join of A(x1, y1) and B(x2, y2) is

x x y y1 2 1 22 2+ +⎛

⎝⎜⎞⎠⎟,

Area of a TriangleArea of ΔABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is

Δ = 1 − + − + −2 1 2 3 2 3 1 3 1 2x y y x y y x y y( ) ( ) ( )

SLOPE OF A LINEIf a line makes an angle θ with the positive direction of x-axis, then tan θ is called the slope or gradient of the line. It is denoted by m.

Slope of a line Value of mParallel to x-axis 0Parallel to y-axis not definedPassing through two points (x1, y1) and (x2, y2)

(y2 – y1)/(x2 – x1)

Straight Lines | Conic Sections | Introduction to Three Dimensional Geometry

STRAIGHT LINES

CONDITION OF PARALLELISM AND PERPENDICULARITYTwo lines having slopes m1 and m2 are said to be(i) Parallel iff m1 = m2(ii) Perpendicular iff m1m2 = –1ANGLE BETWEEN TWO LINESAngle between two lines having slopes m1 and m2 is,

tan ,θ =−

++ ≠

m mm m

m m1 2

1 21 21

1 0


COLLINEARITY OF THREE POINTSThree points A, B, C are said to be collinear, if slope of AB = slope of BCDIFFERENT FORMS OF EQUATION OF LINE

Equation Figure

Equation of x-axis y = 0

Equation of y-axis x = 0

Equation of line parallel to x-axis y = a

Equation of line parallel to y-axis x = b

Point slope form : Equation of line whose slope is m and passing through the point (x1, y1)

y – y1 = m (x – x1)

Two point form : Equation of line passing through A(x1, y1) and B(x2, y2)

y – y1 = y yx x

2 1

2 1

−−

(x – x1)

Slope intercept form : Equation of line having slope m and cuts an intercept c on y-axis.

y = mx + c

Intercept form : Equation of line cuts intercepts a and b on x and y-axes respectively.

xa

yb

+ = 1

Normal form : Equation of line having normal distance from origin p and this normal makes an angle α with the +ve x-axis.

xcosα + ysinα = p

Note : If a line with slope m makes x-intercept d. Then equation of the line is y = m (x – d).

GENERAL EQUATION OF A LINEAn equation of the form Ax + By + C = 0, where A, B, C are constants and A, B are not simultaneously zero is called the general equation of a line.

Different forms of Ax + By + C = 0

Slope - intercept form : y AB

x CB

B= − − ≠, 0

Intercept form : x

C Ay

C BC

−+−

= ≠/ /

,1 0

Normal form : x cos α + y sin α = p


where cos α = ±+

= ±+

A

A B

B

A B2 2 2 2, sinα

p C

A B= ±

+2 2

Distance of a point from a line : Distance of a point (x1, y1) from a line Ax + By + C = 0 is,

dAx By C

A B=

+ +

+

1 12 2

Distance between two parallel lines : Distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0

is, dC C

A B=

−

+

2 12 2

E Q UAT I O N O F FA M I LY O F L I N E S PA S S I N G THROUGH THE POINT OF INTERSECTION OF TWO LINESLet the two intersecting lines be A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0So, equation of line passing through intersection of above lines is A1x + B1y + C1 + λ(A2x + B2y + C2) = 0 where λ is an arbitrary constant called parameter.

SHIFTING OF ORIGINTranslation of axes An equation correspondingto a set of points with reference to a system of coordinate axes may be simplified by taking the s e t o f p oi nt s i n s om e other suitable coordinatesystem such that all geometric properties remain unchanged. We can form a transformation in which the new axes can be transformed parallel to the original axes and origin can be shifted to a new point. This kind of transformation is called a translation of axes. The coordinates of each point of the plane are changed under a translation of axes. To see how coordinates of a point of the plane changed under a translation of axes, let us take a point P(x, y) referred to the axes OX and OY. Let O′X′ and O′Y′ be new axes parallel to OX and OY respectively, where O′ is the new origin. Let (h, k) be the coordinates of O′ referred to the old axes, i.e., OL = h and LO′ = k. Also, OM = x and MP = yThe transformation relation between new coordinates (x′, y′) and old coordinates (x, y) are given by

x = x′ + h, y = y′ + k

DEFINITIONThe locus of a point which moves in a plane such that the ratio of its distance from a fixed point to its perpendicular distance from a fixed straight line is always constant, is known as a conic section or a conic.The f ixed point is cal led focus of the conic, f ixed line is called directrix of the conic and the constant ratio is called eccentricity of the conic.CIRCLE, PARABOLA, ELLIPSE AND HYPERBOLAWhen the plane cuts the nappe (other than the vertex) of the cone, we have the following situations.Relationship/Value of angles

Conic α : Semi ver t ica l angle of right circular cone. β : Angle made by the intersecting plane with the vertical axis of the cone.

β = 90° Circleα < β < 90° Ellipseβ = α Parabola0 < β < α Hyperbola

Degenerated Conic Section :When the plane cuts at the vertex of the cone, we have the following situations.Relationship/Value of angles

Conic Degene-rated conic

α : Semi vertical angle of r ight circular cone. β : Angle made by the intersecting plane with the vertical axis of the cone.

β < β < 90° Pointβ = α Straight

lineParabola

0 < β < α Pair of inter-secting straight lines

Hyper-bola

CIRCLEA circle is the locus of a point which moves in a plane so that its distance from a fixed point remains constant. The fixed point and the constant distance are called centre and radius of the circle respectively.EQUATION OF CIRCLEEquation of circle having centre (a, b) and radius r is (x – a)2 + (y – b)2 = r2

CONIC SECTIONS


PARABOLAA parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point. The fixed line is called the directrix and fixed point is called focus of the parabola.

Standard equation y2 = 4ax y2 = – 4ax x2 = 4ay x2 = – 4ay

Graph

Eccentricity e = 1 e = 1 e = 1 e = 1

Coordinates of focus (a, 0) (–a, 0) (0, a) (0, –a)

Equation of directrix x + a = 0 x – a = 0 y + a = 0 y – a = 0

Equation of axis y = 0 y = 0 x = 0 x = 0

Coordinates of vertex (0, 0) (0, 0) (0, 0) (0, 0)

Extremities of latus rectum (a, ±2a) (–a, ±2a) (±2a, a) (±2a, –a)

Length of latus rectum 4a 4a 4a 4a

Equation of latus rectum x – a = 0 x + a = 0 y – a = 0 y + a = 0

Note : Parabola is symmetric with respect to the axis of the parabola. If the equation has y2 term, then the axis of symmetry is along the x-axis and if the equation has x2 term, then the axis of symmetry is along the y-axis.

ELLIPSEAn ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is constant. The two fixed points are called the foci of the ellipse.


Fundamental terms Horizontal ellipse Vertical ellipse

Standard equation xa

yb

a b2

2

2

2 1+ = >, xa

yb

a b2

2

2

2 1+ = <,

Graph

Coordinates of the centre (0, 0) (0, 0)Coordinates of the vertices (±a, 0) (0, ±b)Length of major axis 2a 2bLength of minor axis 2b 2aCoordinates of foci (±ae, 0) (0, ±be)

Equation of directrices x ae= ± ⎛⎝⎞⎠ y b

e= ± ⎛⎝⎞⎠

Eccentricity e ba

= −12

2 e ab

= −12

2

Length of latus rectum 2 2ba

2 2ab

Ends of latus rectum ± ±⎛⎝⎜

⎞⎠⎟ae b

a,

2± ±

⎛⎝⎜

⎞⎠⎟

ab

be2

,

Focal distance or radii |SP| = (a – ex1) and |S′P| = (a + ex1) |SP| = (b – ey1) and |S′P| = (b + ey1)Sum of focal radii = |SP| + |S′P| 2a 2bDistance between foci 2ae 2be

Note : If length of major axis is 2 a, length of minor axis is 2b and distance between the foci is 2c, then a2 = b2 + c2

Ellipse is symmetric w.r.t. both the coordinate axes.

The foci always lie on the major axis.

If length of semi-major axis and semi-minor axis are equal, then the ellipse become circle.

When distance between foci ( SS′) = a then b = 0. The ellipse is reduced to the line segment SS′.

HYPERBOLAA hyperbola is the set of all points in a plane, the difference of whose distance from two fixed points is constant.


Fundamental terms Hyperbola Conjugate Hyperbola

Standard equation xa

yb

2

2

2

2 1− = yb

xa

2

2

2

2 1− =

Graph

Coordinates of the centre (0, 0) (0, 0)Coordinates of the vertices (±a, 0) (0, ±b)Length of transverse axis 2a 2bLength of conjugate axis 2b 2aCoordinates of foci (±ae, 0) (0, ±be)

Equation of directrices x ae= ± ⎛⎝⎞⎠ y b

e= ± ⎛⎝⎞⎠

Eccentricity e ba

= +12

2 e ab

= +12

2

Length of latus rectum 2 2ba

2 2ab

Ends of latus rectum ± ±⎛⎝⎜

⎞⎠⎟ae b

a,

2± ±

⎛⎝⎜

⎞⎠⎟

ab

be2

,

Difference of focal radii = |SP| – |S′P| 2a 2bDistance between foci 2ae 2be

Note :A hyperbola in which a = b is called an equilateral hyperbola.Hyperbola is symmetric w.r.t. both the axes.

The foci are always on the transverse axis.

INTRODUCTION TO THREE DIMENSIONAL GEOMETRYCOORDINATE AXES AND COORDINATE PLANELet XOX′, YOY′ and ZOZ′ bethree mutually perpendicular lines intersecting at O. The lines XOX′, YOY′ and ZOZ′ are called coordinate axes.Also the planes XOY, YOZ and ZOX are known as coordinate planes.COORDINATES OF A POINT IN SPACELet P be any point in space. Draw PL , PM , PN perpendicular to the YZ, ZX and XY planes respectively, then

(i) LP is called the x-coordinate of P.(ii) MP is called the y-coordinate of P.(iii) NP is called the z-coordinate of P.These three, taken together, are known as coordinates of P. Thus the coordinates of any point in space are the


perpendicular distances of the point from the YZ, ZX and XY planes respectively.Signs of Coordinates in Eight Octants :

The octants XOYZ, X′OYZ, X′OY′Z, XOY′Z, XOYZ′, X′OYZ′, X′OY′Z′ and XOY′Z′ are denoted by I, II, III, ..., VIII respectively.

I II III IV V VI VII VIIIx + – – + + – – +y + + – – + + – –z + + + + – – – –

DISTANCE BETWEEN TWO POINTSDistance between the points P(x1, y1, z1) and Q(x2, y2, z2) is given by

PQ x x y y z z= − + − + −( ) ( ) ( )2 12

2 12

2 12

SECTION FORMULALet M be a point which divides the line joining the points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m : n, then

Types of division Coordinates of M

Internal mx nxm n

my nym n

mz nzm n

2 1 2 1 2 1++

++

++

⎛⎝

⎞⎠, ,

External mx nxm n

my nym n

mz nzm n

2 1 2 1 2 1−−

−−

−−

⎛⎝

⎞⎠, ,

MID POINT FORMULALet M be the mid point of the line joining A (x1, y1, z1) and B (x2, y2, z2). Then coordinates of M is given by,

x x y y z z1 2 1 2 1 22 2 2+ + +⎛

⎝⎜⎞⎠⎟, , .

PROBLEMSVery Short Answer Type

1. Find the slope of the line joining the points (– 2, – 5) and (4, – 6).

2. Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

3. Find the point which is equidistant from points O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c).

4. Find the equation of the line upon which the length of perpendicular p from origin and the angle α made by this perpendicular with the positive direction of x-axis are p = 5, α = 135°.

5. Find the centre and radius of the circle 3x2 + 3y2 – 8x – 10y + 3 = 0

Short Answer Type

6. A line through the point A(2, 0) which makes an angle of 30° with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15°. Find the equation of the straight line in the new position.

7. Find the equation of a circle of radius 5 cm whose centre lies on x - axis and passes through the point (2, 3).

8. Is the triangle, whose vertices are (5, – 6), (1, 2) and (– 7, – 2) a right angled triangle, an acute angled triangle or an obtuse angled triangle ?

9. Find the ratio in which the line joining the points (4, 4, – 10) and (–2, 2, 4) is divided by the XY-plane.

10. Find the equation of the hyperbola whose eccentricity is 2 and the distance between the foci is 16, taking transverse and conjugate axes of the hyperbola as x and y-axes respectively.

Long Answer Type - I

11. Find the equation of the lines which cut off intercepts on the axes whose sum and product are 1 and – 6, respectively.

12. Find the equation of the ellipse with foci at (±5, 0)

and x = 365

as one directrix.

13. If P is any point on a hyperbola and N is the foot of the perpendicular from P on the transverse axis,

then prove that ( )

( )( ).PN

AN A Nba

2 2

2′=

14. Find the equation of the straight line upon which the length of perpendicular from origin is 3 2 units and this perpendicular makes an angle of 75° with the positive direction of x-axis.

15. Prove that the equation of the parabola whose vertex and focus are on the x-axis at distances a′ and a′′ respectively from the origin is

y2 = 4(a′′ – a′)(x – a′).


Long Answer Type - II

16. Find the equations of the two lines which can be drawn through the point (2, 2) to make an angle of 45° with the line x + y = 2.

17. Find the equation of the straight line which passes through the point of intersection of lines 3x – 4y – 7 = 0 and 12x – 5y – 13 = 0 and is perpendicular to the line 2x – 3y + 5 = 0.

18. Find the focus and the equation of the parabola whose vertex is (6, – 3) and directrix is 3x – 5y + 1 = 0.

19. Find the length and equation of major and minor axes, centre, eccentricity, foci, equation of directrices, vertices and length of latus rectum of

the ellipse x y2 2

225 2891+ = .

20. If S and S′ be the foci, C the centre and P be any point on a rectangular hyperbola having lengths of transverse and conjugate axes equal, show that SP·S′P = CP2.

SOLUTIONS

1. Required slope =− − −− −

= −6 5

4 216

( )( )

2. Centre of the given circle, C ≡ (1, 2). Let P ≡ (4, 6)Since point P(4, 6) lies on the circle

∴ Radius of circle = = − + − =CP ( ) ( )1 4 2 6 52 2

Now, the equation of the circle is(x – 1)2 + (y – 2)2 = 52 or x2 + y2 – 2x – 4y – 20 = 0

3. Let P(x, y, z) be the required pointGiven, OP = AP ⇒ OP2 = AP2

∴ x2 + y2 + z2 = (x – a)2 + y2 + z2 ⇒ x = a2

Similarly, OP = BP ⇒ y = a2

and OP = CP ⇒ z = c2

Thus, P a b c≡ ⎛⎝⎜⎞⎠⎟2 2 2

, ,

4. Here p = 5 and α = 135°, therefore, required equation of the line is x cos 135° + y sin 135° = 5⇒ − + =x y

2 25 ⇒ x y− + =5 2 0

5. We have 3x2 + 3y2 – 8x – 10y + 3 = 0

⇒ x x y y2 283

103

1−⎛⎝⎜

⎞⎠⎟ + −⎛

⎝⎜⎞⎠⎟ = −

⇒ −⎛⎝⎜

⎞⎠⎟ + −⎛

⎝⎜⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟x y4

353

43

22 2 2

Hence, centre of the circle is 43

53

,⎛⎝⎜

⎞⎠⎟ and radius is

4 23

units.

6. Given, A ≡ (2, 0), AB is the initial position of the line and AC is its new position.Given, ∠BAX = 30° and ∠BAC = 15° ∴ ∠CAX = 15°Now slope of line AC = tan 15° = 2 – 3Now equation of line AC will be y – 0 = ( )( )2 3 2− −x

or ( )2 3 4 2 3 0− − − + =x y7. Let the coordinates of the centre of the required

circle be C (a, 0). Since it passes through P (2, 3). ∴ CP = radius = 5 cm

⇒ ( ) ( )a − + − =2 0 3 52 2 ⇒ a – 2 = + 4 ⇒ a = 6 or, a = – 2Thus, the coordinates of the centre are (6, 0) or (–2, 0). Hence, the equation of the required circle are (x – 6)2 + (y – 0)2 = 52 and (x + 2)2 + (y – 0)2 = 52

⇒ x2 + y2 – 12x + 11 = 0 and x2 + y2 + 4x – 21 = 08. Let A ≡ (5, – 6), B ≡ (1, 2) and C ≡ (– 7, – 2)

Slope of AB m=− −−

=− =6 2

5 12 1( )say

Slope of BC m= ++

= =2 21 7

12 2 ( )say

Slope of AC m=− +

+=− =

6 25 7

13 3 ( )say

Since m1m2 = – 1, hence AB ⊥ BC ∴ ∠ABC = 90°Thus ΔABC is a right angled triangle.

9. Let A ≡ (4, 4, – 10), B ≡ (– 2, 2, 4)Let the line joining A and B be divided by the XY-plane at point P in the ratio λ : 1

Then, P ≡− +

+++

−+

⎛⎝⎜

⎞⎠⎟

2 41

2 41

4 101

λλ

λλ

λλ

, ,

Since P lies on the XY-plane, therefore z-coordinate of P will be zero.

∴ 4 101

0 52

λλ

λ−+

= ⇒ =

∴ Required ratio is 5 : 2 (internal).

10. Given, 2ae = 16 ⇒ 2·a 2 = 16 ⇒ a = 4 2Also b2 = a2(e2 – 1) = 32(2 – 1) = 32Thus a2 = 32 and b2 = 32

(5, – 6)

(1, 2)

(–7, –2)


∴ The required equation of hyperbola is x y2 2

32 321− = or x2 – y2 = 32

11. Let a and b be the intercepts of the line on x and y axes respectively.∴ Equation of the line will be x

ayb

+ = 1 ...(i)

Given, a + b = 1 ...(ii) and ab = – 6 ...(iii)Putting the value of b from (ii) in (iii), we get a(1 – a) = – 6 ⇒ a2 – a – 6 = 0 ⇒ a = 3, – 2From (ii), when a = 3, b = – 2 and when a = – 2, b = 3∴ Required equation of the lines are

x y3 2

+−

= 1 and x y−

+2 3

= 1

i.e., 2x – 3y – 6 = 0 and 3x – 2y + 6 = 012. Let S ≡ (5, 0) and S′ ≡ (– 5, 0) be the foci.

Centre of the ellipse will be C(0, 0)Clearly, foci S and S′ lie on x-axis. Therefore, major axis of the ellipse will be parallel to x-axis.Let a and b be the length of semi-major and semi-minor axes respectively of the ellipse.Then, ae = 5 ...(i)

Also equation of one directrix is given to be x = 365

∴ ae

= 365

...(ii)

On multiplying (i) and (ii), we get a2 = 36 ∴ a = 6

From (i), e = 56

Now, b2 = a2(1 – e2) = 36 1 2536

11−⎛⎝⎜

⎞⎠⎟ =

∴ The required equation of the ellipse is x y2 2

36 111+ =

13. Let P(α, β) be any point on hyperbola xa

yb

2

2

2

2 1− =

...(i)Then, α β2

2

2

2 1a b

− =

⇒ α β2

2

2

21a b

− =

⇒ ( )( )α α β− +

=a a

a b2

2

2

⇒ ( )( ) ( ) ( )( )( )

AN A Na

PNb

PNAN A N

ba

′ = ⇒′

=2

2

2

2 2

2

14. Let AB be the required lineand OL be perpendicular to it.Given, OL = 3 2and ∠LOA = 75°∴ Equation of line AB will bex cos 75° + y sin 75° = 3 2 (Normal form) ...(i)

Now, cos 75° = cos (30° + 45°) = −3 12 2

and similarly sin 75° = +3 12 2

∴ From (i), equation of line AB is

x y3 12 2

3 12 2

3 2−⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

=

or ( ) ( )3 1 3 1 12 0− + + − =x y15. Since the vertex and focus of

the parabola are on x-axis, therefore, axis of the parabola is x-axis. Let A be the vertex and S the focus of the parabola. Since vertex and focus of the parabola are at distances a′ and a′′ from the origin, therefore A ≡ (a′, 0) and S = (a′′, 0)We know that equation of a parabola whose vertex is (α, β) and axis is parallel to x-axis is (y – β)2 = 4a(x – α)where, a = distance between focus and vertexHere α = a′, β = 0 and a = AS = a′′ – a′∴ Required equation of the parabola will be y2 = 4(a′′ – a′)(x – a′).

16. The given line is x + y = 2 ...(i)Slope of this line = –1Let the slope of the line through (2, 2), which makes an angle of 45° with line (i), be m, then

tan ( )( )

45 11 1

° = − −+ −

mm ⇒ ± = +

−1 1

1m

m⇒ ± (1 – m) = m + 1Case I : 1 – m = m + 1 ⇒ 2m = 0 ⇒ m = 0Hence, one of the required lines through (2, 2) is y – 2 = 0(x – 2) (taking m = 0)⇒ y = 2Case II : – ( 1 – m ) = m + 1 ⇒ – 1 + m = m + 1∴ Slope is not defined.∴ The other required line ought to be vertical.


⇒ The equation of this line is x – 2 = 0 (∵ It passes through the point (2, 2))

17. Given lines are 3x – 4y – 7 = 0 ...(i) 12x – 5y – 13 = 0 ...(ii) 2x – 3y + 5 = 0 ...(iii)Equation of any line through the point of intersection of lines (i) and (ii) is 3x – 4y – 7 + k(12x – 5y – 13) = 0or (3 + 12k) x – (4 + 5k) y – (7 + 13k) = 0 ...(iv)

Slope of line (iv) is 3 124 5++

kk

and slope of line (iii) is 23

If line (iv) is perpendicular to line (iii), then

3 124 5

23

1++

⎛⎝⎜

⎞⎠⎟ ⋅ = −k

k or 6 + 24k = – 12 – 15k

or 39k = – 18 ∴ k = – 613

Putting the value of k in (iv), the equation of required line is

3 7213

4 3013

7 6 0−⎛⎝⎜

⎞⎠⎟ − −⎛

⎝⎜⎞⎠⎟ − − =x y ( )

or 33x + 22y + 13 = 018. Let A(6, – 3) be the vertex

and BC be the directrix of the parabola. Let S be the focus.Equation of directrix BC is 3x – 5y + 1 = 0 ...(i)

Equation of any line perpendicular to line (i) may be taken as 5x + 3y + k = 0 ...(ii)If line (ii) passes through A(6, – 3), then 5 × 6 + 3(– 3) + k = 0 or k = – 21∴ From (ii), equation of line AQ becomes 5x + 3y – 21 = 0 ...(iii)Solving (i) and (iii) by cross multiplication method, we get

x y105 3 5 63

19 25−

=+

=+

∴ x = 3, y = 2

∴ Q ≡ (3, 2). Given A ≡ (6, – 3)Let S ≡ (α, β). Since A is the mid-point of QS

∴ 6 32

9= + ⇒ =α α and − = + ⇒ = −3 22

8β β

∴ S ≡ (9, – 8)Let P(x, y) be an arbitrary point on the parabola, then PS = PN

(6, – 3)

Axis

Tangent at the vertex

90°

( , )

3–

5+

1 =

0

⇒ ( ) ( )x yx y

− + + =− +

+9 8

3 5 19 25

2 2

⇒ 25x2 + 9y2 – 618x + 554y + 30xy + 4929 = 0This is the required equation of the parabola.

19. Equation of ellipse is x y2

2

2

215 171+ =

a = 15, b = 17 so, a < b Length of major axis = 2b = 34Length of minor axis = 2a = 30Equation of major axis is x = 0Equation of minor axis is y = 0Coordinates of centre are (0, 0)Eccentricity of the ellipse

e ab

= − = − = =1 1 225289

64289

817

2

2

Coordinates of foci are given by(0, + be) i.e., (0, + 8)Equation of directrices are

y be

y= ± = ± ⋅or 17 178

or 8y ∓ 289 = 0Coordinates of vertices are given by (0, + b) i.e., (0, + 17)

Length of latus rectum = = × =2 2 22517

45017

2ab

20. Let the equation of the rectangular hyperbola be

xa

yb

2

2

2

2 1− = ...(i)

Given, a = b, therefore (i) becomes x2 – y2 = a2 ...(ii)

Now, e ba

aa

= + = + =1 1 22

2

2

2

Thus e = 2, C ≡ (0, 0), S ≡ ( , ), ( , )2 0 2 0a S a′ ≡ −Let P ≡ (α, β). Since P lies on (ii) ∴ α2 – β2 = a2 ...(iii)Now, SP2 = ( )2 2a −α + β2 = 2a2 + α2 + β2 –2 2 a αand S′P2 = ( )− −2 2a α + β2 = 2a2 + α2 + β2 + 2 2 a α Now, SP2· S′P2 = (2a2 + α2 + β2)2 – 8a2α2

= 4a4 + 4a2 (α2 + β2) + (α2 + β2)2 – 8a2α2

= 4a2(a2 – 2α2) + 4a2(α2 + β2) + (α2 + β2)2 = 4a2(α2 – β2 – 2α2) + 4a2 (α2 + β2) + (α2 + β2)2

[from (iii)] = (α2 + β2)2 = (CP2)2 = CP4

∴ SP ·S′P = CP2


6. Four geometric means are inserted between the numbers 211 – 1 and 211 + 1. The product of these geometric means is(a) 244 – 223 + 1 (b) 244 – 222 + 1(c) 222 – 211 + 1 (d) 222 – 212 + 1.One or More Than One Option(s) Correct Type

7. Let S kn

k k

k

n= −

+

=∑ ( ) .

( )

11

2

1

42 Then Sn can take value(s)

(a) 1056 (b) 1088 (c) 1120 (d) 13328. The largest value of the positive integer m for which

nm + 1 divides 1 + n + n2 + ... + n127 is divisible by(a) 8 (b) 16(c) 32 (d) 64

9. a and c are two distinct positive real numbers such that a, b, c are in G.P. If b – c, c – a, a – b are in H.P., then a + 4b + c is(a) a constant (b) independent of a(c) independent of b (d) independent of c

10. If the first and the (2n – 1)th term of an A.P., a G.P. and an H.P. are equal and their nth terms are a, b and c respectively, then(a) a = b = c (b) a ≥ b ≥ c(c) a + c = b (d) ac – b2 = 0.

11. If log2(5·2x + 1), log4(21 – x + 1) and 1 are in A.P., then x is equal to

(a) loglog

52 (b) log2(0.4)

(c) 1 52

− loglog (d)

loglog

25

Only One Option Correct Type

1. If a1, a2 ........ an are positive real numbers whose product is a fixed number c then the minimum value of a1 + a2 + ...... + an – 1 + 2an is

(a) n c n( )21

(b) ( )n cn+ 11

(c) 21

ncn (d) ( )( )n c n+ 1 21

2. The A.M. between m and n and the G.M. between a

and b are each equal to ma nbm n

++

. Then m =

(a) a ba b+

(b) b aa b+

(c) 2a ba b+

(d) 2b aa b+

3. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2) p2 – 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then a, b, c, d(a) are in A.P. (b) are in G.P.(c) are in H.P. (d) satisfy ab = cd

4. If 1 · 3 + 3 · 32 + 5 · 33 + 7 · 34 + .... upto n terms is equal to 3 + (n – 1) · 3b, then b = (a) n (b) n – 1(c) 2n – 1 (d) n + 1

5. If x = 111….1 (20 digits), y = 333…3 (10 digits) and

z = 222….2 (10 digits), then x yz− 2

=

(a) 1 (b) 2

(c) 12

(d) 3

Total Marks : 80 Time Taken : 60 Min.

This specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four

marks for correct answer and deduct one mark for wrong answer.

Self check table given at the end will help you to check your readiness.

Class XI

Sequences & Series


12. If a, b, c are in H.P., then the value of

1 1 1 1 1 1b c a c a b

+ −⎛⎝⎜

⎞⎠⎟

+ −⎛⎝⎜

⎞⎠⎟ is

(a) 2 1

2bc b− (b)

14

3 2 12 2c ca a

+ −⎛⎝⎜

⎞⎠⎟

(c) 3 22b ab− (d) none of these

13. For 02

< <φ π , if x yn

n

n

n= =

=

∞

=

∞∑ ∑cos , sin2

0

2

0φ φ

and z n n

n=

=

∞∑ cos sin ,2 2

0φ φ then

(a) xyz = xz + y (b) xyz = xy + z(c) xyz = x + y + z (d) xyz = yz + x

Comprehension Type

Let Vr denote the sum of first r terms of an A.P. whose first term is r and the common difference is (2r – 1). Let Tr = Vr + 1 – Vr for r = 1, 2, ...14. The sum V1 + V2 + .... + Vn is

(a) 1

121 3 12n n n n( )( )+ − +

(b) 1

121 3 22n n n n( )( )+ + +

(c) 12

2 12n n n( )− +

(d) 13

2 2 33( )n n− +

15. Tr is always(a) an odd number (b) an even number(c) a prime number (d) a composite number

Matrix Match Type

16. It is given that 1 12

13

14

15

16

2− + − + − + =.... log

and 1 13

15

17

19

111 4

− + − + − + =.... π .

ThenColumn I Column II

A. 11 3

15 7

19 11⋅

+⋅

+⋅

+ ... 1. π/6

B. 11 2

13 4

15 6⋅

+⋅

+⋅

+ ... 2. π/8

C.1

2 11( )r rr −=

∞∑ 3. log2

4. log4

A B C(a) 1 4 3(b) 2 3 4(c) 1 2 4(d) 2 1 3

Integer Answer Type

17. If x, y, z are positive and x + y + z = 1, then

1 1 1 1 1 1x y z−⎛

⎝⎜⎞⎠⎟

−⎛

⎝⎜

⎞

⎠⎟ −⎛⎝⎜

⎞⎠⎟

is greater than or equal to

18. If the function f satisfies the relation f (x + y) = f (x) · f (y) for all natural numbers x, y, f (1) = 2 and

f a rr

n( )+

=∑

1 = 16 (2n – 1), then the natural number

a, is

19. The unit digit of the number of common terms to the sequence 17, 21, 25, ...., 417 and 16, 21, 26,....., 466 is

20. If one G.M. is g and two A.M.s are p and q, are inserted between two numbers a and b, then ( )( )2 2

2p q p q

g− − +

is equal to

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Definition : An equation involving independent variable

x, dependent variable y and the differential coefficients

dydx

d ydx

, , .....2

2 is called differential equation.

Order of a differential equation : The order of

a differential equation is the order of the highest

derivative occurring in the differential equation.

Degree of a differential equation : The degree of

a differential equation is the degree of the highest

order derivative, when differential coefficients are

made free from radicals and fractions.

FORMATION OF DIFFERENTIAL EQUATION Formulating a differential equation from a given

equation representing a family of curves means

finding a differential equation whose solution is

the given equation. The equation so obtained is

the differential equation of order n for the family

of given curves.

(i) Write the given equation involving independent

variable x (say), dependent variable y (say) and

the arbitrary constants.

(ii) Obtain the number of arbitrary constants let

there be n arbitrary constants.

(iii) Differentiate the given relation in n times

with respect to x.

(iv) Eliminate arbitrary constants with the help of

n equations involving differential coefficients

obtained in step (iii) and an equation in step

(i). The equation so obtained is the desired

differential equation.

VARIABLE SEPARABLE FORM If the differential equation of the form

f1(x) dx = f

2(y) dy ...(i)

where f1 and f

2 being functions of x and y only.

Thus, integrating both sides of (i), we get its

solution as

f x dx f y dy c1 2( ) ( )∫ ∫= + ,

where c is an arbitrary constant.

Equations reducible to variable separable form:

(i) D i f f e r e n t i a l e q u a t i o n s o f t h e f o r m

dydx

= f ax by c+ +( ) c a n b e r e d u c e d t o

variable separable form by the substitution

ax + by + c = Z.

∴ a b dydx

dZdx

+ =

∴ dZdx

ab

f Z dZdx

a bf Z−⎛⎝⎜

⎞⎠⎟

= ⇒ = +1( ) ( )

This is variable separable form.

(ii) Differential equation of the form

dydx

ax by cAx By C

aA

bB

K= + ++ +

= = where say)(

∴ = + ++ +

dydx

K Ax By cAx By C( )

Put Ax + By = Z ⇒ + =A B dydx

dZdx

∴ −⎡⎣⎢

⎤⎦⎥

= ++

⇒ = + ++

dZdx

AB

KZ cZ C

dZdx

A B KZ cZ C

1

This is variable separable form and can be solved.

This column is aimed at Class XII students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging.

*ALOK KUMAR, B.Tech, IIT Kanpur

* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91).He trains IIT and Olympiad aspirants.


HOMOGENEOUS DIFFERENTIAL EQUATION If a first order, first degree differential equation is

expressible in the form dydx

f x yg x y

=( , )

( , ) where f(x, y)

and g(x, y) are homogeneous functions of the same

degree, then it is called a homogeneous differential

equation. Such type of equations can be reduced

to variable separable form by the substitution

y = vx.

Equation reducible to homogeneous form : A first

order, first degree differential equation of the form

dydx

a x b y ca x b y c

aa

bb

=+ ++ +

≠1 1 1

2 2 2

1

2

1

2

, where ,

This is non-homogeneous differential equation.

It can be reduced to homogeneous form by certain

substitutions. Put x = X + h, y = Y + k, where h and

k are constants, which are to be determined.

LINEAR DIFFERENTIAL EQUATIONLinear and non-linear differential equations:

A differential equation is a linear differential

equation if it is expressible in the form

P d ydx

P d ydx

P d ydx

P dydx

P y Qn

n

n

n

n

n n n0 1

1

1 2

2

2 1+ + + + + =−

−

−

− −...

where P0, P

1, P

2, .... , Pn – 1

, Pn and Q are either

constants or functions of independent variable x.

Thus, if a differential equation when expressed in

the form of a polynomial involves the derivatives

and dependent variable in the first power and there

are no product of these, and also the coefficient of

the various terms are either constants or functions

of the independent variable, then it is said to be

linear differential equation. Otherwise, it is a non

linear differential equation.

Linear differential equation of first order : The

general form of a linear differential equation of

first order is

dydx

Py Q+ = ...(i)

Where P and Q are functions of x (or constants)

Multiplying both sides by e Pdx∫ , we get

e dydx

Py Q ePdx Pdx∫ +⎛⎝⎜

⎞⎠⎟

= ∫

⇒ d

dxy e Q ePdx Pdx∫{ } = ∫

On integrating both sides w.r.t. x, we get

y e Q e dx cPdx Pdx∫ = ∫ +∫ ...(ii)

which is the required solution, where c is the

constant of integration. e Pdx∫ is called the

integrating factor. The solution (ii) in short may

also be written as y I F Q I F dx c.( . .) .( . .)= +∫Linear differential equation of the form

dxdy

Rx S+ = : Sometimes a linear differential

equation can be put in the form dxdy

Rx S+ = where

R and S are functions of y (or constants). Note that

y is independent variable and x is a dependent

variable.

Equations reducible to linear form (Bernoulli's

differential equation) : The differential equation

of type dydx

Py Qyn+ = ...(i)

where P and Q are functions of x alone or constants

and n is a constant other than zero or unity, can

be reduced to the linear form by dividing by yn

and then putting y–n + 1 = v, as explained below.

Dividing both sides of (i) by yn, we get

y dydx

Py Qn n− − ++ =1

Putting y–n + 1 = v so that ( ) ,− + =−n y dydx

dvdx

n1

we get

1

11 1

− ++ = ⇒ + − = −

ndvdx

Pv Q dvdx

n Pv n Q( ) ( )

which is a linear differential equation.

Differential equation of the form

dydx

P y Q y+ =φ ψ( ) ( )

where P and Q are functions of x alone or

constants.

Dividing by ψ(y), we get 1

ψφψ( )

( )

( )ydydx

yy

P Q+ =

Now put φψ

( )

( )

yy

v= , so that d

dxyy

dvdx

φψ

( )

( )

⎧⎨⎩

⎫⎬⎭

=

or dvdx

ky

dydx

= ⋅ 1

ψ( ), where k is constant

We get dvdx

kP v kQ+ =

which is linear differential equation.

APPLICATION OF DIFFERENTIAL EQUATION Differential equation is applied in various practical

fields of life. It is used to define various physical

laws and quantities. It is widely used in physics,

chemistry, engineering etc.


Some important fields of application are

Rate of change

Geometrical problems etc.

MISCELLANEOUS DIFFERENTIAL EQUATION A special type of second order differential equation :

d ydx

f x2

2= ( ) ...(i)

Equation (i) may be re-written as d

dxdydx

f x⎛⎝⎜

⎞⎠⎟

= ( )

⇒ ⎛⎝⎜

⎞⎠⎟

=d dydx

f x dx( )

Integrating, dydx

f x dx c= +∫ ( ) 1

i.e. dydx

F x c= +( ) 1 ...(ii)

Where F x f x dx( ) ( )= ∫ From (ii), dy = F (x)dx + c

1dx

Integrating, y F x dx c x c= + +∫ ( ) 1 2

∴ y = H(x) + c1x + c

2

where H x F x dx( ) ( )= ∫ and c1 and c

2 are arbitrary

constants.

PROBLEMSSingle Correct Answer Type

1. The degree and order of the differential equation

x d ydx

dydx

y x2

2

3 42

⎛

⎝⎜

⎞

⎠⎟ + ⎛

⎝⎜⎞⎠⎟

+ = respectively is

(a) 3, 2 (b) 1, 1 (c) 4, 3 (d) 4, 4

2. If m and n are the order and degree of the

differential equation d ydx

d ydxd ydx

d ydx

x2

2

5

2

2

3

3

3

3

3

24 1

⎛

⎝⎜

⎞

⎠⎟ +

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟

+ = − ,

then (m, n) is

(a) (3, 5) (b) (3, 1) (c) (3, 3) (d) (3, 2)

3. The order and degree of the differential equation

ρ =

+ ⎛⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

1

23 2

2 2

dydx

d y dx

/

/

are respectively

(a) 2, 2 (b) 2, 3

(c) 2, 1 (d) none of these

4. The differential equation for all the straight lines

which are at a unit distance from the origin is

(a) y x dydx

dydx

−⎛⎝⎜

⎞⎠⎟

= − ⎛⎝⎜

⎞⎠⎟

2 2

1

(b) y x dydx

dydx

+⎛⎝⎜

⎞⎠⎟

= + ⎛⎝⎜

⎞⎠⎟

2 2

1

(c) y x dydx

dydx

−⎛⎝⎜

⎞⎠⎟

= + ⎛⎝⎜

⎞⎠⎟

2 2

1

(d) y x dydx

dydx

+⎛⎝⎜

⎞⎠⎟

= − ⎛⎝⎜

⎞⎠⎟

2 2

1

5. The differential equation whose solution is

y = c1cosax + c

2sinax is (where c

1, c

2 are arbitrary

constants)

(a) d ydx

y2

2

20+ = (b)

d ydx

a y2

2

20+ =

(c) d ydx

ay2

2

20+ = (d)

d ydx

a y2

2

20− =

6. The differential equation of the family of curves

y2 = 4a(x + a), where a is an arbitrary constant, is

(a) y dydx

x dydx

1 2

2

+ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

(b) y dydx

x dydx

1 2

2

− ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

(c) d ydx

dydx

2

22 0+ = (d)

dydx

dydx

y⎛⎝⎜

⎞⎠⎟

+ + =3

3 0

7. The differential equation of all parabolas whose

axes are parallel to y-axis is

(a) d ydx

3

30=

(b)

d xdy

c2

2=

(c)

d ydx

d xdy

3

3

2

20+ =

(d)

d ydx

dydx

c2

22+ =

8. Family of curves y = ex(Acosx + Bsinx), represents

the differential equation

(a)

d ydx

dydx

y2

22= −

(b)

d ydx

dydx

y2

22 2= −

(c) d ydx

dydx

y2

22= −

(d)

d ydx

dydx

y2

22= +

9. y = aemx + be–mx satisfies which of the following

differential equation

(a) dydx

my− = 0 (b) dydx

my+ = 0


(c) d ydx

m y2

2

20+ = (d)

d ydx

m y2

2

20− =

10. Differential equation of y = sec(tan–1x) is

(a) ( )12+ = +x dy

dxy x (b) ( )1

2+ = −x dydx

y x

(c) ( )12+ =x dy

dxxy (d) ( )1

2+ =x dydx

xy

11. If x = sint, y = cospt, then

(a) (1 – x2)y2 + xy

1 + p2y = 0

(b) (1 – x2)y2 + xy

1 – p2y = 0

(c) (1 + x2)y2 – xy

1 + p2y = 0

(d) (1 – x2)y2 – xy

1 + p2y = 0

12. The solution of the differential equation

3extanydx + (1 – ex)sec2ydy = 0 is

(a) tany = c(1 – ex)3 (b) (1 – ex)3 tany = c(c) tany = c(1 – ex) (d) (1 – ex) tany = c

13. The solution of the equation dydx

e x ex y y= +− −2 is

(a) e e x cy x= + +3

3 (b) e e x cy x= + +2

(c) e e x cy x= + +3 (d) y = ex + c


dydx

e x x xx= + + +cos tan is

(a) y e x x x cx= + + + +sin log cos

2

2

(b)

y e x x x cx= + + + +sin log sec

2

2

(c) y e x x x cx= − + + +sin log cos

2

2

(d)

y e x x x cx= − + + +sin log sec

2

2


(sinx + cosx)dy + (cosx – sinx)dx = 0 is

(a) ex(sinx + cosx) + c = 0

(b) ey(sinx + cosx) = c

(c) ey(cosx – sinx) = c(d) ex(sinx – cosx) = c16. The solution of the differential equation

dydx

x y xy= + + +1 is

(a) log( )12

2

+ = + +y x x c

(b)

( )12

22

+ = + +y x x c

(c) log(1 + y) = log(1 + x) + c(d) none of these


( )x yx dydx

y xy2 2 2 20− + + =

is

(a) logxy x y

c⎛⎝⎜

⎞⎠⎟

= + +1 1 (b) log

yx x y

c⎛⎝⎜

⎞⎠⎟

= + +1 1

(c) log xyx y

c( ) = + +1 1 (d) log( )xy

x yc+ + =1 1


x y dydx

sec = 1 is

(a) xsecy tany = c (b) cx = secy + tany

(c) cy = secx tanx (d) cy = secx + tanx

19. The solution of differential equation x dydx

y y+ = 2

is

(a) y = 1 + cxy (b) y = log{cxy}

(c) y + 1 = cxy (d) y = c + xy

20. If dydx

xy yxy x

= ++

, then the solution of the differential

equation is

(a) y = xex + c (b) y = ex + c(c) y = Axex – y (d) y = x + A21. The general solution of the equation

(ey + 1)cosxdx + eysin xdy = 0 is

(a) (ey + 1)cosx = c (b) (ey – 1)sinx = c(c) (ey + 1)sinx = c (d) none of these


x2dy = –2xydx is

(a) xy2 = c (b) x2y2 = c (c) x2y = c (d) xy = c

23. The solution of the equation sin− ⎛⎝⎜

⎞⎠⎟

= +1 dydx

x y is

(a) tan(x + y) + sec(x + y) = x + c(b) tan(x + y) – sec(x + y) = x + c(c) tan(x + y) + sec(x + y) + x + c = 0

(d) none of these


dydx

x yx y

= − +− +

3

2 5( ) is

(a) 2(x – y) + log(x – y) = x + c(b) 2(x – y) – log(x – y + 2) = x + c(c) 2(x – y) + log(x – y + 2) = x + c(d) none of these



(x – y2x)dx = (y – x2y)dy is

(a) (1 – y2) = c2(1 – x2)

(b) (1 + y2) = c2(1 – x2)

(c) (1 + y2) = c2(1 + x2)

(d) none of these

26. The solution of (cosecx logy)dy + (x2y)dx = 0 is

(a) log

( )cos siny x x x c

22 2

2+ − + =

(b) log

( )cos siny x x x x c

22 2

22⎛

⎝⎜⎞⎠⎟

+ − + =

(c) (log )

( )cos siny x x x x c

22 2

22+ − + =

(d) none of these


xy dydx

y x xx

= + + ++

( )( )

( )

1 1

1

2 2

2is

(a) 1

21

2 1log( ) log tan+ = − +−y x x c

(b) 1

21

2 1log( ) log tan+ = + +−y x x c

(c) log(1 + y2) = logx – tan–1x + c

(d) log(1 + y2) = logx + tan–1x + c

28. The solution of ( ) ( )x y dx y x dy1 1 02 2+ + + = is

(a) 1 12 2+ + + =x y c

(b) 1 12 2+ − + =x y c

(c) ( ) ( )/ /

1 12 3 2 2 3 2+ + + =x y c

(d) none of these


(1 + x2)(1 + y)dy + (1 + x)(1 + y2)dx = 0 is

(a) tan–1x + log(1 + x2) + tan–1y + log(1 + y2) = c

(b) tan log( ) tan log( )− −− + + − + =1 2 1 21

21

1

21x x y y c

(c) tan log( ) tan log( )− −+ + + + + =1 2 1 21

21

1

21x x y y c

(d) none of these


cosy log(secx + tanx)dx = cosx log(secy + tany)dy is

(a) sec2x + sec2y = c (b) secx + secy = c

(c) secx – secy = c (d) none of these

31. The solution of the equation dydx

y yx x

= − −+ −

2

2

2

2 3 is

(a)

1

3

2

1

1

4

3

1log log

yy

xx

c−+

= +−

+

(b)

1

3

1

2

1

4

1

3log log

yy

xx

c+−

= −+

+

(c)

42

13

1

3log log

yy

xx

c−+

= −+

+

(d) none of these

32. Solution of ydx – xdy = x2ydx is

(a) ye cxx22= (b) ye cxx− =

22

(c) y e cxx2 22

= (d) y e cxx2 22− =

33. The solution of dydx

y x= −2 is

(a) 2x + 2y = c (b) 2x – 2y = c

(c)

1

2

1

2x y c− = (d) x + y = c


34. The curve whose subtangent is ‘n’ times the

abscissa of the point of contact and passes through the

point (2, 3), then

(a) for n = 1 equation of the curve is 2y = 3x(b) for n = 1 equation of the curve is 2y2 = 9x

(c) for n = 2 equation of the curve is 2y = 3x(d) for n = 2 equation of the curve is 2y2 = 9x

35. If the solution of the equation d xdt

dxdt

x2

24 3 0+ + =

given that for t = 0, x = 0 and dxdt

= 12 is in the form

x = Ae–3t + Be–t then

(a) A + B = 0 (b) A + B =12

(c) |AB| = 36 (d) |AB| = 49

36. If differential equation is formed to the family of all

the central conics centred at origin, then

(a) order = 2 (b) order = 3

(c) degree = 1 (d) degree = order = 3

37. Solution of the differential equation :

x y dydx

y x dydx

x x yy

+

−= +sin ( )

2 2 2

3 is

(a) − + = ⎛⎝⎜

⎞⎠⎟

+cot( )x y xy

c2 22

(b) y

x y cx y

2

2 2

2 2

+= − +tan( )


(c) − + = ⎛⎝⎜

⎞⎠⎟ +cot( )x y y

xc2 2

2

(d) y x c

xx y

2 2

2

2 2 2+ = − +tan ( )

38. Let C be a curve such that the normal at any point

P on it meets x-axis and y-axis at A and B respectively.

If BP : PA = 1 : 2 (internally) and the curve passes

through the point (0, 4) then which of the following

alternative(s) is/are correct?

(a) The curves passes through ( , )10 6−(b) The equation of tangent at ( , )4 4 3 is 2x + 3y = 20

(c) The differential equation for the curve is yy′ + 2x = 0

(d) The curve represent a hyperbola

39. If solution of x d ydx

x dydx

y22

29 21 0− + = is of the

form y = c1xm + c

2xn (c

1, c

2 are arbitrary constants,

m, n ∈ N) then values of m, n can be

(a) m = 3, n = 1 (b) m = 3, n = 7

(c) m = 7, n = 3 (d) m = 1, n = 7

40. If the solution of y d ydx

dydx

x2

2

2

+ ⎛⎝⎜⎞⎠⎟ = , y(0) = y(1) = 1

is given by y2 = f(x) then

(a) f(x) is monotonically increasing ∀ ∈ ∞x ( , )1

(b) f(x) = 0 has only one real root

(c) f(x) is neither even nor odd

(d) f(x) has 3 real roots

41. Solution of the differential equation

(3 tanx + 4 coty – 7) sin2y dx

– (4 tanx + 7 coty – 5)cos2xdy = 0 is

(a) 3

27

7

25 4

2 2cot cot tan tan cot tanx x y y x y c− + − + ⋅ =

(b)

3

27

7

25 4

2 2tan tan cot cot tan cotx x y y x y c− + − + ⋅ =

(c) 3 tan2y – 14cotx⋅tan2y + 7cot2x – 10 tany cot2x

+ 8 cotx·tany + 2c cot2x tan2y = 0.

(d) 3 cot2y – 14cotx·cot2y + 7 cot2x + 10 coty tan2x+ 8 tanx·coty = 0.

Comprehension Type

Paragraph for Q. No. 42 to 44Consider a tank which initially holds V

0 ltr. of brine

that contains a lb of salt. Another brine solution,

containing b lb of salt/ltr., is poured into the tank at

the rate of e ltr./min while, simultaneously, the well-

stirred solution leaves the tank at the rate of f ltr./min.

The problem is to find the amount of salt in the tank

at any time t.

Let Q denote the amount of salt in the tank at any time.

The time rate of change of Q, dQ/dt, equals the rate

at which salt enters the tank minus the rate at which

salt leaves the tank. Salt enters the tank at the rate of

be lb/min. To determine the rate at which salt leaves

the tank, we first calculate the volume of brine in the

tank at any time t, which is the initial volume V0 plus

the volume of brine added et minus the volume of brine

removed ft. Thus, the volume of brine at any time is

V0 + et – ft ... (a)

The concentration of salt in the tank at any time is

Q(V0 + et – ft), from which it follows that salt leaves

the tank at the rate of f QV et ft0 + −⎛⎝⎜

⎞⎠⎟

lb / min

Thus, dQdt

be f QV et ft

= −+ −

⎛⎝⎜

⎞⎠⎟0

... (b)

ordQdt

fV et ft

Q be++ −

=0

42. A tank initially holds 100 ltr. of a brine solution

containing 20 lb of salt. At t = 0, fresh water is poured

into the tank at the rate of 5 ltr./min, while the well-

stirred mixture leaves the tank at the same rate. Then

the amount of salt in the tank after 20 min.

(a) 20/e (b) 10/e (c) 40/e2 (d) 5/e43. A 50 ltr. tank initially contains 10 ltr. of fresh

water. At t = 0, a brine solution containing 1 lb of

salt per gallon is poured into the tank at the rate of

4 ltr./min, while the well-stirred mixture leaves the

tank at the rate of 2 ltr./min. Then the amount of time

required for overflow to occur is

(a) 30 min (b) 20 min (c) 10 min (d) 40 min

44. In the above question, the amount of salt in the

tank at the moment of overflow is

(a) 20 lb (b) 50 lb

(c) 30 lb (d) None of these

Paragraph for Q. No. 45 to 47Newton’s law of cooling states that the rate at which

a substance cools in moving air is proportional to the

difference between the temperatures of the substance

and that of the air. If the temperature of the air is

290 K.

We can write it as dTdt

k T k= − − >( ), ,290 0 a constant.

where T is temperature of substance.

45. The substance cools from 370 K to 330 K in 10 min,

then


(a) T = 290 + 160e–kt (b) T = 290 + 80e–kt (c) T = 290 + 40e–kt (d) T = 290 + 20e–kt

46. In the above question, the value of k must be

(a) ln2 (b) ln2

40 (c)

ln2

20 (d)

ln2

10

47. The time taken when temperature be 295 K is

(a) 10 min (b) 20 min

(c) 40 min (d) 80 min

Matrix – Match Type


Column-I Column-IIA. If order and degree of the

differential equation formed by

differentiating and eliminating

the constants from y = a sin2x

+ b cos2x + c sin2x + d cos2x,

where a, b, c, d are arbitrary

constants are represented by O and D, then

P. O + 2D = 5

B. The order and degree of the

differential equation, whose

general solution is given by

y = (c1 + c

2) sin(x + c

3) – c

4

ex c c+ +5 6 , where c1, c

2, c

3, c

4, c

5,

c6 are arbitrary constants, are

O and D, then

Q. 2O + 3D = 5

C. T h e ord e r and d e g re e of

t he d i f fe rent i a l e qu at i on

satisfying ( ) ( )1 12 2+ + +x y

= + + +A x y y x( ( ) ( ))1 12 2

are O and D, then

R. O = D

S. OD + DO = 4


Column-I Column-IIA. If the function y = e4x + 2e–x is a

solution of the differential equation

d ydx

dydx

yK

3

313−

= , then the value of

K/3 is

P. 3

B. Number of straight lines which

satisfy the differential equation

dydx

x dydx

y+ ⎛⎝⎜

⎞⎠⎟− =

2

0 is

Q. 4

C. If re a l va lue of m for w hich

the subst itut ion, y = um wi l l

transform the differential equation,

2 44 4 6x y dy

dxy x+ − into a homoge-

neous equation, then the value of

2m is

R. 2

Integer Answer Type

50. If the solution of the differential equation

e3x(p – 1) + p3e2y = 0 is in the form ey = cex + ck then

k = __ (where p dydx

= and ‘c’ is arbitrary constant)


y2(y – xp) = x4p2 is in the form 1

ycx

ck= + then

k = __ (where p dydx



y = 2px + y2p3 is in the form y2 = 2cx + ck then

k = __(where p dydx


53. The solution of x dydx

x y e yx3 24+ =tan sec

satisfying y(1) = 0 is siny = ex(x – 1)x–k then k =

54. Differential equation, satisfying

y = (sin–1x)2 + A(cos–1x) + B, where A and B are

arbitrary constants is

( )p x d ydx

xdydx

q p q− − = + =22

2 then __

55. A curve passing through the point (1, 1) has the

property that the perpendicular distance of the origin

from normal at any point ‘P’ of the curve is equal to the

distance of P from the x-axis is a circle with radius ___.

SOLUTIONS

1. (a) : Given differential equation,

x d ydx

dydx

y x2

2

3 42

⎛

⎝⎜

⎞

⎠⎟ + ⎛

⎝⎜⎞⎠⎟

+ =

Hence, order = 2 and degree = 3

2. (d) : The order (m) of the given equation is 3 and

degree (n) of the given equation is 2. Therefore m = 3

and n = 2.

3. (a) : ρ.

/

d ydx

dydx

2

2

23 2

1= + ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥


⇒ ρ.d ydx

dydx

2

2

2 23

1⎛

⎝⎜

⎞

⎠⎟ = + ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

∴ Order = 2, degree = 2.

4. (c) : Since the equation of lines whose distance

from origin is unit, is given by

xcosα + ysinα = 1 ...(i)

Differentiate w.r.t. x, we get

cos sinα α+ =dydx

0 ...(ii)

On solving (i) and (ii),

sinα αy x dydx

y x dydx

−⎛⎝⎜

⎞⎠⎟

= ⇒ −⎛⎝⎜

⎞⎠⎟

=1 cosec ...(iii)

Also (ii) ⇒ = − ⇒ ⎛⎝⎜

⎞⎠⎟

=dydx

dydx

cot cotα α2

2 ...(iv)

Therefore by (iii) and (iv), 1

2 2

+ ⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟

dydx

y x dydx

.

5. (b) : Since, y = c1cosax + c

2sinax

Differentiate w.r.t. x, we get

dydx

c a ax c a ax= − +1 2sin cos

⇒ d ydx

c a ax c a ax2

2 12

22= − −cos sin

d ydx

a c ax c ax d ydx

a y2

2

21 2

2

2

2= − + ⇒ = −( cos sin )

ord ydx

a y2

2

20+ = .

6. (b) : Given y2 = 4a(x + a) ...(i)

differentiating, we get

2 4y dydx

a⎛⎝⎜

⎞⎠⎟

= ...(ii)

Eliminating a from (i) and (ii),

Required differentially equation is

y dydx

x dydx

1 2

2

− ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= .

7. (a) : The equation of a member of the family of

parabolas having axis parallel to y-axis is

y = Ax2 + Bx + C ...(i)

where A, B, C are arbitrary constants.

Differentiating (i) w.r.t. x, we get dydx

Ax B= +2 ...(ii)

Differentiating (ii) w.r.t. x gives

d ydx

A2

22= ...(iii)

Differentiating (iii) w.r.t. x again, we get d ydx

3

30= .

8. (b) : Given, y = exAcosx + exBsinx

⇒ dydx

Ae x Ae x Be x Be xx x x x= − + +cos sin sin cos

dydx

A B e x B A e xx x= + + −( ) cos ( ) sin

⇒ d ydx

A B e x e A B xx x2

2= + − +( ) cos ( ) sin

+ (B – A)ex sinx + (B – A)ex cosxd ydx

Be x Ae xx x2

22 2= −cos sin .

Hence,d ydx

dydx

y2

22 2= − .

9. (d) : y = aemx + be–mx

⇒ dydx

mae mbemx mx= − −

⇒ d ydx

m ae m be m ymx mx2

2

2 2 2= + =−

or d ydx

m y2

2

20− = .

10. (c) : y = sec(tan–1x)

dydx

x xx

xyx

=+

=+

− −sec(tan )tan(tan ).

1 1

2 2

1

1 1

⇒ + =( ) .12x dy

dxxy

11. (d) : x = sint, y = cospt dxdt

t dydt

p pt dydx

p ptt

= = − ⇒ =−

cos ; sinsin

cos

d ydx

t p pt dt dx p pt t dt dx

t

2

2

2

2=− −cos cos ( / ) sin sin ( / )

cos

⇒ − − + =( )1 02

2

2

2x d ydx

x dydx

p y

or (1 – x2)y2 – xy

1 + p2y = 0.

12. (a) : Given differential equation can be written in

the form of

sec

tan

2

31

yy

dy ee

dxx

x= −−

On integrating,


sec

tan

2

31

yy

dy ee

dxx

x∫ ∫= −−

⇒ log(tany) = 3log(1 – ex) + logc

⇒ tany = c(1 – ex)3.

13. (a) : dydx

e x e e e xx y y y x= + = +− − −2 2( )

⇒ eydy = (x2 + ex)dxOn integrating both sides, we get

e x e cy x= + +3

3.

14. (b) : dydx

e x x xx= + + +cos tan

On integrating both sides, we get

y e x x x cx= + + + +sin log sec .

2

2

15. (b) : dydx

x xx x

= − −+

cos sin

sin cos

⇒ = − −+

⎛⎝⎜

⎞⎠⎟

dy x xx x

dxcos sin

sin cos


⇒ y = –log(sinx + cosx) + logc

⇒ =+

⎛⎝⎜

⎞⎠⎟⇒ + =y c

x xe x x cy

logsin cos

(sin cos ) .

16. (a) : dydx

x y xy= + + +1

⇒ = + + ⇒+

= +dydx

x y dyy

x dx( )( ) ( )1 11

1

On integrating, we get log( ) .12

2

+ = + +y x x c

17. (a) : The given equation is

( )x yx dydx

y xy2 2 2 20− + + = ⇒

1 10

2 2

−+ + =

yy

dy xx

dx

⇒ −⎛

⎝⎜

⎞

⎠⎟ + +

⎛⎝⎜

⎞⎠⎟

=1 1 1 10

2 2y ydy

x xdx

On integrating, we get the required solution

log .xy x y

c⎛⎝⎜

⎞⎠⎟

= + +1 1

18. (b) : We have, x y dydx

ydy dxx

sec sec= ⇒ =1


log(secy + tany) = logx + logc ⇒ secy + tany = cx.

19. (a) : We have, x dydx

y y x dydx

y y+ = ⇒ = −2 2

⇒−

= ⇒−

−⎡

⎣⎢

⎤

⎦⎥ =

dyy y

dxx y y

dy dxx2

1

1

1

On integrating, we get, log(y – 1) – logy = logx + logc

⇒−

= ⇒ = +y

yxc y cxy1

1 .

20. (c) : dydx

xy yxy x

yy

dy xx

dx=++

⇒+⎛

⎝⎜⎞⎠⎟

= +⎛⎝⎜

⎞⎠⎟

1 1


logy + y = logx + x + logA

⇒ ⎛⎝⎜

⎞⎠⎟

= − ⇒ = −log

yAx

x y y Axex y

21. (c) : (ey + 1)cosxdx + eysinxdy = 0

⇒+

+ =e dye

xx

dxy

y1

0cos

sin


log(ey + 1) + log(sinx) = logc ⇒ (ey + 1)sinx = c

22. (c) : We have, x dy xydxy

dy xx

dx2

22

1 2= − ⇒ = −

On integrating, logy = –2logx + logc⇒ logy = logx–2 + logc ⇒ logyx2 = logc or yx2 = c.

23. (b) : Here,dydx

x y= +sin( ) ...(i)

Put x + y = v ⇒ dydx

dvdx

= −1

∴ (i) becomes dv

vdx

1+=

sin

Now on integrating both sides, we get

tanv – secv = x + c or tan(x + y) – sec(x + y) = x + c.

24. (c) : Here, dydx

x yx y

=− +− +

3

2 5( )

Let x – y = v and dydx

dvdx

= −1

Thus the differential equation reduces to dvdx

vv

= ++

2

2 5

⇒ 2 5

22

1

2

vv

dv dxv

dv dx++

= ⇒ ++

⎡⎣⎢

⎤⎦⎥

=∫ ∫ ∫ ∫( )

⇒ + + = +2 2v v x clog( )

or 2(x – y) + log(x – y + 2) = x + c.

25. (a) : Given equation can be written as

xx

dx yy

dy1 1

2 2−=

−

On integrating we get

− − = − − +1

21

1

21

2 2log( ) log( ) logx y c


⇒ − − − = − ⇒ −−

= −log( ) log( ) log1 1 2

1

1

2 22

2

2x y c xy

c

Hence, (1 – y2) = c2(1 – x2).

26. (c) : (cosecx logy)dy + (x2y)dx = 0

⇒ = −1 2

yydy x xdxlog sin


(log )[ ( cos ) cos ]

y x x x xdx c2

2

22+ − + =∫

⇒ + − + =(log )

( )cos sin .y x x x x c

22

22 2

27. (b) : xy dydx

y x xx

=+ + +

+( )( )

( )

1 1

1

2 2

2

⇒+

= + ++

= ++∫∫ ∫ ∫

ydyy

x xx x

dxx

dx dxx1

1

1

1

12

2

2 2

( )

( )

⇒ + = + +−1

21

2 1log( ) log tan .y x x c

28. (a) : Given equation is,

( ) ( )x y dx y x dy1 1 02 2+ + + =

⇒ + = − +x y dx y x dy1 12 2

⇒+

= −+

∫ ∫x

xdx y

ydy

1 12 2

⇒ + + + =1 12 2x y c.

29. (c) : Given equation is,

(1 + x2)(1 + y)dy + (1 + x)(1 + y2)dx = 0

⇒+

+= − +

+( )

( )

( )

( )

1

1

1

12 2

yy

dy xx

dx

⇒+

++

⎡

⎣⎢⎢

⎤

⎦⎥⎥

++

++

⎡

⎣⎢

⎤

⎦⎥ − =∫ ∫

1

1 1

1

1 10

2 2 2 2yyy

dyx

xx

dx c

⇒ + + + + + =− −tan log( ) tan log( ) .

1 2 1 21

21

1

21y y x x c

30. (d) : cosylog(secx + tanx)dx = cosxlog(secy + tany)dy

⇒ sec log(sec tan )y y y dy+∫= +∫ sec log(sec tan )x x x dx

⇒ [log(sec tan )] [log(sec tan )]

.x x y y c+

=+

+2 2

2 2

31. (c) : dydx

y yx x

=− −+ −

2

2

2

2 3

⇒− +

=+ −

dyy y

dxx x( )( ) ( )( )2 1 3 1

⇒− +

=+ −∫ ∫

dyy y

dxx x( )( ) ( )( )2 1 3 1

⇒−

−+

⎛⎝⎜

⎞⎠⎟

=−

−+

⎛⎝⎜

⎞⎠⎟∫ ∫

1

3

1

2

1

1

1

4

1

1

1

3y ydy

x xdx

⇒−+

= −+

+42

13

1

3log log .

yy

xx

c

32. (c) : Given equation can be written as 1

2−⎛

⎝⎜

⎞

⎠⎟ =x

xdx dy

y

On integration, we get log log logx x y c− = +2

12

⇒ − + = ⇒ =log log log logx y c x cxy

x2 2 22

2

2

⇒ = ⇒ =cxy

e cx y ex x2

2

2 22 2

.

33. (c) : Given, ordydx

dy dxy xy

x y x= = =−2

2

2 2 2

Integrating both sides, dy dx

y x2 2

= ∫∫⇒ –2–ylog2 = –2–xlog2 + c

1

⇒ log log

log.

2

2

2

2

1

2

1

2 21

1

x y x ycc

c− = ⇒ − = =

34. (a, d)35. (a, c) : x = Ae–3t + Be–t ⇒

dxdt

Ae Bet t= − −− −3

3

At t = 0, x = 0 ⇒ 0 = A + B ...(i)

At t dxdt

A B= = ⇒ = − −0 12 12 3, ...(ii)

Solving (i) and (ii), we get A = –6, B = 6

36. (b, c) : Equation of such conics are

ax2 + by2 + cxy = 1 ⇒ Order = 3

Degree = 1(no parameters is being repeated)

37. (a, b) : xdx ydyydx xdy

x x yy

+−

=+sin ( )

2 2 2

3

⇒ + + =⎛⎝⎜

⎞⎠⎟

1

2

2 2 2 2 2cos ( ) ( )ec x y d x y x

yd x

yOn integrating, we get

⇒ − + =⎛⎝⎜

⎞⎠⎟

+1

2

1

2

2 22

cot( )x y xy

K

⇒ − + = +cot( ) ,x y xy

c2 22

2 where c = 2K

⇒+

= − +y

x cyx y

2

2 2

2 2tan( )


38. (a, d) : The equation of normal at P(x, y) is

( ) ( )Y y dydx

X x− = − −1

∴ +⎛⎝⎜

⎞⎠⎟

A x y dydx

,0 and B y xdydx

0, +

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

Now,

( )1 2 0

1 2

x y dydx x

+⎛⎝⎜

⎞⎠⎟

+

+= ⇒ x y dy

dxx+ = 3

⇒ y dydx

x= 2 ...(i)

⇒ = ⇒ = +∫∫ ydy xdx y x C22

22

Also (0,4) satisfy it, so C = 8

∴ y2 = 2x2 + 16 which represents a hyperbola.

Alsodydx

⎤⎦⎥

= =( , )

( )

4 4 3

2 4

4 3

2

3

39. (b, c) : y = c1xm + c

2xn

⇒ y1 = mc

1xm – 1 + nc

2xn – 1

⇒ y2 = m(m – 1)c

1xm – 2 + n(n – 1)c

2xn – 2

∴ On substituting in given equation, we get

m2 – 10m + 21 = 0, n2 – 10n + 21 = 0

⇒ m = 3 or 7 and n = 3 or 7

40. (a, b, c) : Given d

dxy dy

dxx y dy

dxx c⎛

⎝⎜⎞⎠⎟

= ⇒ = +2

2

⇒⎛

⎝⎜

⎞

⎠⎟ = +d

dxy x c

2 2

2 2

⇒ = + +y x x23

3α β where α = 2c

given y(0) = 1, y(1) = 1

⇒ = = −β α11

3,

∴ = = − + ⇒ ′ = − > >y f x x x f x x x23 2

3

3

3 1

30 1( ) ( ) for

∴ f(x) ↑

f f1

3

1

30

⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

>

∴ f(x) = 0 has only one real root.

41. (b, c) : Dividing throughout by cos2x sin2y, the

given differential equation becomes

(3 tanx + 4 coty – 7)sec2x dx – (4tanx + 7 coty – 5)

cosec2y dy = 0

⇒ 3tanx sec2x dx – 7sec2x dx – 7coty cosec2y dy +

5cosec2ydy + 4coty sec2x dx – 4tanx cosec2y dy = 0

⇒ 3 tanx d(tanx) – 7d(tanx) + 7coty d(coty)

– 5d( coty) + 4d(tanx coty) = 0

On integrating, we obtain

3

27

7

25 4

2 2tan tan cot cot tan cotx x y y x y c− + − + ⋅ =

42. (a) : Here V0 = 100, a = 20, b = 0, and e = f = 5

HencedQdt

Q+ =1

200

The solution of this linear equation is Q = ce–t/20 ...(i)

At t = 0, we are given that Q = a = 20

Substituting these values into equation (i) we find

that c = 20, so that equation (i) can be rewritten as

Q = 20e–t/20. For t = 20, Q = 20/e43. (b) : Here a = 0, b = 1, e = 4, f = 2, and V

0 = 10

The volume of brine in the tank at any time t is given

as V0 + et – ft = 10 + 2t

We require t when 10 + 2t = 50, hence, t = 20 min.

44. (d) : For the equationdQdt t

Q++

=2

10 24

This is a linear equation; its solution is

Q t t ct

= + ++

40 4

10 2

2 ...(i)

At t = 0, Q = a = 0. Substituting these values into

equation (i), we find that c = 0. We require Q at the

moment of overflow, i.e. t = 20.

Thus, lbQ =( ) + ( )

+ ( ) =40 20 4 20

10 2 2048

2

45. (b) : ∵ dTdt

k T= − −( )290 ⇒−( ) = −dT

Tk dt

290On integrating, we get

⇒ ln(T – 290) = –kt + c ...(i)

Initially, T = 370 K and t = 0, then

ln (80) = c ...(ii)

From Eq. (i) and (ii), ln (T – 290) = –kt + ln 80

ln

T kt−⎛⎝⎜

⎞⎠⎟

= −290

80 ⇒ − = −T e kt290

80

or T = 290 + 80e–kt

46. (d) : For t =10 min, T = 330

Then, 330 – 290 = 80e–10k

⇒ = −1

2

10e k ⇒ e10k = 2 ⇒ 10k = ln2 or k = ln2

10

47. (c) : At T = 295 K

295 – 290 = 80e–kt ⇒ 16 = ekt

⇒ = = ×4 22

10ln

lnkt t ∴ t = 40 min


48. A – P, S; B – P, S; C – Q, R49. A – Q ; B – R ; C – PA. y e e y e ex x x x= + = −− −4

14

2 4 2; ;

y e e y e ex x x x

24

34

16 2 64 2= + = −− −;

Now, y3 – 13y

1 = (64e4x – 2e–x) – 13(4e4x – 2e–x)

= 12e4x + 24e–x ⇒ y3 – 13y = 12(e4x + 2e–x) = 12y

∴ K = 12 and K/3 = 4

B. Since equation is of 2 degree, two lines are possible.

C. y u dydx

mu dudx

m m= ⇒ = −1

Substituting the value of y and dydx

in 2 44 4 6x y dy

dxy x+ = we have

2 44

2

4 1 4 66 4

4 2 1x u mu du

dxu x du

dxx u

m x um m m

m

m−

−+ = ⇒ = −

For homogeneous equation, 4 63

2m m= ⇒ =

and 2 1 23

2m m− = ⇒ =

50. (3)

51. (2) : Put andxX

yY

= =1 1

dxX

dX dyY

dY= − = −1 1

2 2and

⇒ = = =p dydx

XY

dYdX

XY

P2

2

2

2

dYdX

P=⎛⎝⎜

⎞⎠⎟( )say

∴ The given equation becomes

1 1 1 1

2

2

2 4

4

4

2

Y Y XXY

PX

XY

P− ⋅⎛

⎝⎜

⎞

⎠⎟ = ⋅

⇒ Y – XP = P2 or Y = PX + P2

Which is the Clairaut’s form

∴ The solution is Y = cX + c2 or1 2

ycx

c= +

52. (3) : The given equation is y = 2px + y2p3 ...(i)

Solving for x x yp

y p, = −2

1

2

2 2

Differentiating w.r.t y, we get

dxdy p p

yp

dpdy

yp y p dpdy

= = − ⋅ − − ⋅1 1

2 22

2 2

or 2 2 24 2 3p p y dp

dyyp y p dp

dy= − − −

or p yp y dpdy

yp( ) ( )1 2 1 2 03 3+ + + =

or ( )1 2 03+ +⎛⎝⎜

⎞⎠⎟

=yp p y dpdy

Neglecting the first factor which does not involvedpdy

,

we have

p y dpdy

+ = 0 ⇒ + =dpp

dyy

0

On integrating we get, logp + logy = logcor logpy = logc ⇒ py = c ...(ii)

Eliminating p between (i) and (ii)

y x cy

y cy

= ⋅ + ⋅22

3

3 or y cxy

cy

= +23

or y2 = 2cx + c3 which is the required solution

53. (4) : Put sin cosy t y dydx

dtdx

dtdx x

t ex

x= ⇒ = ∴ + =4

3

I.Fln

.( / )= ∫ = =e e xx dx x4 4 4

∴ The solution is tx xe dxx4 = +∫ λ⇒ tx4 = ex(x – 1) + λ i.e., siny = ex(x – 1)x–4

λ = 0 due to initial condition.

54. (3)55. (1) : Equation of normal at the point p(x, y) is

Y y dxdy

X x− = − −( )

Let, m dydx

X mY x my= ⇒ + − +( ) = 0 ...(i)

Distance of perpendicular from the origin to line (i)

isx my

my

+

+=

12

⇒ = −dydx

y xxy

2 2

2

This is homogeneous equation. Let, y = zx

⇒ = + ⇒+

= −dydx

z x dzdx

zz

dz dxx

2

12

Integrating both sides, 2

12

zz

dz dxx+

= −∫ ∫⇒ log(1 + z2) = –logx + c ⇒ (x2 + y2) = x · ec

This curve passes through (1, 1)

⇒ 1 + 1 = 1 · ec ⇒ ec = 2

The required equation of the curve is

⇒ x2 + y2 = 2x

ANSWER KEYMPP-5 CLASS XI

1. (a) 2. (d) 3. (b) 4. (d) 5. (a)

6. (a) 7. (a,d) 8. (a,b,c,d) 9 . (a,b,c,d)

10. (b,d) 11. (b,c) 12. (a,b,c) 13. (b) 14. (b)

15. (d) 16. (b) 17. (8) 18. (3) 19. (0)

20. (1)


VECTORQuantities which have both

magnitude and direction are

called vector.

It is generally represented by AB , where A is the initial

point and B is the terminal point.

The arrow indicates the direction of the vector.

MAGNITUDE OF A VECTORThe magnitude of a vector AB is the length of the line

segment representing it. So, magnitude will be non-

negative quantity. It is represented by AB .

POSITION VECTORIf P is a point in space having coordinates (x, y, z) w.r.t.

the origin O(0, 0, 0). Then the vector OP is called the

HIGHLIGHTS

CLASS XII Series 7

Previous Years Analysis

2016 2015 2014Delhi AI Delhi AI Delhi AI

VSA 2 2 2 2 2 2

SA 1 1 1 1 1 1

LA - - - - - -

Vector Algebra

position vector of the point P. The magnitude of OP is,

OP x y z= + +2 2 2

DIRECTION COSINESIf a vector a makes angles

α , β , γ with the posit ive

direction of X, Y and Z axes

respectively, then cos α, cos β,

cos γ are called direction

cosines of vector a and are

usually denoted by l, m, n.

Vectors lie in the same plane or parallel to the same plane.

Vectors having opposite direction.

Type

s of

Vec

tors A vector whose magnitude is 0

and direction is in determinate.

Representation : 0

Vectors with same initial point.

Coinitial Vectors

Vectors having equal magnitudes and same direction.

Vectors having same

magnitude but opposite

in direction.

A vector whose magnitude is unity.Representation : a a

a^ =

Unlike Vectors

Coplanar Vectors

Zero or Null Vector Equal Vectors

Unit Vector

Vectors parallel to the same line irrespective of their magnitudes and directions.

Collinear or Parallel Vectors

Negative of a VectorVectors having same direction.

Like Vectors


COMPONENTS OF A VECTORLet O be the origin and P(x, y, z) be any point in space. Let i j k, , be unit vectors along X, Y and Z-axes

respectively. Then OP xi y j zk= + +

Here x, y, z are scalar components and xi y j zk, , are vector components of OP .

OPERATIONS ON VECTORS

Operations Properties

Addition of

two vectors

Triangle Law - InΔOAB,

let position vectors of

OA and AB be a and

b respectively. Then

OB a b= +

Commutative Law - a b b a+ = +Associative Law - a b c a b c+ + = + +( ) ( )

Existence of Additive Identity - a a a+ = = +0 0

i.e., 0 is the additive identity

Existence of Additive Inverse - a a a a+ − = = − +( ) ( )0

i.e., ( )−a is the additive inverse of a .

Parallelogram Law -

In parallelogram OABC,

let position vectors of

OA and AB be a and

b respectively. Then

OB a b= +

Multiplication

of a vector by

a scalar

Let a be any vector. Then for any scalar

m, the product m a is defined as the

vector whose magnitude is |m| times to

that of a and the direction is same as

that of a , if m is positive and opposite

to that of a if m is negative.

Commutative Law - m a a m ma( ) ( )= =Associative Law - m na n ma mn a( ) ( ) ( )= =

Distributive Law - ( )m n a ma na+ = + and

m a b ma mb( )+ = +

Multiplication

of two vectors

Scalar Product or Dot Product - Let

a and b be any two non-zero vectors,

then scalar or dot product of a and b

is denoted and defined by,

a b a b⋅ = cos ,θ where θ is the angle

between a and b .

Commutative Law - a b b a⋅ = ⋅Distributive Law - a b c a b a c⋅ + = ⋅ + ⋅( )

Vector Product or Cross Product - Let

a and b be any two non-zero vectors,

then vector or cross product of a

and b is denoted and defined by,

a b a b n× = sin ,^θ where θ is the angle

between a and b and n^ is the unit

vector along a b× .

Distributive Law - a b c a b a c× + = × + ×( )

Vector product is not commutative.

Vector product is not associative.


SOME IMPORTANT RESULTSScalar Product Vector Product

a b⋅ is a scalar quantity. a b× is a vector quantity.

a b a b⋅ = ⇔0 ⊥ or a b= =0 0or a b a b× = ⇔0 | |

If θ = 0, then a b a b⋅ =

If θ = π, then a b a b⋅ = −

i i j j k k⋅ = ⋅ = ⋅ = 1

i j j k k i⋅ = ⋅ = ⋅ =0

Angle between two non-zero vectors a

and b is cos cosθ θ= ⋅ = ⋅⎛

⎝⎜

⎞

⎠⎟

−a ba b

a ba b

or1

If θ π=2

, then a b a b× =

i i j j k k× = × = × = 0

i j k j k i k i j× = × = × =, ,

j i k k j i i k j× =− × =− × =−, ,

Angle between two non-zero vectors a and b is

sinθ = ×a ba b

Area of ΔABC = 1

2a b× , where a b, are adjacent sides.

Area of parallelogram ABCD = a b× , where a b, are

adjacent sides.

VECTOR JOINING TWO POINTSLet A ≡ (x1, y1, z1) and B ≡ (x2, y2, z2). Then

AB x x i y y j z z k= − + − + −( ) ( ) ( )2 1 2 1 2 1

Also, AB x x y y z z= − + − + −( ) ( ) ( )2 12

2 12

2 12

SECTION FORMULALet A and B be any two points with position vectors a

and b respectively. Let C ( )c divides AB in the ratio

m : n, then

Internal division : c mb nam n

= ++

External division : c mb nam n

= −−

MID POINT FORMULA

Let C is the mid point of AB then, c a b= +2

PROJECTION OF A VECTOR ON A LINEProjection of a given vector a along a directed line

l is given by a b

p⋅

| |, where p is a vector along the

direction of the line.

Projection of a vector a on other vector b is given

by a b a bb

a bb

⋅ ⋅ ⋅or

( )or

( )

SCALAR TRIPLE PRODUCTLet a b c, , be any three vectors. Then the scalar

product of a b× and c is called scalar triple product

of a b c, and . It is denoted by ( )a b c× ⋅ or [ ]a b c or

[ , , ]a b c .

Remarks -

a b ca a ab b bc c c

⋅ × =( ) ,

1 2 3

1 2 3

1 2 3

where a a i a j a k= + +1 2 3 , b b i b j b k= + +1 2 3

and c c i c j c k= + +1 2 3

[ ] [ ] [ ]a b c b c a c a b= =

[ ] [ ]a b c a c b= −

[ ]a a b = 0

Volume of parallelopied whose coterminous edges

are a b c, , is [ ( )]a b c⋅ ×COPLANARITY OF THREE VECTORSThree vectors a b, and c are said to be coplanar iff

a b c⋅ × =( ) 0 i.e., [ ]a b c = 0

PROBLEMSVery Short Answer Type

1. If r i j k r i j k1 23 2 2 4 3= − + = − −, and

r i j k3 2 2= − + + , find the modulus of 2 3 51 2 3r r r− − .


2. Let a i j b i j= − = +2 3 3 2and . Is a b= ? Are the

vectors a and b equal ?

3. If a band are two vectors such that a b= =10 15,

and a b⋅ = 75 2, find the angle between a band .

4. Find a unit vector in the direction of the resultant of

vectors i j k i j k i j+ + − + + +2 3 2 3, .and

5. If a b a b i j k= = × = + +2 7 3 2 6, ( ) ,and find the

angle between a band .

Short Answer Type6. Vectors drawn from the origin O to the points

A, B and C are respectively a b, and 4 3a b− .

Find AC BCand .

7. If a i j k b i j k= + − = − +2 3 3 2and , show that the

vectors a b a b+ −and are perpendicular to each

other.

8. Find the projection of the vector 2 3 6i j k− − on the

line joining the points (5, 6, – 3) and (3, 4, – 2).

9. If a i j k= + +4 3 2^ ^ ^

and b i k= +3 2^ ^

, find | | .b a× 2

10. Find the area of the parallelogram whose adjacent

sides are a i j k b i j k= + + = − +2 3 3 2and .

Long Answer Type-I

11. The scalar product of the vector i j k+ + with a unit

vector along the sum of the vectors 2 4 5i j k+ − and

λi j k+ +2 3 is equal to one. Find the value of λ.

12. If a i j k= − −^ ^ ^,3 b i j k= − +4 3

^ ^ ^ and c i j k= − +2 2

^ ^ ^,

verify that a b c a b a c× + = × + ×( ) .

13. Prove that [ ] [ ]a b b c c a a b c+ + + = 2 .

14. (i) Find the volume of the parallelopiped whose

coterminous edges are represented by the vectors

v i j k v i j k v i j k1 2 32 2 3 3 2= + − = + + = − +, .and

(ii) Show that the vectors a i j k= − +2 3 ,

b i j k= − + −2 3 4 and c i j k= − +3 5 are coplanar.

15. Prove that a b c b c a c a b× + + × + + × + =( ) ( ) ( ) .0

Long Answer Type-II

16. Prove that four points 4 5i j k j k+ + − +, ( ),

3 9 4i j k+ + and 4( )− + +i j k are coplanar.

17. Using vector method, prove that in a ΔABC,a

Ab

Bc

Csin sin sin= =

where a, b, c are the lengths of the sides opposite to

the angles A, B and C respectively of ΔABC.

18. (i) If a b⋅ = 0 and a b× = 0, prove that a = 0

or .b = 0

(ii) Prove that ( ) ( ) ( ).a b a b a b− × + = ×2

19. For any three vectors a b c, , show that the vectors

( ),( ),a b b c− − ( )c a− are coplanar.

20. Using vectors find the area of ΔABC whose vertices

are A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1).

SOLUTIONS

1. 2 3 51 2 3r r r− −

= − + − − − − − + +2 3 2 3 2 4 3 5 2 2( ) ( ) ( )i j k i j k i j k= − +5 2i j k∴ 2 3 5 5 2 1 301 2 3

2 2 2r r r− − = + − + =( ) .

2. a i j= − = + − =2 3 2 3 132 2

( )

and b i j= + = + =3 2 3 2 132 2 ∴ a b=

Since the corresponding components of the given

vectors a and b are not equal therefore, a b≠ .

3. Let θ be the angle between the vectors a band , then

0 ≤ θ ≤ π and a b a b⋅ =| || | cos θ

⇒ 75 2 10 1575 2

150= × ⇒ =cos cosθ θ

⇒ cos .θ θ= ⇒ = °1

245

4. Let a be the resultant of given vectors. Then,

a i j k i j k i j i j k= + + − + + + + = + +( ) ( )2 3 2 3 3 5 4

∴ a = + + =3 5 4 5 22 2 2

Now unit vector along a aa

i j k= = + +3

5 2

5

5 2

4

5 2

5. Given, a b a b i j k= = × = + +2 7 3 2 6, and

∴ a b× = + + = =3 2 6 49 72 2 2

Let θ be the angle between a band

Now, sinθ = ×a ba b

=×

=7

2 7

1

2 ∴ θ π=

6

6. In ΔOAC, we have OA AC OC+ =

4–

3

⇒ AC OC OA= − = − −4 3a b a = −3( )a bIn ΔOBC, we have

OB BC OC+ =⇒ BC OC OB= − = − − = −4 3 4a b b a b( ).


7. We have, a i j k b i j k= + − = − +2 3 3 2and ,

∴ a b i j k+ = + −4 and a b i j k− = − + −2 3 5

∴ ( ) ( ) ( ) ( ) ( )( )a b a b+ ⋅ − = − + × + − −4 2 1 3 1 5 = 0.

Thus the dot product of two non-zero vectors

a b a b+ −and is zero.

Therefore, the vectors a b a b+ −and are

perpendicular to each other.

8. Let a i j k= − −2 3 6 , A ≡ (5, 6, – 3), B ≡ (3, 4, – 2)

∴ AB i j k= − + − + − +( ) ( ) ( )3 5 4 6 2 3

= − − +2 2i j kNow the projection of a ABon

= ⋅ =− + −

+ += −a AB

AB4 6 6

4 4 1

4

3

9. 2 8 6 4a i j k= + +^ ^ ^ and b i k= +3 2

^ ^

∴ b i k a i j k= + = + +3 2 2 8 6 4^ ^ ^ ^ ^

and

Now, b ai j k

× =2 3 0 2

8 6 4

^ ^ ^

= − + +12 4 18i j k^ ^ ^

∴ | | ( ) ( )b a× = − + + =2 12 4 18 222 2 2

10. Area of the parallelogram = ×a b

Now, a bi j k

i j k× =−

= + −1 2 3

3 2 1

8 8 8

∴ a b× = + + − =8 8 8 8 32 2 2

( ) sq. units.

11. Let a i j k b i j k= + − = + +2 4 5 2 3and λThen, a b i j k+ = + + −( )2 6 2λUnit vector along the sum of these vectors is

( )

( ) ( )

2 6 2

2 6 22 2 2

+ + −

+ + + −

λ

λ

i j k

Given, ( )( )

( ) ( )

i j k i j k+ + ⋅

+ + −

+ + + −=

2 6 2

2 6 2

12 2 2

λ

λ

⇒ 1 × (2 + λ) + 1 × 6 + 1 × (–2) = ( )2 402+ +λ

⇒ λ + 6 = ( )2 402+ +λ ⇒ λ = 1.

12. Given, a i j k= − −^ ^ ^,3 b i j k= − +4 3

^ ^ ^ and c i j k= − +2 2

^ ^ ^

Now, b c i j k i j k+ = − + + − +( ) ( )^ ^ ^ ^ ^ ^

4 3 2 2

= − +6 4 3i j k^ ^ ^

∴ a b ci j k

× + = − −−

( )

^ ^ ^

1 1 3

6 4 3

= − − +15 21 2i j k^ ^ ^ ...(i)

Also, a bi j k

× = − −−

^ ^ ^

1 1 3

4 3 1

= − − +10 13i j k^ ^ ^

and a ci j k

× = − −−

^ ^ ^

1 1 3

2 1 2

= − − +5 8i j k^ ^ ^

∴ a b a c i j k i j k× + × = − − + + − − +( ) ( )^ ^ ^ ^ ^ ^

10 13 5 8

= − − +15 21 2i j k^ ^ ^ ...(ii)

From (i) and (ii), we have

a b c a b a c× + = × + ×( )

13. L.H.S. = + + +[ ]a b b c c a = + ⋅ + × +( ) [( ) ( )]a b b c c a

= + ⋅ × + × + × + ×( ) [ ]a b b c b a c c c a

= + ⋅ × + × + ×( ) [ ]a b b c b a c a [ ]∵ c c× = 0

= ⋅ × + ⋅ × + ⋅ × + ⋅ ×a b c a b a a c a b b c( ) ( ) ( ) ( ) + ⋅ × + ⋅ ×b b a b c a( ) ( ) = + + +[ ] [ ] [ ] [ ]a b c a b a a c a b b c

+ +[ ] [ ]b b a b c a

= + = =[ ] [ ] [ ] R.H.Sa b c b c a a b c2

14. (i) [ ]v v v1 2 3

2 1 11 2 33 1 2

=−

− = 14 + 7 + 7 = 28.

Hence, the required volume = 28 cubic units.

(ii) Given vectors a b c, , are coplanar if [ ]a b c = 0

Now, [ ]a b c =−

− −−

1 2 32 3 4

1 3 5= 3 – 12 + 9 = 0

Hence, the given vectors a b c, and are coplanar.15. We have,

a b c b c a c a b× + + × + + × +( ) ( ) ( )= × + × + × + × + × + ×( ) ( ) ( ) ( ) ( ) ( )a b a c b c b a c a c b

[by distributive law]= × − × + × − × + × − ×( ) ( ) ( ) ( ) ( ) ( )a b c a b c a b c a b c= 0Hence, a b c b c a c a b× + + × + + × + =( ) ( ) ( ) .0


16. Let the given points be A, B, C, D respectively.Let O be the origin, then OA i j k OB j k= + + = − +4 5 , ( ),

OC i j k OD i j k= + + = − + +3 9 4 4, ( )

∴ AB OB OA j k i j k= − = − + − + +( ) ( )4 5 = − − −4 6 2i j k

AC OC OA i j k i j k= − = + + − + +( ) ( )3 9 4 4 5 = − + +i j k4 3

AD OD OA i j k i j k= − = − + + − + +4 4 5( ) ( )

= − − +8 3i j k

Now, [ , , ]AB AC AD =− − −−− −

4 6 21 4 38 1 3

= – 60 + 126 – 66 = 0Hence the points A, B, C, D are coplanar.

17. In triangle ABCLet BC a CA b AB c= = =, ,then a b c+ + = 0 ...(i)Taking vector product with a, on both sides of (i) we have a a b c a× + + = ×( ) 0⇒ a a a b a c× + × + × = 0⇒ a b a c× + × = 0 ⇒ a b c a× − × = 0

⇒ a b c a× = × ...(ii)Similarly, taking vector product with b on both sides of (i) we have a b b c× = × ...(iii)From (ii) and (iii), we get a b b c c a× = × = ×⇒ a b b c c a× = × = ×or ab sin (π – C) = bc sin (π – A) = ca sin (π – B)or ab sin C = bc sin A = ca sin BDividing throughout by abc, we getsin sin sinC

cA

aB

b= = or a

Ab

Bc

Csin sin sin= =

18. (i) Let a b a b⋅ = × =0 0and .⇒ a b a b= = ⊥0 0or orand a b a b= =0 0or or || ⇒ a b= =0 0or[ and || can never hold simultaneously].∵a b a b⊥

Hence, a b a b⋅ = × =0 0and⇒ a b= =0 0or .

(ii) We have ( ) ( )a b a b− × += × + × − × − ×a a a b b a b b [by distributive law]= × − ×a b b a [ and ]∵ a a b b× = × =0 0= × + × = × − × = ×( ) ( ) ( ) [ ( ) ( )]∵a b a b a b b a a b2Hence, ( ) ( ) ( ).a b a b a b− × + = ×2

19. Consider, [ ]a b b c c a− − − = − ⋅ − × −( ) [( ) ( )]a b b c c a = − ⋅ × − × − × + ×( ) [ ( ) ( ) ]a b b c b a c c c a [by distributive law] = − ⋅ × + × + ×( ) [ ]a b b c a b c a = ⋅ × + ⋅ × + ⋅ ×a b c a a b a c a( ) ( ) ( ) − ⋅ × − ⋅ × − ⋅ ×b b c b a b b c a( ) ( ) ( )

= + + −[ ] [ ] [ ] [ ]a b c a a b a c a b b c− −[ ] [ ]b a b b c a

= −[ ] [ ]a b c b c a = − = =[ ] [ ] ( [ ] [ ])∵a b c a b c b c a a b c0Hence, the vectors ( ), ( ), ( )a b b c c a− − − are coplanar.

20. We have,Position vector of A i j k= + +( ),2 3

Position vector of B i j k= − +( )2 4 andPosition vector of C i j k= + −( )4 5

(1, 2, 3) (2, –1, 4)

(4, 5, –1)

∴ AB = − + − + + = − +( ) ( )2 4 2 3 3i j k i j k i j k

AC = + − − + + = + −( ) ( )4 5 2 3 3 3 4i j k i j k i j k

Now, AB ACi j k

× = −−

1 3 13 3 4

= − − − − + +( ) ( ) ( )12 3 4 3 3 9i j k = + +9 7 12i j k

∴ Area of ΔABC = 12

AB AC×

= + + = ⋅ + +12

9 7 12 12

9 7 122 2 2( ) ( ) ( ) ( )i j k

= ⋅ + + =12

81 49 144 12

274( ) sq. units


1. Vertex A of the acute triangle ABC is equidistant from the circumcentre O and the orthocentre H. The possible value of measure of ∠A is(a) 30° (b) 60° (c) 75° (d) 90°

2. If A, B, C, D are four points in space such that ∠ABC = ∠BCD = ∠CDA = ∠DAB = π/2, then(a) A, B, C, D are coplanar points(b) A, B, C, D are non-coplanar points(c) line AB is skew to line CD(d) nothing definite can be said about their relative

positions3. Given P, Q, R, S are points on the sides of the quadri-

lateral ABCD such that

APPB

BQQC

CRRD

DSSA

k= = = =

and the area of the quadrilateral PQRS is exactly 52% of the area of quadrilateral ABCD then k =(a) 1/3 (b) 2/3 (c) 2/1 (d) 1/2

4. The volume of the region in space defined by |x + y + z| + |x + y – z| ≤ 8, (x, y, z) ≥ 0 is(a) 16 (b) 24 (c) 32 (d) 48

5. Let P be a point in the interior of a tetrahedron ABCD such that its projections A1, B1, C1, D1 onto the planes (BCD), (CDA), (DAB), (ABC) respective-ly are all situated in the interior of the faces. If Δ is the total area and r, the inradius of the tetrahedron, such that Δ Δ Δ Δ ΔBCD CDA DAB ABC

PA PB PC PD r1 1 1 1+ + + =

then

point P is(a) centroid of tetrahedron ABCD(b) circumcentre of tetrahedron ABCD(c) incentre of tetrahedron ABCD(d) orthocentre of tetrahedron ABCD

6. If ‘a’ is a real constant and α, β, γ are variable angles and ( )tan tan ( )tana a a a2 24 4 6− + + + =α β γ then the least value of tan2α + tan2β + tan2γ is(a) 3 (b) 4 (c) 7 (d) 12

7. If vectors a c x i j k= − +( log ) ^ ^ ^2 6 3 and

b x i j c x k= + +(log ) ( log )^ ^ ^2 22 2 make an obtuse angle

for any x ∈ (0, ∞) then c ∈

(a) −⎛⎝⎜⎞⎠⎟

43

0, (b) (–2, 0)

(c) −⎛⎝⎜⎞⎠⎟

43

1, (d) − −⎛⎝⎜

⎞⎠⎟2 4

3,

8. The maximum value of | | | | | |a b b c c a− + − + −2 2 2, where a b c, , are unit vectors is(a) 6 (b) 9 (c) 12 (d) 24

9. If v^ is the unit vector along the incident ray, w^ is the unit vector along the reflected ray and a^ is the unit vector along the outward normal to the plane mir-ror at the point of incidence then w^ =(a) v a a v^ ^ ^ ^( )− ⋅2 (b) v a a v^ ^ ^ ^( )− ⋅(c) a v a v^ ^ ^ ^( )− ⋅2 (d) a v a v^ ^ ^ ^( )+ ⋅2

10. Arc AC of a circle subtends a right angle at the centre O. Point B divides the arc in the ratio 1 : 2. If OA a= and OB b= then OC =(a) 3 2a b+ (b) − +3 2a b(c) 3 2a b− (d) 2 3a b−

11. If (l, m, n) are the direction cosines of a line, then maximum value of product (lmn) is

(a) 3 (b) 13

(c) 3 3 (d) 13 3

12. The four planes lx + my = 0, nz + lx = 0, my + nz = 0 and lx + my + nz = p form a tetrahedron whose vol-ume is

(a) 23

3plmn

(b) 32

3plmn

(c) 32 3lmnp

(d) 23 3lmnp

13. If the lines x = ay + b, z = cy + d and x = a′y + b′, z = c′y + d′ are perpendicular then(a) aa′ + cc′ + 1 = 0 (b) aa′ – bb′ = 0(c) aa′ = dd′ (d) bc′ + b′c + 1 = 0

By : Tapas Kr. Yogi, Mob : 9533632105.


14. A uni-modular tangent vector on the curve

x = t2 + 2, y = 4t – 5, z = 2t2 – 6t at t = 2 is λ( )^ ^ ^

2 2i j k+ +

where λ =

(a) 1 (b) 1/2 (c) 1/3 (d) 1/4

15. A rigid body rotates about an axis through the origin

with an angular velocity of 10 3 radians/sec. If ω

point in the direction of i j k^ ^ ^+ + then the equation

to the locus of the points having tangential speed of

20 m/sec is x2 + y2 + z2 – xy – yz – zx + λ = 0 where λ =(a) 1 (b) –1 (c) 2 (d) –2

16. A rod of length 2 units whose one end is

(1, 0, –1) and the other end moves on the plane

x – 2y + 2z + 4 = 0 encloses a volume (in cubic units) of

(a) π (b) 2π (c) 3π (d) 4π17. Let T be the largest of the areas of the four faces of a

tetrahedron and Y be the total surface area of the

tetrahedron. Which of the following statements is true?

(a) 2 4< ≤Y

T (b)

3 4< ≤Y

T(c)

5

2

9

2< ≤Y

T (d)

7

2

11

2< <Y

T18. Given ΔABC and ΔAEF such that B is the midpoint

of EF. Also AB = EF = 1, BC = 6, CA = 33 and

AB AE AC AF⋅ + ⋅ = 2. The cosine of the angle between

EF BC and is

(a) 1/3 (b) 2/3 (c) 3/4 (d) 3/5

SOLUTIONS1. (b) : Let circumcentre be origin in our vector

system then | | | | | |A A A H− = = −0 and circumradius

R A B C= = =| | | | | | and H A B C= + + and let

a B C= −| |

⇒ R A A H B C2 2 2 2= = − = +| | | | | |

So, R2 = 4R2 – a2

i e aR

A. ., sin= ⇒ =33

2 (by sine rule).

So, A = 60°

2. (a) : Let B = (0, 0, 0), A = (v, 0, 0), C = (0, w, 0),

D = (a, b, c).

Since, ∠BAD = 90°. So a = v and ∠BCD = π/2 gives

b = w.

Now, v2 + w2 = CA2 = CD2 + DA2

= v2 + c2 + w2 + c2

⇒ c = 0

i.e. Point D is also in the XY plane along with the other

points; i.e. coplanar.

3. (b) : Using section formula,

PR kBD AC

k= +

+1 and QS kAC BD

k= − +

+1

So, area of PQRS = [PQRS] = ( )PR QS× ⋅ 12

= +

+⎛⎝⎜

⎞⎠⎟ × − +

+⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

2 1 1

kBD ACk

kAC BDk

= +

+⋅ ×( )

( )( )

kk

AC BD2

2

1

2 1

[ ]( )

( )[ ]P Q R S k

kABCD= +

+

2

2

1

1

⇒ = ++

52

100

1

1

2

2

kk( )

Solving, k = 2

3

3

2 or

4. (c) : Let a = x + y, so, |a + z| + |a – z| ≤ 8

⇒ 8 ≥ |a + z| + |a – z| ≥ |a + z + a – z| = 2ai.e. a = x + y ≤ 4

and 8 ≥ |a + z| + |a – z| ≥ (a + z) – (a – z) = 2zi.e. z ≤ 4

So, 0 ≤ x + y ≤ 4, 0 ≤ z ≤ 4.

So, a right prism formed with vertices (0, 0), (4, 0),

(0, 4) and height = 4.

So, volume = (base area) × height =

4 4

24 32

× × =

5. (c) : Using Cauchy-Schwarz inequality,

ab

ab

ab

ab

a b a b a b a b1

1

2

2

3

3

4

41 1 2 2 3 3 4 4+ + +⎛

⎝⎜⎞⎠⎟

+ + +( )

≥ (a1 + a

2 + a

3 + a

4)2

With a1 = ΔBCD, a

2 = ΔCDA, a

3 = ΔDAB, a

4 = ΔABC,

b1 = PA

1, b

2 = PB

1, b

3 = PC

1, b

4 = PD

1 gives,

31

2VPABCDΔ

Δ≥∑ , where V is volume.

So,

Δ Δ ΔΔ

ΔBCDPA V r r1

2 2

3≥ = =∑

and equality occurs if ab

a b ii

ii i= ⋅ =λ ( , , , )1 2 3 4

i e bi. . = 1

λ. So, PA

1 = PB

1 = PC

1 = PD

1

i.e. P is the incentre of tetrahedron ABCD.

6. (d) : Consider the two vectors

P a i a j a k= − + + +( ) ( )^ ^ ^2 2

4 4 and

Q i j k= + +(tan ) (tan ) (tan )

^ ^ ^α β γ then

( ) | | | |P Q P Q⋅ ≤2 2 2 gives the required least value = 12


7. (a) : a b⋅ < 0 (obtuse angle)

⇒ c · (log2x)2 + 6c(log

2x) – 12 < 0

So, c < 0 and 36c2 + 48c < 0 i e c. . ,∈ −⎛⎝⎜

⎞⎠⎟

4

30

8. (b) : Using, | | ( ) /a b c a b+ + ≥ ⇒ ⋅ ≥ −∑20 3 2

So, | | | | | | ( ) ( )a b b c c a a b− + − + − = − ⋅ =∑2 2 22 3 2 9

Equality occurs when each of the vectors are inclined

at 120° to each other.

9. (a) : Let θ be the incident angle to normal to the

mirror then | | | | | |^ ^ ^ ^ ^ ^w v w v w v− = + − ⋅2 2 2

2

= 1 + 1 – 2 · 1 · 1 cos(π – 2θ)

⇒ − =| | cos^ ^w v 2 θ and a

w v

w v

^^ ^

^ ^| |

=−

−So, w v a a v^ ^ ^ ^ ^

( )= − ⋅2

10. (b) : Let c xa yb= +

| | | | | |a b c r= = =

⇒

a b r a c⋅ = ⋅ =3

20

2

,

and b c r⋅ =

2

2

Considering c c⋅ and simplifying we have

x y i e c a b= − = = − +3 2 3 2, . .

11. (d) : Use A.M. ≥ G.M. on l2, m2, n2 and

l2 + m2 + n2 = 1

12. (a) : On solving the four plane equations three at a

time, we have the vertices of the tetrahedron as (0, 0, 0),

pl

pm

pn

pl

pm

pn

pl

pm

pn

, , , , , , , ,−⎛

⎝⎜⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

So, required volume =−

−−

1

6

0 0 0 1

1

1

1

p l p m p np l p m p n

p l p m p n

/ / /

/ / /

/ / /

= 2

3

3plmn

13. (a) : Rewrite the given line equation as

x ba

y z dc

− ′′

= − = − ′′

0

1 and

x ba

y z dc

− = − = −0

1

So, perpendicular condition gives

aa′ + 1⋅1 + cc′ = 0.

14. (c) : The position vector at any point t is

r t i t j t t k= + + − + −( ) ( ) ( )

^ ^ ^2 4 5 2 6

2 2

So, drdt

i j kt=

= + +2

4 4 2^ ^ ^

and

drdt t

⎤⎦⎥

= + + ==2

16 16 4 6

So, unit tangent vector = 4 4 2

6

i j k^ ^ ^+ +

15. (d) : n

i j kn i j k^

^ ^ ^^ ^ ^ ^

, | | ( )=+ +

= = + +3

10ω ω

v r i j k x i y j z k= × = + + × + +ω 10( ) ( )

^ ^ ^ ^ ^ ^

and | |v = 20 gives locus as

x2 + y2 + z2 – xy – zx – yz – 2 = 0.

16. (c) : The rod sweeps out a cone

The distance of (1, 0, –1) from the plane is | |1 2 4

91

− + =

The slant height of the cone = 2

So, radius of base = − =4 1 3

So, volume = ⋅ =π π( )3 1 32

17. (a) : Clearly YT≤ 4, with equality holding for regular

tetrahedron. On the other hand since the sum of the

areas of any three faces of a tetrahedron is greater than

that of the fourth so, YT

> 2.

18. (b) : From given data,

2 = ⋅ + ⋅ = ⋅ + +AB AE AC AF AB AB BE AC( ) ⋅ +( )AB BF

i e AB AB BE AC AB AC BF. .,2

2+ ⋅ + ⋅ + ⋅ = ...(1)

and AB AC AB21 33 1

33 1 36

2 33 11= ⋅ = ⋅ ⋅ + −

⋅ ⋅= −,

( )

and so, eqn.(1) becomesBE BF= −

1 1 2 2+ ⋅ − − = ⇒ ⋅ =BF AC AB BF BC( )

i e BF BC. ., | || | cosθ = 2

Hence, cosθ = 2

3


Only One Option Correct Type

1. x xdxsec2

2∫ =

(a) x2

tan 2x + 1

4ln sin 2x + c

(b) x2

tan 2x + 1

4ln cos 2x + c

(c) x2

sec 2x + 1

4 ln sin 2x + c

(d) x2

sec 2x + 1

4 ln cos 2x + c

2. dx

x x2 4

3

41( )+

=∫

(a) ( )x

xc

4

1

41+ + (b) − + +( )xx

c4

1

41

(c) x

xc

41+ + (d) − + +x

xc

41

3. cos

(cos sin )

2

2

x dxx x+∫ =

(a) 1

2 4lnsin x c+⎛

⎝⎜⎞⎠⎟

+π (b)

1

2 4ln cos x c+⎛

⎝⎜⎞⎠⎟

+π

(c) ln cos x c+⎛⎝⎜

⎞⎠⎟

+π4

(d) ln cos x c−⎛⎝⎜

⎞⎠⎟

+π4

4. x

x x xdx

2

3 4 2

1

2 2 1

−

− +=∫

(a) 2 2 1

4 2

2

x xx

c− + + (b) 2 2 1

4 2

3

x xx

c− + +

(c) 2 2 1

2 2x xx

c− + + (d) 2 2 1

2

4 2

2

x xx

c− + +

5. e x xx

dxxln

. ( )1

1

22

4

1

1

+⎛

⎝⎜⎞

⎠⎟∫−

+=

(a) ln xx

c2

2

1+⎛⎝⎜

⎞⎠⎟

+ (b) ln 112

+⎛⎝⎜

⎞⎠⎟

+x

c

(c) 1

2

12

2ln x

xc+⎛

⎝⎜⎞⎠⎟

+ (d) 1

21

12

ln +⎛⎝⎜

⎞⎠⎟

+x

c

6. dx

x x( )4 3 3 42 2+ −

∫ =

(a) 1

5

2

3 4

1

2tan

−

−+x

xc

(b) 1

10

5

2 3 4

1

2tan

−

−

⎛

⎝⎜

⎞

⎠⎟ +x

xc

(c) 1

5

5

2 3 4

1

2tan

−

−+x

xc

(d) 1

10

5

3 4

1

2tan

−

−+x

xc

Indefinite Integration

This specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four

marks for correct answer and deduct one mark for wrong answer.

Self check table given at the end will help you to check your readiness.

Total Marks : 80 Time Taken : 60 Min.

Class XII


One or More Than One Option(s) Correct Type

7. I f3 4

2 42

3

xx x

dx x K f x c+− −

= − + +∫ log | | log ( ) ,

then

(a) K = –(1/2)

(b) f (x) = x2 + 2x + 2

(c) f (x) = |x2 + 2x + 2|

(d) K = 1/4

8. If the antiderivative of sin–1

x

x +1 is

x sin–1

x

xx fog x c

+− + +

1( ) then

(a) f (x) = sin–1x (b) g x x( ) = +1

(c) f (x) = tan–1x (d) g x x( ) =

9. If cos cos

cos

x xx

dx−−∫

3

31

= f(x), then f(x) =

(a) 3

2sin

–1(cos

3/2x) + c

(b) − cos–1

(cos3/2x) + c

(c) − 2

3sin

–1 (cos

3/2x) + c

(d) − sin–1

(cos3x) + c

10. cos cos

sin sin

3 5

2 4

x xx x

dx++∫ is equal to

(a) sin x – 2 cosec x + 5 tan–1

(cos x) + c(b) sin x – 2 cosec x – 6 tan

–1(sin x) + c

(c) –[cosec x + cot x cos x + 6 tan–1

(sin x)] + c(d) sin x – 6 tan

–1(sin x) + c.

11. dxx xsin cos+

=∫3

(a) 1

2 6 6log sec tanx x c−⎛

⎝⎜⎞⎠⎟ + −⎛

⎝⎜⎞⎠⎟ +π π

(b) 1

2 3 3log cosec cotx x c+⎛

⎝⎜⎞⎠⎟ − +⎛

⎝⎜⎞⎠⎟ +π π

(c) − +⎛⎝⎜

⎞⎠⎟ + +⎛

⎝⎜⎞⎠⎟ +1

2 3 3log cosec cotx x cπ π

(d) 1

2 2 6log tan .

x c+⎛⎝⎜

⎞⎠⎟ +π

12. If cos ( ( )) ,ec(2 ) thenx dx kf g x c∫ = +(a) range of g(x) = (–∞, ∞)

(b) domain of f (x) = R – {0}

(c) g ′(x) = sec2x

(d) f ′(x) = 1/x for all x ∈ (0, ∞).

13. log( ) log

( )

x xx x

dx+ −+

=∫1

1

(a) log( ) (log )x x+ −⎡⎣⎢

⎤⎦⎥

11

2

2

2

+ c

(b) –1

2[log (x + 1) – (log x)]

2 + c

(c) cx

− +⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

21

12

log

(d) none of these

Comprehension Type

If the integrand is a rational function of x and fractional

power of a linear fractional function of the form

ax bcx d

++

⎛⎝⎜

⎞⎠⎟

i.e., f x ax bcx d

ax bcx d

dxm n r s

, , ..., .

/ /++

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

++

⎛⎝⎜

⎞⎠⎟

⎞⎠⎟∫


In this form substitute ax bcx d

tm++

= where m is the LCM

of the denominators of fractional powers of ax bcx d

++

.

On the basis of above information, answer the following

questions:

14. The value of dx

x x( ) ( )1 11 2 1 3+ − +∫ is

(a) 2 3 6 6 11 2 1 3 1 6 1 6λ λ λ λ/ / / /ln+ + + − + c

(b) 2 3 6 6 11 2 1 3 1 6 1 6λ λ λ λ/ / / /ln− + + − + c

(c) 2 3 6 6 11 2 1 3 1 6 1 6λ λ λ λ/ / / /ln+ − + − + c

(d) 2 3 6 6 11 2 1 3 1 6 1 6λ λ λ λ/ / / /ln+ + − − + c

where λ = (1 + x)

15. The value of 1

1

1 2 2 3

1 3

+ −+∫

x xx

dx is

(a) 3

4

6

7

6

52

4 3 7 6 5 6 1 2x x x x x/ / / /+ + + +

– 6x1/6 + 6 tan

–1(x1/6

) + c

(b) − + + + −3

4

6

7

6

52

4 3 7 6 5 6 1 2x x x x x/ / / /

+ 6x1/6 – 6 tan

–1(x1/6

) + c

(c) 3

4

6

7

6

52

4 3 7 6 5 6 1 2x x x x x/ / / /− + − +

– 6x1/6 + 6 tan

–1(x1/6

) + c

(d) − + + − + −3

4

6

7

6

52 6

4 3 7 6 5 6 1 2 1 6x x x x x x/ / / / /

+ 6 tan–1

(x1/6) + c

Matrix Match Type

16. Match the columns:

Column I Column II

(P) ( )log loge e dxa x x a+∫ (1)

1

2

1

2

12

tan− −⎛

⎝⎜⎞

⎠⎟+x

xc

(Q) e

xx

dxx

log 11

2

2

2

1

+⎛

⎝⎜⎞

⎠⎟

+∫ (2) tan

–1 (tanx + 2) + c

(R)dx

x x x

x

sin sin cos

cos

2

2

4

5

+

+

∫ (3)xa

aa

ca x+

++ +

1

1 log

P Q R

(a) 2 1 3

(b) 2 3 1

(c) 3 1 2

(d) 1 2 3

Integer Answer Type

17. If dx

x kx

5 4

2 5 2 4

3

1

+= +⎡

⎣⎢⎤⎦⎥

−∫ sintan

( tan / ), then k is

equal to

18. If dx

x x4 3 21( )+∫ = a x

xbx

cx

dln ,1

1

3

3 3 3

+⎛

⎝⎜⎞

⎠⎟+ +

++

then a + b + c =

19. If x x x

x xdx

3 2

2

6 11 6

4 3

− + −

+ +∫

= + + + + ++ +

∫( )Ax Bx C x x dx

x x2 2

24 3

4 3

λ

then the value of 27A – 6B – C is

20. If the integral

5

22

tan

tanln |sin cos |

xx

dx x a x x k−

= + − +∫ , then a

is equal to

Keys are published in this issue. Search now!

Check your score! If your score is> 90% EXCELLENT WORK ! You are well prepared to take the challenge of final exam.

90-75% GOOD WORK ! You can score good in the final exam.

74-60% SATISFACTORY ! You need to score more next time.

< 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

No. of questions attempted ……

No. of questions correct ……

Marks scored in percentage ……


JEE MAIN

1. A triangle is inscribed in a rectangular hyperbola.

Which one of the following points of the triangle

lies on the hyperbola ?

(a) centroid (b) incentre

(c) circumcentre (d) orthocentre

2. ABCD is an isosceles trapezium, AB = 4, BC = DA = 3,

CD = 2. Circles of radii 2 are drawn with centres A

and B and circles of radii 1 are drawn with centres C

and D. The radius of the circle touching these four

circles externally is

(a) 3

7 (b)

4

7 (c)

5

7 (d)

2

7

3. Let f x a xrr

r( ) ,=

=∑

0

6

where a0 = 64, a5 = – 12, a6 = 1.

If the roots of f(x) = 0 are positive, then f ′′(1) =

(a) 12 (b) 24 (c) 30 (d) 60

4. Let f x e dtt x( ) ,

| |= −∫0

4

0 ≤ x ≤ 4. If the range of f(x) is

[a, b], then b a− is

(a) e2 (b) e2

+ 1 (c) e2 – 1 (d) e2

/2

5. Let P(x1, y1) and Q(x2, y2), y1 > 0, y2 > 0 be the

ends of latus rectum of the ellipse x2 + 2y2

= 2. The

equations of parabolas with latus rectum are

(a) x2 – 2y = 1 – 2 (b) x2

– 2y = 2 + 2

(c) x2 + 2y = 2 – 2 (d) None of these

JEE ADVANCED

6. Let z1 = 3 – 4i, z2 = 2 + i, α ≠ 1, α6 = 1.

The sum S z z r

r= +

=∑ 1 2

0

5 2

α is divisible by

(a) 2 (b) 3 (c) 5 (d) 7

COMPREHENSION

Let y = f(x) such that xy = x + y + 1, x ∈ R – {1} and

g(x) = x f(x)

7. The minimum value of g(x) =

(a) 3 2− (b) 3 2+(c) 3 2 2− (d) 3 2 2+

8. The area bounded by the curve y = g(x) and the

x-axis is

(a) 3/2 + ln 4 (b) 3/2 – ln 4

(c) 1/2 + ln 4 (d) ln 4 – 1

INTEGER MATCH

9. Let PQ and PR are the focal chords of the ellipse

x2 + 2y2

= 2. If p =⎛⎝⎜

⎞⎠⎟

11

2, and QR = d, then [d2

] is

MATRIX MATCH

10. Column I Column II(a)

limx

t xx

xe dt→∞

− =∫2 2

0

(p) 1

3

(b) Let f (x) be differentiable

such that the function

f x x e f x t dttx

( ) | | ,= + −−∫2

0

then f (1) =

(q) 1

2

(c)

If f x x t dt x( ) | | , [ , ]= − ∈∫ 0 1

0

1

then f x dx( ) =∫0

1

(r) 1

(d)1 22 2

0

2++

⋅ =∫ cos( cos )

/x

xdx

π (s) 23

(t) 43

Maths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material.

During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India.Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

See Solution Set of Maths Musing 166 on page no. 83


CATEGORY-IFor each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.

1. The middle term of an arithmetic progression

consisting of n number of terms is m, the sum of

the terms of the series will be

(a) m + n (b) mn (c) m – n (d) m/n2. The 5th term of a G.P. is 2, then the product of the

first 9 terms is

(a) 256 (b) 128

(c) 512 (d) none of these

3. If 1 1 1

ab ac bc ba ca cb+ + +, , are in H.P., then a, b, c

are in

(a) A.P. (b) G.P.

(c) H.P. (d) none of these

4. If a, x, b are in A.P. and a, y, z, b are in G.P., then the

value of y z

xyz

3 3+ is

(a) 1 (b) 2

(c) 1/2 (d) none of these

5. If x be the arithmetic mean of two positive numbers

and y and z are two geometric means between them,

then the harmonic mean of the numbers will be

(a) xyz

(b) yzx

(c) 2yz

x (d)

2xyz

6. If S nP n n Qn = + −( ),

1

2 where Sn denotes the sum

of the first n terms of an A.P., then the common

difference is

(a) P + Q (b) 2P + 3Q(c) 2Q (d) Q

7. If a, b, c, d are four numbers such that the first three

are in A.P., while the last three are in H.P., then

(a) ab = cd (b) ac = bd(c) ad = bc (d) none of these

8. The sum of integers from 1 to 100 which are divisible

by 2 or 5, is

(a) 3000 (b) 3010 (c) 3150 (d) 3050

9. If pth, qth and rth terms of a G.P. are x, y, z respectively then r q–r × yr–p × zp–q =

(a) 0 (b) 1

(c) xyz (d) none of these

Time: 1 hr 15 min.

The entire syllabus of Mathematics of WB-JEE is being divided into eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given below.

Series-4

Unit No.

Topic Syllabus In Details

UN

IT N

O. 4

Mathematical Induction Principle of mathematical induction and its simple applicationBinomial Theorem & it’s Simple Applications

Binomial Theorem of positive integral index, general term and middle term, properties of Binomial coefficients and simple applications

Sequences and Series Definition of series and sequence. Arithmetic, geometric and harmonic progressions. Arithmetic, geometric & harmonic means between two given numbers, relation between A.M., G.M. and H.M., Sum upto n terms of special series: ∑n, ∑n2, ∑n3. Arithmetic- Geometric progression.

Co-ordinate geometry-2D

Conic Sections: Sections of cones, equation conic sections( parabola, ellipse and hyperbola) in standard forms, condition for y = mx + c to be a tangent and point(s) of tangency

Time : 1 hr 15 min Full marks : 45

By : Sankar Ghosh, S.G.M.C, Kolkata, Ph: 09831244397.


10. The sum of n terms of the series 2

3

8

9

26

27

80

81+ + + + ...

is

(a) n n+ −1

33 1( ) (b) n n− −1

33 1( )

(c) n n− −−1

23 1( ) (d) n n− − −1

21 3( )

11. If 540 is divided by 11, then remainder is

(a) 2 (b) 3 (c) 5 (d) 1

12. The number of terms in the expansion of (a + b + c)n,

where n ∈ N is

(a) ( )( )n n+ +1 2

2 (b) n + 1

(c) n + 2 (d) (n + 1)n/2

13. The term independent of x in x

x3

3

22

10

+ ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

is

(a) 1 (b) 5/12

(c) 10C1 (d) none of these

14. If the coefficient of x7 and x8 in 23

+⎛⎝⎜

⎞⎠⎟

x n are equal,

then n is(a) 56 (b) 55 (c) 45 (d) 15

15. Given the integers r > 1, n > 2 and coefficients of

(3r)th and (r + 2)th terms in the binomial expansion

of (1 + x)2n are equal, then

(a) n = 2r (b) n = 3r(c) n = 2r + 1 (d) none of these

16. The digit at unit’s place in the number

172017 + 112017 – 72017 is

(a) 0 (b) 1 (c) 2 (d) 3

17. If z i i= +⎛

⎝⎜⎞

⎠⎟+ −⎛

⎝⎜⎞

⎠⎟3

2 2

3

2 2

5 5

then

(a) Re(z) = 0 (b) Im(z) = 0

(c) Re(z) > 0, Im(z) > 0

(d) Re(z) > 0, Im(z) < 0

18. The number of integral terms in the expansion of

( )2 5 76 642+ is

(a) 105 (b) 107 (c) 321 (d) 108

19. If the coefficient of the second, third and fourth

terms in the expansion of (1 + x)n are in A.P., then

n equal to

(a) 2 (b) 7

(c) 9 (d) none of these

20. The focus of the parabola whose vertex is (3, 2) and

whose directrix is x – y + 1 = 0 is

(a) (4, 1) (b) (1, –1)

(c) (8, 7) (d) (–4, 1)

21. If the slope of a chord OP of the parabola y2 = 4ax

(where O is the vertex) is tan α, then the length of

the chord is

(a) 4a tanα secα (b) 4a tanα cosα(c) 4a cotα cosecα (d) 4a sinα tanα

22. M is the foot of the perpendicular from a point

P on y2 = 4ax to its directrix. If S is the focus and

the ΔSPM is equilateral, then its area is equal to

(a) 4 32a (b) 2 3

2a(c) 8 3

2a (d) 16a2

23. Two conics xa

yb

x ab

y2

2

2

2

21− = = − and intersect if

(a) 0

1

2< <b

(b) 0

1

2< <a

(c) a2 < b2 (d) a2 > b2

24. An ellipse has eccentricity 1/2 and one focus at

S 1

21, .

⎛⎝⎜

⎞⎠⎟ Its one directrix is the common tangent,

(nearer to S) to the circle x2 + y2 = 1 and x2 – y2 = 1.

The equation of the ellipse in standard form is

(a)

91

312 1 1

22x y−⎛

⎝⎜⎞⎠⎟

+ − =( )

(b) 121

39 1 1

22x y−⎛

⎝⎜⎞⎠⎟

+ − =( )

(c)

x y−⎛⎝⎜

⎞⎠⎟

+ − =

1

2

12

1

91

2

2( )

(d)

31

24 1 1

22x y+⎛

⎝⎜⎞⎠⎟

+ − =( )

25. The eccentricity of the hyperbola x2 – 3y2 + 1 = 0 is

(a) 3 (b) 2 (c) 3/2 (d) 4/3

26. If C is the centre and A, B are two points on the conic

4x2 + 9y2 – 8x – 36y + 4 = 0 such that ∠ =ACB π2

,

then (CA)–2 + (CB)–2 is equal to

(a) 13

36 (b)

36

13 (c)

16

33 (d)

33

16


27. If e1 is the eccentricity of the conic 9x2 + 4y2 = 36

and e2 is the eccentricity of the conic 9x2 – 4y2 = 36,

then

(a) e12 + e2

2 = 2 (b) 3 < e12 + e2

2 < 4

(c) e12 + e2

2 > 4 (d) none of these

28. A tangent having slope –4/3 to the ellipse x y2 2

18 321+ =

intersects the major and minor axes at A and B. If O is

the origin, then the area of the ΔOAB is (in sq. units)

(a) 48 (b) 9 (c) 24 (d) 16

29. Tangents at any point on the hyperbola xa

yb

2

2

2

21− =

cut the axes at A and B respectively. If the rectangle

OAPB (where O is origin) is completed then the

locus of point P is given by

(a) ax

by

2

2

2

21− =

(b)

ax

by

2

2

2

21+ =

(c)

ay

bx

2

2

2

21− =

(d) none of these

CATEGORY-IIEvery correct answer will yield 2 marks. For incorrect response, 25% of full mark (1/2) would be deducted. If candidates mark more than one answer, negative marking will be done.

30. Let p ≥ 3 is a positive integer and α and β are two

roots of the equation x2 – (p + 1)x + 1 = 0, then

αn + βn (where n ∈ N) =

(a) 0 (b) an integer

(c) a proper fraction (d) none of these

31. The middle term in the expansion of xx

n+⎛

⎝⎜⎞⎠⎟

12

is

(a)

1 3 5 2 1⋅ ⋅ ⋅ ⋅ − ⋅... ( )

!

nn

xn

(b)

1 3 5 2 12

⋅ ⋅ ⋅ ⋅ + ⋅... ( )

!

nn

n

(c)

1 3 5 2 1

12

⋅ ⋅ ⋅ ⋅ −+

⋅... ( )

( )!

nn

n

(d) 1 3 5 2 1

2⋅ ⋅ ⋅ ⋅ − ⋅... ( )

!

nn

n

32. The sum to n-terms of the series

1

2

3

4

7

8

15

16+ + + + ⋅⋅⋅⋅⋅⋅

(a) 2n – n – 1 (b) 1 – 2–n

(c) n + 2–n – 1 (d) 2n – 1

33. The value of c for which the line y = 3x + c touches

the ellipse 16x2 + y2 = 16 is

(a) 5 (b) 1 (c) 4 (d) 3

CATEGORY-IIIIn this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If, candidates marks one correct and one incorrect answer then no marks will be awarded. But if, candidate makes only correct, without making any incorrect, formula below will be used to allot marks.2×(no. of correct response/total no. of correct options)

34. Let n ∈ N then x(xn – 1 – nan – 1) + an(n – 1) is

divisible by (x – a)2 when

(a) n = 2 (b) n = 3

(c) ∀ n ∈ N (d) none of these

35. The 6th term of the expansion

2 210 1010 3 2 35log ( ) log− −( )+⎡⎣

⎤⎦

x xm

is 21 and the

coefficient of 2nd, 3rd and 4th terms of it are

respectively 1st, 3rd and 5th term of an A.P. Find x.

(a) 0 (b) 1 (c) 2 (d) 3

36. A chord PP′ of a parabola cuts the axis of the

parabola at O. The feet of the perpendiculars

from P and P′ on the axis are M and M′ respectively.

If V is the vertex then VM, VO, VM′ are in

(a) A.P. (b) G.P.

(c) H.P. (d) none of these

37. A focus of the hyperbola 25x2 – 36y2 = 225 is

(a) ( , )61 0 (b) 1

261 0,

⎛⎝⎜

⎞⎠⎟

(c) ( , )− 61 0 (d) −⎛⎝⎜

⎞⎠⎟

1

261 0,

SOLUTIONS1. (b) : Given that, the middle term of an A.P., consisting

of n terms is m. Since there is only one middle term,

therefore, n is odd. Let n = 2p + 1

Thus tp + 1 = m ⇒ a + pd = m [where a and d are

respectively the first term and the common difference]

Now, S p a p dn = + + + −2 1

22 2 1 1{ ( ) }

= (2p + 1)(a + pd) = m(2p + 1) = mn∴ Sn = mn2. (c) : Let the first term be a and the common ratio

be r.

∴ t5 = 2 ⇒ ar4 = 2

Now a ar ar ar ar a r⋅ ⋅ ⋅ ⋅ = + + + +2 3 8 9 1 2 3 8...

...

= = =( )ar4 9 92 512


3. (c) : Given that

1 1 1

ab ac bc ba ca cb+ + +∈, , H.P.

⇒ ab + ac, bc + ba, ca + cb ∈ A.P.

⇒ –bc, –ca, –ab ∈ A.P. ⇒ bc, ca, ab ∈ A.P.

⇒ ∈bcabc

caabc

ababc

, , A.P.

⇒ ∈1 1 1

a b c, , A.P.

∴ a, b, c ∈ H.P.

4. (b) : Given that a, x, b ∈ A.P. ∴ = +x a b2

...(i)

Again a, y, z, b ∈ G.P. ∴ y2 = az and z2 = by

⇒ = =a yz

b zy

2 2

and

Now from (i), we get

yz

zy

x y zyz

x2 2 3 3

2 2+ = ⇒ + =

⇒ + =y zxyz

3 3

2

5. (b) : Let the two given numbers be a and b

∴ = +x a b2

And y az by a yz

b zy

2 22 2

= = ⇒ = = and and z

Now, H.M. of a and b is

H ab

a b

yz

zy

xyzx

=+

=× ×

=22

2

2 2

6. (d) : Here S nP n n Qn = + −( )1

2

⇒ S Q n Q P nn = ⎛⎝⎜

⎞⎠⎟

− −⎛⎝⎜

⎞⎠⎟

1

2

1

2

2

∴ Common difference = 1

22Q Q⎛

⎝⎜⎞⎠⎟

× =

7. (c) : Given that a, b, c ∈ A.P. ∴ 2b = a + c

Again b, c, d ∈H.P. ∴ =+

c bdb d2

⇒ − =+

22b a bd

b d ⇒ (2b – a)(b + d) = 2bd

⇒ 2b2 – ab = ad ⇒ b(2b – a) = ad⇒ b(a + c – a) = ad ⇒ bc = ad8. (d) : Sum of the integers from 1 to 100 which are

divisible by 2 or 5 is

(2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... + 100)

– (10 + 20 + 30 + ... + 100)

= + + + − +50

22 100

20

25 100

10

210 100( ) ( ) ( ) = 3050

9. (b) : Let the first term and common ratio of the

G.P. respectively be m and n.

Then, tp = x ⇒ mnp – 1 = x tq = y ⇒ mnq – 1 = y tr = z ⇒ mnr – 1 = zNow, x y zq r r p p q− − −⋅ ⋅ = ⋅ ⋅− − − − − −

( ) ( ) ( )mn mn mnp q r q r p r p q1 1 1

= ⋅− + − + − − − + − − + − −m nq r r p p q p q r q r p r p q( )( ) ( )( ) ( )( )1 1 1

= m0n0 = 1

10. (d) : Let Sn = + + + +2

3

8

9

26

27

80

81....

= − + − + − + − +3 1

3

3 1

9

3 1

27

3 1

81

2 3 4

...

= + + − + + + +⎛⎝⎜

⎞⎠⎟

( ... to terms) ...1 11

3

1

3

1

3

1

32 3 4

n to termsn

= −−⎛

⎝⎜⎞⎠⎟

−= − − −n n

n n

1

31

1

3

11

3

1

21 3( )

11. (d) : Here, 540 = (52)20 = (22 + 3)20, so the remainder

is same as 320.

Now 320 = (32)10 = (11 – 2)10, so the remainder is

same as 210.

But 210 = 1024 = 11 × 93 + 1.

So, the remainder is 1.

12. (d) : (a + b + c)n = an + nC1an – 1(b + c)

+ nC2 an – 2(b + c)2 + nC3 an – 3(b + c)3 + ...

+ nCn(b + c)n

Further expanding each term of R.H.S., first term gives

one term, second term gives two terms, third term

gives three terms and so on.

∴ Total number of terms = + + + + = +1 2 3

1

2...

( )n n n

13. (d) : Let tr + 1th term is independent of x.

∴ =⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜⎞

⎠⎟+

−

t C xxr r

r r

110

10

23

3

2

= ⋅ ⋅

−( ) −10

1

210 3

5

23

1

2C xr

r rr/

For independent of x terms, 10 3

20

10

3

− = ⇒ =r r

This is not an integer. Therefore, there will be no

constant term.

14. (b)15. (a) : The coefficient of (3r)th and (r + 2)th terms

will be 2nC3r – 1 and 2nCr + 1 respectively.


By the problem, 2nC3r – 1 = 2nCr + 1

⇒ 3r – 1 + r + 1 = 2n

⇒ 4r = 2n ⇒ n = 2r

16. (b) : 172017 + 112017 – 72017

= (7 + 10)2017 + (1 + 10)2017 – 72017

= + ⋅ ⋅ + ⋅ ⋅ +[ ...7 7 10 7 102017 2017

12016 2017

22015 2C C

... ]+ ⋅20172017

201710C

+ + ⋅ + ⋅ + + −[ ... ]1 10 10 10 72017

12017

22 2017 2017C C

= ⋅ ⋅ + ⋅ ⋅ +[ ...2017

12016 2017

22015 2

7 10 7 10C C

... ]+ ⋅20172017

201710C

+ ⋅ + ⋅ + + +[ ... ]2017

12017

22 2017

10 10 10 1C C

= a multiple of 10 + 1

Thus unit’s place digit is 1

17. (b)

18. (d) : The general term is.

t C Cr rr r

rr

r r

+− − −

= = ⋅1642 642 6 642 642

3212 62 5 7 2 5 7( ) ( )

The above term will be integral if and only if r is a

multiple of 6.

Thus r takes the values 0, 6, 12, 18, …, 642.

Therefore the number of integral terms in the given

expansion is 108.

19. (b) : According to the given problem we have nC1, nC2, nC3 are in A.P. ⇒ nC1 + nC3 = 2nC2

⇒ + − − = ⋅ −n n n n n n( )( ) ( )1 2

62

1

2

⇒ + − − = −11 2

61

( )( )n n n

⇒ 6 + (n – 1)(n – 2) = 6n – 6

⇒ n2 – 3n + 2 = 6n – 12

⇒ n2 – 9n + 14 = 0 ⇒ n = 7, 2

But n = 2 is rejected as nC3 is not possible.

20. (a) : The equation of the axis of the parabola is

y – 2 = –1(x – 3) ⇒ x + y – 5 = 0

Hence the point of intersection of axis and directrix

of the parabola is Z = (2, 3)

A (vertex) is the mid-point of ZS (S is the focus)

∴ S = (4, 1)

21. (c) : Slope of OP atat t

= = =2 2

2tanα (given).

Length of the chord OP a t a t at t= + = +2 4 2 2 24 4

= + =2 44 4

2

a atan tan

cotα α

α αcosec

22. (a) : From the diagram, we have

From ΔNSM MNNS

, tan= °30

⇒

2

2

1

3

aat

=

∴ = ⇒ = + =t MP a at a3 42

Area of ΔSPM = ×1

2MP SN

= = =1

24 2 4 4 3

2 2( )( )a at a t a

23. (b) : Eliminating x, we haveyb

yab

2

21 0+ + =

The given curves will intersect each other if

1 40

14

1

4

1

22 2 2 2

2

a b b aa a− > ⇒ > ⇒ < ⇒ <

Hence the conics intersect if 01

2< <a .

24. (a) : For the circle x2 + y2 = 1 and rectangular

hyperbola x2 – y2 = 1, one common tangent is evidently

x = 1, the other being x = –1.

We require in standard form the ellipse with focus at

S 1

21,

⎛⎝⎜

⎞⎠⎟ and directrix x = 1 which is

x y x−⎛⎝⎜

⎞⎠⎟

+ − = ⎛⎝⎜

⎞⎠⎟

−1

21

1

21

22

22

( ) ( )

⇒ − + − =3

4 21 0

22x x y( )

⇒ −⎛⎝⎜

⎞⎠⎟

+ − =3

4

1

31

1

12

22x y( )

⇒ −⎛⎝⎜

⎞⎠⎟

+ − =91

312 1 1

22x y( )


25. (b) : The given equation of the conic is

x y y x2 22 2

3 1 01

3

11− + = ⇒ − = ,

which is the conjugate hyperbola .

∴ = −11

31

2( )e , which gives e = 2.

26. (a) : The equation can be written as

4(x – 1)2 + 9(y – 2)2 = 36

which is an ellipse centred at (1, 2).

If CA makes an angle θ with the major axis, then

A CA CA≡ + +⎡⎣ ⎤⎦1 2cos , sin ,θ θ

B CB CB≡ + +⎛

⎝⎜⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

12

22

cos , sinπ θ π θ

As A and B are two points on the conic

∴ CA2(4cos2θ + 9sin2θ) = 36 and

CB2(4sin2θ + 9cos2θ) = 36

giving ( ) ( )CA CB− −+ =2 2 13

36

27. (b) : The given equations in standard form are

x y x y2 2 2 2

4 91

4 91+ = − = and .

Now, e e12

229 4

9

5

9

4 9

4

13

4= − = = + =,

and but e e12

22 137

363 4+ = > <

28. (c) : Any point on the ellipse

x y2

2

2

23 2 4 2

1( ) ( )

+ =

can be taken as ( cos , sin )3 2 4 2θ θ and slope of the

tangent

= − = − = −b xa y

2

2

32 3 2

18 4 2

4

3

( cos )

( sin )cot

θθ

θ

... (i)

Given slope of the tangent = − 4

3 ... (ii)

From equation (i) and (ii), cotθ θ π= ⇒ =14

Hence equation of tangent is

x y x y⋅+

⋅= ⇒ + =

1

2

3 2

1

2

4 21

6 81

Hence A ≡ (6, 0), B = (0, 8)

Area of ΔOAB = × × =1

26 8 24 sq.units.

29. (a) : Equation of tangent at the point θ is

xa

yb

A a B bsec tan( cos , ), ( , cot )

θ θθ θ− = ⇒ −1 0 0

Let P be (h, k) ⇒ h = acosθ, k = –bcotθ

⇒ = − ⇒ = − ⇒ + =kh

ba

bhak

b ha k

hasin

sinθ

θ2 2

2 2

2

21

⇒ + = ⇒ − =bk

ah

ah

bk

2

2

2

2

2

2

2

21 1.

Thus the locus of P is ax

by

2

2

2

21− = .

30. (b) : Given that α and β are the roots of the equation

x2 – (p + 1)x + 1 = 0

∴ α + β = p + 1 and αβ = 1

If n = 1, then αn + βn= α + β = p + 1, which is an

integer (∵ p ≥ 3 is a positive integer)

If n = 2, then αn + βn = α2 + β2 = (α + β)2 – 2αβ = (p + 1)2 – 2 · 1 = p2 + 2p – 1

which is an integer (∵ p ≥ 3 is a positive integer)

Let p(n) : αn + βn = an integer

∴ p(1) and p(2) are true.

So, it can be prove by the principle of mathematical

induction that the statement p(n) is true.

31. (d) : The required middle term is

t C xx

C xxn

nn

n nn

nn

nn+

−= ⋅ ⋅ ⎛⎝⎜

⎞⎠⎟

= ⋅12 2 21 1

= = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ −2 2 1 2 3 4 5 6 2 1 2nnC n

n nn n

n n...... ( )

= ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅( ....( ))( ... )1 3 5 2 1 2 4 6 2n n

n n

= ⋅ ⋅ − = ⋅ ⋅ − ⋅( ....( )) ( ....( ))1 3 5 2 1 2 1 3 5 2 1

2n n

n nn

n

nn

32. (c) : Let the sum of the first n terms of the series

be S and its nth term be tn.

∴ = + + + + +S tn1

2

3

4

7

8

15

16.....

S t tn n= + + + + +−

1

2

3

4

7

81......

_______________________

(Subtracting) 01

2

1

4

1

8

1

16= + + + + −... to termsn tn


⇒

t nn n= + + + + = −1

2

1

4

1

8

1

161

1

2..... to terms

Thus, S n= −⎛⎝⎜

⎞⎠⎟

+ −⎛⎝⎜

⎞⎠⎟

+ −⎛⎝⎜

⎞⎠⎟

+ + −⎛⎝⎜

⎞⎠⎟

11

21

1

21

1

21

1

22 3

.....

= − + + + +⎛⎝⎜

⎞⎠⎟

= − −⎛⎝⎜

⎞⎠⎟

= − + −n n nn nn1

2

1

2

1

2

1

21

1

21 2

2 3.... .

33. (a) : The given equation of the ellipse is

16x2 + y2 = 16 ⇒ + =x y2 2

1 161

The line which touches the above ellipse is y = 3x + cHere, a = 1, b = 4, m = 3

∴ = + = ⋅ + =c a m b2 2 2 2 2 21 3 4 5

34. (a, b, c) : If n = 1, then

x x na a n x a an n n( ) ( ) ( )

− −− + − = − ⋅ + −( ) ⋅1 1 01 1 1 1 1

= x · 0 + 0 · a = 0

which is divisible by (x – a)2

If n = 2, then

x x na a n x x a an n n( ) ( ) ( ) ( )

− −− + − = − + −1 1 21 2 2 1

= x2 – 2xa + a2 = (x – a)2


If n = 3, then

x x na a n x x a an n n( ) ( ) ( )

− −− + − = − +1 1 2 2 31 3 2

= x3 – 3a2x + 2a3 = x3 – a2x – 2a2x + 2a3

= x(x2 – a2) – 2a2(x – a) = (x – a)(x2 + ax – 2a2)

= (x – a)(x – a)(x + 2a) = (x – a)2(x + 2a)


∴ For any n ∈ N, x(xn – 1 – nan – 1) + an(n – 1) is

divisible by (x – a)2.

35. (a, c) : The coefficients of 2nd, 3rd and 4th terms of

the given expansion are respectively mC1, mC2, mC3.

Also these are respectively the 1st, 3rd and 5th term of

an A.P. Again the 1st, 3rd and 5th terms of an A.P. are

also in A.P.

Therefore, 2(mC2) = mC1 + mC3 ⇒ m = 2, 7

∵ The 6th term of the given expansion is 21,

∴ m ≠ 2, so, m = 7

Now the 6th term of the expansion is

t Cx x

67

510 3

7 52 35

5

2 210 10= { } ⋅{ }−−

−( )log ( ) log

⇒ ⋅ ⋅ =− −75

10 3 2 32 2 2110 10C

x xlog ( ) ( )log

⇒ = =− ⋅ −2 1 210

210 3 3 0log {( ) }

x x

⇒ {(10 – 3x)·3x – 2} = 1

Solving we get x = 0, 2

36. (b) : Let the parabola be

y2 = 4axand P(at1

2, 2at1), P′ (at2

2, 2at

2)

∴ = ′ =VM at VM at12

22

,

and VO = k

where are collinear)

at at

at atk

P P O12

1

22

2

2 1

2 1

0 1

0= ∴ ′( , ,

∴ VM.VM′ = (at1t2)2 = k2 = VO2

37. (b, d) : The given equation of the hyperbola is

25 36 2259 5

2

12 2

2 2

2x y x y− = ⇒ −

⎛⎝⎜

⎞⎠⎟

=

∴ = = ⎛⎝⎜

⎞⎠⎟⇒ = + =a b e2 2

2

95

21

25

36

61

6 and

∴ ± = ±⎛

⎝⎜⎞

⎠⎟ Focus is ( , ) ,ae 0

61

20


SECTION-1This section contains eight questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened, 0 In all other cases.

1. How many solutions are there for the equation,

|cos x – sin x| = 2 cosx in the interval [0, 2π]?

2. The number of solutions of the system

log2x logy 2 + 1 = 0 and sinx cosy = 1 – cosx siny,

which satisfy the condition x + y < 8 is

3. tan100° + tan125° + tan100° tan125° =

4. If cos

cos

cos,

tan

tan2

3 2 1

3 2

2

α ββ

αβ

= −−

⎛⎝⎜

⎞⎠⎟

then is equal to

5. If A + B + C = 180°,

sin sin sin

sin sin sinsin sin sin ,

2 2 2

2 2 2

A B CA B C

k A B C+ ++ +

=

then the value of k is equal to

6. The expression tan55° · tan65° · tan75° simplifies to

cotx°, where x ∈ (0, 90) then x equals

7. The maximum value of y = 2 sin2x – 3sinx + 1 ∀ x ∈R is

8. Let x and y be positive real numbers and θ be an angle

such that θ π≠ ∈n n I2

, . Suppose sin cosθ θ

x y= and

cos sin sin,

4

4

4

4 3 3

97 2θ θ θx y x y y x

+ =+

then the value of

xy

yx

+⎛⎝⎜

⎞⎠⎟

is equal to

PAPER-I

SECTION-2This section contains ten questions. Each question has four options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubbles corresponding to all the correct options in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened, 0 If none of the bubbles is darkened. –2 In all other cases.

9. The vectors a b c, , are of the same length and

the angle between any two of them is the same. If

a i j b j k= + = +^ ^ ^ ^, then c is

(a) − − +1

34( )

^ ^ ^i j k (b) i k^ ^+

(c) 1

35( )

^ ^ ^− + +i j k (d) ( )^ ^ ^i j k+ +2 3

10. The lines whose vector equations are

r i j k i p j k= − + + + +2 3 7 2 5^ ^ ^ ^ ^ ^

( )λ and

r i j k i p j p k= + + + − +^ ^ ^ ^ ^ ^( )2 3 3μ

are perpendicular for all values of λ and μ if (a) p = –6 (b) p = –1

(c) p = 1 (d) p = 6

11. The vectors a x i j k= − +^ ^ ^2 5 and b i y j z k= + −^ ^ ^

are collinear if

(a) x = 1, y = –2, z = –5

(b) x = 1/2, y = –4, z = –10

(c) x = –1/2, y = 4, z = 10

(d) x = –1, y = 2, z = 5

12. Vector equation of the line 6x – 2 = 3y + 1 = 2z – 2

is

(a) r i j k i j k= − + + + +^ ^ ^ ^ ^ ^( )3 2 3λ

By : Vidyalankar Institute, Pearl Centre, Senapati Bapat Marg, Dadar (W), Mumbai - 28. Tel.: (022) 24306367


(b) r i j k i j k= + + + − +⎛⎝⎜

⎞⎠⎟

^ ^ ^ ^ ^ ^2 3

1

3

1

3λ

(c) r i j k i j k= − + + + +1

3

1

32 3

^ ^ ^ ^ ^ ^( )λ

(d) r i j k i j k= − + − + + +2 2 6 3 2^ ^ ^ ^ ^ ^

( )λwhere, λ being a parameter.

13. The equations of the line of shortest distance between

the lines x y z2 3 1

=−

= and x y z− = −

−= +2

3

1

5

2

2 are

(a) 3(x – 21) = 3y + 92 = 3z – 32

(b) x y z− = + = −( / )

/ /

( / )

/

62 3

1 3

31

1 3

31 3

1 3

(c) x y z− = + = −21

1 3

92 3

1 3

32 3

1 3/

( / )

/

( / )

/

(d)

x y z− = + = −2

1 3

3

1 3

1

1 3/ / /

14. If Δ denotes the area of the triangle ABC, then Δ is

equal to

(a) absinC (b)

1

2

2a B CB C

sin sin

sin( )+

(c) 1

2( )a b c r+ + (d)

1

2( )a b c r+ −

15. cos sin sin cos2 1 2 2x x x x+ + = + if

(a) x n= −π π3

(b) x = 2nπ

(c) x n= −π π4

(d) x n= ± −⎛⎝⎜⎞⎠⎟

−2

1

5

1π cos

16. If A and B are angles such that A B+ = π3

and

y = tanA tanB, then y can be equal to

(a) 0 (b) 2 (c) 4 (d) 5

17. The equation sin4x + cos4x = a has a solution for

(a) all values of a (b) a = 1

(c) a = 1/2 (d) a = 3/4

18. If θ θ θ= − ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

− −tan ( tan ) tan tan ,

1 2 12

1

3 then θ =

(a) 0 (b) π/4

(c) tan–1(–2) (d) none of these

SECTION - 3This section contains TWO questions. Each question contains two columns, Column I and Column II. Column I has four entries (A), (B), (C) and (D). Column II has five entries (P),

(Q), (R), (S) and (T). Match the entries in Column I with the entries in Column II. One or more entries in Column I may match with one or more entries in Column II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one as shown.For each entry in Column I, darken the bubbles of all the matching entries. For example, i f entry (A) in Column I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). Marking scheme : For each entry in Column I, +2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened; 0 If none of the bubbles is darkened; –1 In all other cases.


Column I Column IIA. Maximum value of

5 33

1cos cosθ θ π+ +⎛⎝⎜

⎞⎠⎟ −

is

P. 1

B. tan ,−

=

∞ ⎛⎝⎜

⎞⎠⎟

=∑ 1

21

1

2it

i

then tant = Q. 6

C. If 3cosθ – 4sinθ = 5, then

3sinθ + 4cosθ =

R. 2

D. If sinA sinB sinC + cosA cosB =

1, then 2cos(A – B) =

S. 0

20. Match the following:

Column I Column IIA. In a ΔABC,

cosA + cosB + cosC is

equal toP.

a b c2 2 2

2

+ +Δ

B.r r

ar r

br r

c1 2 3−

+−

+−

= Q. 1+ rR

C.rbc

rca

rab

1 2 3+ + = R.r r r

s1 2 3+ +

D.1 1 1 1

12

22

32 2r r r r

+ + + S.1 1

2r R−


PAPER-II

SECTION-1This section contains eight questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme : +4 If the bubble corresponding to the answer is darkened, 0 In all other cases.

1. If l is the shortest distance between the lines

x y z− = +−

= −3

2

15

7

9

5 and

x y z+ = − = −−

1

2

1

1

9

3,

then determine [l] ([x] denotes G.I.F)

2. Let a b and be unit vectors. If c is a vector such

that c c a b+ × =( ) , then the maximum value of

|( ) |a b c× ⋅ is A × 10–1. Find A.

3. The straight lines whose direction cosines are

given by the relations al + bm + cn = 0 and

fmn + gnl + hlm = 0 are perpendicular then the

value of fa

gb

hc

+ + is

4. If the distance of the point B i j k( )^ ^ ^+ +2 3 from the

line which is passing through A i j k( )^ ^ ^

4 2 2+ + and

which is parallel to the vector c i j k= + +2 3 6^ ^ ^

is λ

then value of λ2 – 1 is

5. Find the number of real solutions of the equation

(tan ) (cot ) .− −+ =1 2 1 2

25

8x x π

6. Number of roots of the equation cos(sin ) ,x = 1

2

(0 < x < π) is

7. Let x, y, z be real numbers such that cosx + cosy +

cosz = 0 and cos3x + cos3y + cos3z = 0 then find the

maximum value of cos2x cos2y cos2z.

8. If x + y = 2π/3 and sinx/siny = 2, then the number

of values of x ∈ [0, 4π] is

SECTION-2This section contains eight questions. Each question has four options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct options in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened, 0 if none of the bubbles is darkened, –2 In all other cases.

9. Y-coordinate of the point of intersection of the

curves is, y = cosx, y = sin3x, − ≤ ≤π π2 2

x is

(a)

cosπ8

⎛⎝⎜

⎞⎠⎟

(b)

1

2

(c) sinπ8

⎛⎝⎜

⎞⎠⎟

(d) −1

2

10. The satisfactory value of x, where x ∈ (–π, π) for the

given equation,

( sin cos )( sin cos )

3 43 2 2 2x x x x+ =− + is

(a) π3

(b) − 5

3

π (c) − 2

5

π (d) − π

3

11. If coscos

( tan )( sin ) ,2

2

211 2 3 3 4x

xy z+⎛

⎝⎜⎞⎠⎟

+ + = then

(a) x may be a multiple of π

(b) x cannot be an odd multiple of π(c) y can be a multiple of π/2

(d) z can be a multiple of π/2

12. The values of 'x' which satisfy the equation

2(cosx + cos2x) + sin2x(1 + 2 cosx) = 2 sinx, and

–π ≤ x ≤ π are

(a) π, –π (b) –(π/3), (π/3)

(c) –π/2 (d) none of these

13. For θ ∈ [0, 2π], the values of ‘θ’ and ‘y’ satisfying

the inequality, 2 2 2 2

1

22sin θ ⋅ − + ≤y y are

(a) π4

1, −⎛⎝⎜

⎞⎠⎟ (b)

π2

1,⎛⎝⎜

⎞⎠⎟

(c) 3

21

π,

⎛⎝⎜

⎞⎠⎟ (d)

π4

1,⎛⎝⎜

⎞⎠⎟

14. The positive integral solutions (x, y) of

tan cos sin− − −+

+

⎛

⎝⎜

⎞

⎠⎟ = ⎛

⎝⎜⎞⎠⎟

1 1

2

1

1

3

10x y

y

are

(a) (1, 2) (b) (2, 1)

(c) (2, 7) (d) (2, 2)

ANSWER KEYMPP-5 CLASS XII

1. (b) 2. (b) 3. (d) 4. (d) 5. (c)

6. (b) 7. (a,b,c) 8. (c,d) 9 . (b,c) 10. (b,c)

11. (a,b,c,d) 12. (a,b,c,d) 13. (b,c) 14. (a) 15. (d)

16. (c) 17. (3) 18. (0) 19. (0) 20. (2)


15. Solving the equation 3

2

2sin cos cosx x x− = , we get

(a) x = (2n + 1)π (b) x n= +23

π π

(c) x n= −23

π π (d) none of these

16. The roots of the equation sin

cot cotπ6

3 2⎛⎝⎜

⎞⎠⎟ =

+x x

sin ( ) cos( ) sin2

22

π π π− − − × +⎛⎝⎜

⎞⎠⎟x x x

are

(a) x n= + −π tan1 7

5

(b) x n= − −π tan1 7

5

(c) x = −tan

1 5

7

(d) none of these

SECTION-3This section contains two paragraphs. Based on each paragraph, there will be two questions. Each question has four options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened ; 0 If none of the bubbles is darkened ; – 2 In all other cases.

Paragraph for Q. No. 17 & 18Let ‘k’ be the length of any edge of a regular tetrahedron.

(A tetrahedron whose edges are all equal in length is

called a regular tetrahedron). The angle between a line

and a plane is equal to the complement of the angle

between the line and the normal to the plane whereas

the angle between two planes is equal to the angle

between the normals. Let 'D' be the origin be one of

the vertex and A, B and C vertices with position vectors

a b c, and respectively of the regular tetrahedron.

17. The angle between any edge and a face not

containing the edge is

(a) cos–1(1/2) (b) cos–1(1/4)

(c) cos ( / )−1

1 3 (d) π/3

18. The angle between any two faces is

(a) cos ( / )−1

1 3 (b) cos–1(1/4)

(c) π/3 (d) none of these

Paragraph for Q. No. 19 & 20The circle touching BC and the two sides AB and AC produced of a triangle ABC is called the described

circle opposite angle A. Its radius is denoted by r1.

Similarly, r2 and r

3 denote the radii of the described

circles opposite angles B and C, respectively. They

are known as the ex-radii of triangle ABC. Here Δ is

the area of a triangle, Δ = − − −s s a s b s c( )( )( ) where

2s = a + b + c. Also R is the radius of the circumcircle

of a triangle ABC and r is the radius of the circle

inscribed in triangle ABC.

19. Suppose in a triangle ABC, a : b : c = 4 : 5 : 6. The

ratio of the radius of the circumcircle to that of the

incircle is

(a) 1

3 (b)

16

7 (c)

3

14 (d)

8

19

20. Suppose in a triangle ABC, 8R2 = a2 + b2 + c2, then

the triangle ABC is(a) right angled (b) isosceles

(c) equilateral (d) none of these

ANSWER KEYPAPER-I

1. (2) 2. (2) 3. (1) 4. (2) 5. (8)

6. (5) 7. (6) 8. (4) 9. (a, b) 10. (b, d)

11. (a, b, c, d) 12. (c) 13. (a, b, c)

14. (b, c) 15. (b, c) 16. (a, c, d)

17. (b, c, d) 18. (a, b)

19. (A) → (Q), (B) → (P), (C) → (S), (D) → (R)

20. (A) → (Q), (B) → (R), (C) → (S), (D) → (P)

PAPER-II1. (6) 2. (5) 3. (0) 4. (9) 5. (1)

6. (2) 7. (0) 8. (4) 9. (a, b, c)

10. (a, b) 11. (a, c) 12. (a, b, c)

13. (b, c) 14. (a, c) 15. (a, b) 16. (a, b) 17. (c)

18. (d) 19. (b) 20. (a)

For detailed solution to the Sample Paper, visit our website www.vidyalankar.org

Solution Sender of Maths MusingSET-165

1. S Ahmed Thawfeeq Kerala

2. Gouri Sankar Adhikary W.B.

3. N. Jayanthi Hyderabad

SET-166

1. N. Jayanthi Hyderabad


1. Ifwhen

whenthenf x

x xx x( ) =

≤[ ] >⎧⎨⎪

⎩⎪

,

,,

2

2

(a) limx

f x→ −

( ) = −2

2 (b) limx

f x→ +

( ) = −2

2

(c) limx

f x f→ +

( ) = ( )2

2

(d) lim does not existx

f x→

( )2

2. If then limn N xex

n

x∈ =→∞

, 0

(a) when n is even only (b) for no value of n(c) for all values of n (d) when n is odd only

3. If theny xx

dydx

= −+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=−sin ,

12

2

1

1

(a) −+

2

12x

(b) 2

12+ x

(c) 1

22+ x

(d) 2

22− x

4. The derivative of tan− + −⎛

⎝⎜⎜

⎞

⎠⎟⎟

12

1 1xx

with respect

to tan− −

−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

12

2

2 1

1 2

x xx

at x = 0 is

(a) 1

8 (b)

1

4 (c)

1

2 (d) 1

5. If ye x

exxx

=( )( )

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

++−

⎛⎝⎜

⎞⎠⎟

− −tan

log /

logtan

log

log

1

2

2

1 3 2

1 6

then d ydx

2

2is

(a) 2 (b) 1 (c) 0 (d) –1

6. Let f x xe xx

x x( ) = ≠=

⎧

⎨⎪

⎩⎪

− +⎛

⎝⎜⎞

⎠⎟1 1

0

0 0

;

;

Test whether f(x) is differentiable at x = 0. Is it

continuous at x = 0?Justify

7. A triangle ABC, right angle at C, with CA = b and

CB = a, moves such that the angular points A and B

slide along x-axis and y-axis respectively. Find the locus

of C.

8. Prove that sin .sec tan tanθ θ θ θ31

23= −( ) and

hence find the sum to 'n' terms of the series

sinθ⋅sec3θ + sin3θ⋅sec32θ + sin32θ⋅sec33θ + …

9. Consider a real valued function f(x) satisfying

2f(xy) = (f(x))y + (f(y))x∀ x, y ∈ R and f(1) = p where

p ≠ 1 then find p f rr

n−( ) ( )

=∑1

1

.

10. A line makes angles α, β, γ, δ with four diagonals of

a cube. Prove that cos

, , ,

2 4

3r

r( ) =

∈{ }∑α β γ δ

.

SOLUTIONS

1. (c) : lim limx h

f x h f→ →+

( ) = +[ ] = = ( )2 0

2 2 2

lim limx h

f x h→ →−

( ) = − =2 0

2 2

Hence limit exists

2. (c) : Use L′ Hospital rule, lim!

x

n

xxe

ne→∞ ∞= = 0

Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of IIT-JEE Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for IIT-JEE. In every issue of MT, challenging problems are offered with detailed solution. The readers’ comments and suggestions regarding the problems and solutions offered are always welcome.

BESTPROBLEMS10

By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021


3. (a) : Put x = tanθ ⇒ = −y π θ2

2

4. (b) : p xx

x= + −⎛

⎝⎜⎜

⎞

⎠⎟⎟

=−tan tan

12

1 1Put θ

∴ p dpdx x

= ⇒ =+

θ2

1

2 12

( ) ...(i)

For q x xx

x= −−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=−tan sin

12

2

2 1

1 2Put φ

∴ dqdx x

=−

2

12

...(ii)

From (i) and (ii)

dpdq

x

x

=+( )−

1

2 1

2

1

2

2

∴ ==

dpdq x 0

1

4

5. (c) : In given function put logx2 = tanθ

Therefore, tany = + −π4

31

. Hence,d ydx

2

20=

6. f xxe

x

xxx

x

( ) =>=<

⎧

⎨

⎪⎪

⎩

⎪⎪

−2

0

0

0

0

,

,

,

∴ f(0) = 0

L.H.D= ff h f

hh′ =

−( )− ( )−

−→

( ) lim00 0

0= −

−=

→limh

hh0

1

R.H.D = ff h f

hh′ =

+( )− ( )+→

( ) lim00 0

0

= = =→

−

−∞limh

hheh

e0

2

0

∵ L. H. D ≠ R. H. D

∴ f is not differentiable at x = 0

∵ L. H. D and R. H. D are exist

∴ f is continuous at x = 0

7. Let A = (p, 0), B = (0, q)

Let C(h, k) be any point on the locus

CB a h k q= = + −( )2 2 ...(i)

CA b h p k= = −( ) +2 2 ...(ii)

AB p q= +2 2 ...(iii)

∵ ∠C = 90°

∴ AB2 = AC2 + BC2 ⇒ p2 + q2 = a2 + b2 ...(iv)

From (i) and (ii)

q k a h= ± −2 2, p h b k= ± −2 2

∴ From (iv), locus is bx ± ay = 0

8. sin

cos

cos .sin

cos .cos

θθ

θ θθ θ3 3

=

=−( )

= −( )1

2

3

3

1

23

sin

cos .costan tan

θ θθ θ

θ θ

∴ −

=∑ sin .sec3 3

1

1

r r

r

nθ θ

= −( ) = −⎡⎣

⎤⎦

−

=∑1

23 3

1

23

1

1

tan tan tan tanr r

r

nnθ θ θ θ

9. 2f(xy) = (f(x))y + (f(y))x ∀ x, y ∈ RPut y = 1

⇒ 2f(x) = f(x) + (f(1))x ⇒ ( ) =f x px

∴ ( ) = =−−= =

+

∑ ∑f r p p ppr

nr

r

n n

1 1

1

1

⇒ p f r p pr

nn−( ) ( ) = −( )

=

+∑1

1

1

10. d c s OP. ’ , ,of =⎛⎝⎜

⎞⎠⎟

1

3

1

3

1

3

d c s AR. ’ , ,of = −⎛⎝⎜

⎞⎠⎟

1

3

1

3

1

3

d c s BS. ’ , ,of = −⎛⎝⎜

⎞⎠⎟

1

3

1

3

1

3

d c s CQ. ’ , ,of = −⎛⎝⎜

⎞⎠⎟

1

3

1

3

1

3

Let l, m, n be the d.c’s of required line

∴ cos2α + cos2β + cos2γ + cos2δ

= + +( ) + − +( ) + + −( ) + − + +( )( )1

3

2 2 2 2l m n l m n l m n l m n

= 4

3


SOLUTION SET-166

1. (c) : Let f(x, y) = a1x + a2y + a3x2 + a4xy + a5y2

+ a6x3 + a7x2y + a8xy2

+ a9y3.

f(± 1, 0) = 0 ⇒ a3 = 0, a6 = – a1

f(0, ± 1) = 0 ⇒ a5 = 0, a9 = – a2

Also, a8 = 0, a7 = – a4

f (2, 2) = 0 ⇒ a a a4 1 2

3

2= − +( )

Now f(x, y) = ax2

(1 – x)(2 + 2x – 3y)

+ a y2

2 (2 – 3x + 3x2 – 2y2)

f x y y x( , ) ( )= = +02

31if and

2 3 38

91 0

21

19

2 2− + − + = ⇒ + =x x x a b( )

2. (b) : Let C = π2

in triangle ABC.

Δ = rs ⇒ rc = ab – (a + b)r.

On squaring, we get

(a – 2r)(b – 2r) = 2r2 = 2·2011

2.

Number of triangles is one half of the number of divisors

of 21

· 20112, which is

1

21 1 2 1 3( )( ) .+ + =

3. (c) : C P= ⎛⎝⎜

⎞⎠⎟

=1

20 2, , ( cos ,sin )θ θ

CP22

2 22

1

2

5

42= −⎛

⎝⎜⎞⎠⎟

+ = − +cos sin cos cosθ θ θ θ

Derivative = 2 2 0sin sin cosθ θ θ− =

⇒ CP is min. for cosθ = 1

2 ∴ Largest radius is

3

2.

4. (c) : In triangle EDB, 11

3

5

sin sinθ θ=

⇒ = =sin ,cosθ θ1

5

2

5

In triangle BCD,

BC = BD sin2θ

= ⋅ ⋅ ⋅ =11 21

5

2

5

44

5

5. (c) : Differentiating the given curve, we get

2y′ + 25x4 – 30x2

+ 1 = 0, y′(0) = − 1

2

The normal at (0, – 3) is y = 2x – 3.

Eliminating y with the given equation, we get

x(x2 – 1)

2 = 0 ⇒ x = 0, 1, – 1

6. (a,c,d) : 151

2

15

2

715

2

13

N rr

rr

= ( ) = ( )= =∑ ∑

∴ 30 N = 215

– 2(1 + 15) = 215

– 32

⇒ 5 N = 5456 = 24 ·11·31

7. (d) 8. (a)

9. (3) : The differential equation satisfied by

(x – c)2 + y2

= 1 is y2(y′)2

= 1 – y2

Replacing by we get′ −′

yy1

( )( )11

12 2

2

2− − ′= ⇒

−= −

y yy

yy

dy dx

Integrating we get

x c yy

y+ = − − +

+ −⎛

⎝⎜

⎞

⎠⎟1

1 12

2

ln

x = 0, y = 1 ⇒ c = 0

∴ y x= ⇒ = − +3

2

1

23ln ⇒ e2x + 1

= 3

10. (b) : (P) → 4; (Q) → 2; (R) → 3; (S) → 3

(P) tdt

tdt

t tt

ue

x

e

x

1 1

1

21

21+

++

=⎛⎝⎜

⎞⎠⎟∫ ∫

/

tan

/

cot

( ),

=

++

+=

+=∫ ∫ ∫

tdtt

uduu

tdtte

x

x

e

e

e

1 1 11

21

2 21/

tan

tan /( ).

(Q) dx

x xx t

( sin cos ), (tan )

/

+=∫ 4

0

2π

=

+= −

⎛⎝⎜

⎞⎠⎟

=∞ ∞

∫ ∫dt

t u udu

( ).

12

1 1 1

340

3 41

(R) I xx

dx d1

1

0

1

0

42

2= =

−

∫ ∫tan

sin

/ θθ

θπ

= = ⇒ =∫

1

2

1

2

1

22

1

20

2 xx

dx I IIsin

/π

(S) F e tt

dt tt

dt tu

e e( )

ln ln,

/

=+

++

=⎛⎝⎜

⎞⎠⎟∫ ∫

1 1

1

1 1

1

=+

++

=+

+⎛⎝⎜

⎞⎠⎟ =∫ ∫ ∫

ln ln

( )

ln

( ).

tt

dt uu u

du tt t

dte e e

1 1 11

1 1

21 1 1


1. Find the number of real solutions of the equations

x x y x y2 22 1 1− + = + =, ?

Abhishek, KolkataAns. We will consider the four cases separately:

(0, 1)

( , )1 00 2–1

–1

–2

–2

(a) x2 – 2x ≥ 0, y ≥ 0 : Here, we have to solve

x2 – 2x + y = 1, x2 + y = 1

The unique solution is x = 0, y = 1.

(b) x2 – 2x ≤ 0, y ≥ 0 : Here, we have to solve

–x2 + 2x + y = 1, x2 + y = 1

We have –2x2 + 2x = 0 and (x, y) = (0, 1) or (1, 0).

The new solution is (1, 0).

(c) x2 – 2x ≥ 0, y ≤ 0 : Here, we have to solve

x2 – 2x + y = 1, x2 – y = 1

Here x2 – x – 1 = 0, x = y. Solving the quadratic in

x we get x = ±1 5

2. y ≤ 0,

⇒ = − −⎛⎝⎜

⎞⎠⎟( , ) ,x y 1 5

2

1 5

2

(d) x2 – 2x ≤ 0, y ≤ 0 : Here, we have to solve

–x2 + 2x + y = 1, x2 – y = 1

Do you have a question that you just can’t get answered?Use the vast expertise of our MTG team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough.The best questions and their solutions will be printed in this column each month.

Here again the unique solution is (1, 0). Thus there

are three solutions.

2. Let ABC be a triangle with incentre I and inradius

r. Let D, E, F be the feet of the perpendiculars

from I to the sides BC, CA and AB respectively. If

r1, r2, r3 are the radii of the circles inscribed in the

quadrilaterals AFIE, BDIF and CEID respectively,

then prove that

rr r

rr r

rr r

r r rr r r r r r

1

1

2

2

3

3

1 2 3

1 2 3−+

−+

−=

− − −( )( )( ).

Vishal, U.P.Ans.

The exaggerated view of the quadrilateral AFIE and its incircle is shown below where IF = r and

MI1 = NI1 = r1.

From the figure, we have

∠ =∠ =MAI NI I A1 1 2/

and IN = IF – NF = IF – MI1 = r – r1

Therefore, we have

cotA NI

NIr

r r2

1 1

1

⎛⎝⎜

⎞⎠⎟

= =−

Similarly, we have

cotB r

r r2

2

2

⎛⎝⎜

⎞⎠⎟

=−

and cotC r

r r2

3

3

⎛⎝⎜

⎞⎠⎟

=−

Now, we have

A B C2 2 2 2

+ + = π

i.e. cot cot cot

cot cot

A B C

A B2 2 2

2 2

⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ccotC2

⎛⎝⎜

⎞⎠⎟

i.e. rr r

rr r

rr r

r r rr r r r r r

1

1

2

2

3

3

1 2 3

1 2 3−+

−+

−=

− − −( )( )( )

which is the desired result.

Y UASKWE ANSWER

contents 69 77 84

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