S1. Ans.(d)

Sol.

3 bowlers can be selected from the five players and 8 players can be selected from 12 players (17-5) = 12

is the number of ways of selecting the cricket team of 11 players.

P = C(5, 3) × C(12, 8)

S2. Ans.(b)

Sol.

log𝑎 27 + log8 32 log 27

log 9+

log 32

log 8

log 33

log 32 +log 25

log 23

3 log 3

2 log 3+

5 log 2

3 log 2

3

2+

5

3 =

9+10

6 =

19

6

S3. Ans.(a)

S4.Ans.(d)

Sol.

|a– x c b

c b– x ab a c– x

| = 0

c1 → c2 + c2 + c3

|a + b + c – x a + b + c – x a + b + c – x

c b– x ab a c– x

| = 0

(a + b + c – x) |1 1 1c b – x ab a c – x

| = 0

⇒ x = a + b + c = 0

[ ∵ a + b + c = 0]

S5. Ans.(b)

Sol.

Matrix A have an inverse iff |A| ≠ 0.

Consider |A| = 0.

⇒ 2x + 32 = 0

x =–32

2

x = –16

S6. Ans.(b)

Sol.

From the system of equations

2x + y – 3z = 5

3x – 2y + 2z = 5 and

5x – 3y – z = 16 𝑎1

𝑎2=

2

3≠

1

–2=

𝑏1

𝑏2

𝑎2

𝑎3=

3

5≠

2

3=

𝑏2

𝑏3

𝑎3

𝑎1=

2

5≠

1

–3=

𝑏3

𝑏1

Hence the given system of equations is consistent, with a unique solution.

S7. Ans.(d)

Sol.

Cube root of unity lie on the unit circle |z| = 1.

S8. Ans.(a)

Sol.

u = arp–1

v = arq–1

w = arr–1

consider |ln 𝑢 𝑝 1ln 𝑣 𝑞 1ln 𝑤 𝑟 1

|

= |

ln(arp–1) p 1

ln(𝑎𝑟𝑞−1) p 1

ln(arr–1) r 1

|

= |

(p– 1) ln(ar) p 1(q– 1) ln(ar) q 1(r– 1) ln(ar) r 1

| [∵ ln ab = b ln a]

= ln(ar) |p– 1 p 1q– 1 q 1r – 1 r 1

|

= ln(ar) ||p p 1q q 1r r 1

| – |1 p 11 q 11 r 1

||

ln(ar) [0 – 0] = 0 [∵ determinant have two columns are identical∴ its value is zero

]

S9. Ans.(c)

Sol.

The middle term of the expansion (1 + 𝑥)2𝑛

= (2𝑛

2+ 1)

𝑡ℎ

𝑡𝑒𝑟𝑚

The coefficient of the middle term of (1 + 𝑥)2𝑛

= 𝐶𝑛+1 =∝ 2𝑛

The middle terms of the expansion (1 + 𝑥)2𝑛−1

= (2𝑛−1+1

1)

𝑡ℎ

𝑡𝑒𝑟𝑚 𝑎𝑛𝑑 (2𝑛−1+1

2+ 1)

𝑡ℎ

𝑡𝑒𝑟𝑚

The coefficients of the middle term of the expansion (1 + 𝑥)2𝑛−1 = 𝐶𝑛 2𝑛−1 𝑎𝑛𝑑 𝐶𝑛+1

2𝑛−1

Given that 𝐶𝑛 = 𝛽 2𝑛−1

And 𝐶𝑛+1 2𝑛−1 = 𝛾

Consider,

α = β + γ

𝐶𝑛+1 2𝑛 = 𝐶𝑛

2𝑛−1 + 𝐶𝑛+1 2𝑛−1

(2𝑛)!

(𝑛+1)!(2𝑛−𝑛−1)!=

(2𝑛−1)!

(2𝑛−1−𝑛)!𝑛!+

(2𝑛−1)!

(𝑛+1)!(2𝑛−1−𝑛−1)!

(2𝑛)!

(𝑛+1)!(𝑛−1)!= (2𝑛 − 1)! [

1

(𝑛−1)!𝑛!+

1

(𝑛+1)!(𝑛−2)!]

= (2𝑛 − 1)! [(𝑛+1)!(𝑛−2)!+(𝑛−1)!𝑛!

(𝑛−1)!𝑛!(𝑛+1)!(𝑛−2)!]

= (2𝑛 − 1)! [(𝑛+1)𝑛!(𝑛−2)!+(𝑛−1)(𝑛−2)!𝑛!

(𝑛−1)!𝑛!(𝑛+1)!(𝑛−2)!]

= (2𝑛 − 1)! 𝑛! (𝑛 − 2)! [𝑛+1+𝑛−1

(𝑛−1)!𝑛!(𝑛+1)!(𝑛−2)!]

=2𝑛! 𝑛! (𝑛−2)!

(𝑛−1)! 𝑛! (𝑛+1)! (𝑛−2)!

=(2𝑛)!

(𝑛−1)!(𝑛+1)!

= 𝐿. 𝐻. 𝑆 = 𝑅. 𝐻. 𝑆.

Q10. Ans.(d)

Sol.

A function 𝑓(𝑥1 … . 𝑥𝑛) has the property, that for one set of values (𝑣1 … . 𝑣𝑛) there is at most one result. If

you compare. Your f(0)=1, but there are 2 values for y s.t 𝑦2 + 𝑥2 = 1 | 𝑥 = 0, namely {1, −1}.

S11. Ans.(c)

Sol.

Given Tm =1

n

a + (m– 1)d =1

n …(i)

and Tn =1

m

a + (n– 1)d =1

m …(ii)

solving (i) and (ii), we gets

d =1

mn and a =

1

mn

Now,

Tmn = a + (mn– 1)d

= 1

mn+ (mn– 1)

1

mn

= 1

S12. Ans.(b)

Sol.

𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 , 𝑓 > 0 ⇒ 𝑎 > 0.

𝑓(𝑥) + 𝑓′(𝑥) + 𝑓"(𝑥) = 𝑎𝑥2 + (2𝑎 + 𝑏)𝑥 + 𝑐 + 𝑏 + 2𝑎 = 𝑔(𝑥)

Reformulating g in terms of x + 1 gives

𝑎(𝑥 + 1)2 + (2𝑎 + 𝑏)(𝑥 + 1) + 𝑐 + 𝑏 + 2𝑎 − (2𝑎𝑥 + 𝑎) − (2𝑎 + 𝑏)

= 𝑎(𝑥 + 1)2 + (2𝑎 + 𝑏)(𝑥 + 1) + 𝑐 − 2𝑎(𝑥 + 1) + 𝑎

= 𝑎(𝑥 + 1)2 + 𝑏(𝑥 + 1) + 𝑐 + 𝑎 = 𝑔(𝑥)

So g(x) = f(x + 1) + a, so g(x) is f(x) translated by 1 to the left and by a upwards.

𝑓 > 0 ⇒ 𝑔 > 0.

S13. Ans.(c)

Sol.

1 and 3 only

S14. Ans.(b)

Sol.

First of all adding (11011)₂ & (10110110)₂ 1 1 0 1 1

1 0 1 1 0 1 1 0

1 1 0 1 0 0 0 1

Now by adding (11010001)₂ & (10011x 0y)₂

We get (101101101)₂

i.e. 1 1 0 1 0 0 0 1

1 0 0 1 1 𝑥 0 𝑦

1 0 1 1 0 1 1 0 1

⇒ y = 0 & x = 1

S15. Ans.(b)

Sol.

Given B = adj A

AB = A (adj A)

= |A|In where In is the identity matrix of A.

= kℓ [ ∵ |A| = k & In= ℓ]

S16. Ans.(c)

Sol.

Given that

ar = 2 …(i)

and 𝑎

1–𝑟= 8 …(ii) [∴ If number of terms is infinite then sum of the terms is S =

𝑎

1–𝑟, |r| < 1]

Solving (i) and (ii), we get

a = 4 and r = ½

∴ The G.P. is

4, 2, 1,1

2,

1

22 , …

Q17. Ans.(c)

Sol.

I. Let a, b, c are in A.P.

∴ 2b = a + c. ⇒ a = 2b – c

Consider 𝑎−𝑏

𝑏−𝑐=

2𝑏−𝑐−𝑏

𝑏−𝑐=

𝑏−𝑐

𝑏−𝑐= 1

II. Let a, b, c, are in G.P.

⇒ 𝑏2 = 𝑎𝑐. ⇒ 𝑐 =𝑏2

𝑎

Consider 𝑎−𝑏

𝑏−𝑐

=𝑎−𝑏

𝑏−𝑏2

𝑎

=(𝑎−𝑏)

𝑎𝑏−𝑏2

𝑎

=(𝑎−𝑏).𝑎

𝑏(𝑎−𝑏)

=𝑎

𝑏

III. Let a, b, c are in H.P

⇒ b = 2𝑎𝑐

𝑎+𝑐

Consider 𝑎−𝑏

𝑏−𝑐

= 𝑎−

2𝑎𝑐

𝑎+𝑐2𝑎𝑐

𝑎+𝑐−𝑐

= 𝑎2+𝑎𝑐−2𝑎𝑐

2𝑎𝑐−𝑎𝑐−𝑐2

=𝑎2−𝑎𝑐

𝑎𝑐−𝑐2 =𝑎(𝑎−𝑐)

𝑐(𝑎−𝑐)=

𝑎

𝑐

S18. Ans.(a)

Sol.

345 + 354 + 435 + 453 + 534 + 543 = 2664

S19. Ans.(b)

Sol.

Given, −𝑏+√𝑏2−4𝑎𝑐

−𝑏−√𝑏2−4𝑎𝑐=

−𝑞+√𝑞2−4𝑝𝑟

−𝑞−√𝑞2−4𝑝𝑟

−𝑏+√𝐷1

−𝑏−√𝐷1=

−𝑞+√𝐷2

−𝑞−√𝐷2

𝑏𝑞 + 𝑏√𝐷2 − 𝑞√𝐷1 − √𝐷1𝐷2 = 𝑏𝑞 − 𝑏√𝐷2 + 𝑞√𝐷1 − √𝐷1𝐷2

2𝑏√𝐷2 = 2𝑞√𝐷1

𝑏

𝑞=

√𝐷1

√𝐷2

⟹𝐷1

𝐷2=

𝑏2

𝑞2

S20. Ans.(b)

Sol.

𝐴 = sin2 𝜃 + cos4 𝜃

= sin2 𝜃(sin2 𝜃 + cos2 𝜃) + cos4 𝜃

= sin4 𝜃 + 2 sin2 𝜃 cos2 𝜃 + cos4 𝜃 − sin2 𝜃 cos2 𝜃

= (sin2 𝜃 + cos2 𝜃)2 − sin2 𝜃 cos2 𝜃

= 1 − sin2 𝜃 cos2 𝜃

= 1 −1

4(sin 2𝜃)2

∵ −1 ≤ sin 2𝜃 ≤ 1

0 ≤ sin2 2𝜃 ≤ 1

0 ≤1

4sin2 2𝜃 ≤

1

4

0 ≥ −1

4sin2 2𝜃 ≥ −

1

4

1 ≥ 1 −1

4sin2 2𝜃 ≥ 1 −

1

4

1 ≥ 1 −1

4sin2 2𝜃 ≥

3

4

S21. Ans.(d)

S22. Ans.(d)

S23. Ans.(c)

Sol.

Here, 𝑚1 = –ℓ

𝑚

𝑚2 =(–ℓ1)

𝑚1

Angle between two lines is

tan–1 θ = |𝑚1–m2

1+m1m2|

= |–ℓ

𝑚+

ℓ1

𝑚1

1+ℓ

𝑚

ℓ1

𝑚1

|

= |–ℓm1+ℓ1m

mm1+ℓℓ1|

= |ℓ𝑚1–ℓ1𝑚

𝑚𝑚1+ℓℓ1|

S24. Ans.(d)

S25. Ans.(a)

Sol.

Intersection of the lines 𝑥

2+

𝑦

3 = 1 and

𝑥

3+

𝑦

2 = 1 is (

6

5,

6

5)

Slope of the line 4x + 5y – 6 = 0 is –4

5

Slope of line passing through (6

5,

6

5) is same as the slope of the line 4x + 5y – 6 as they are same.

∴ the equation of line passing through (6

5,

6

5) and having slope is

–4

5

(𝑦–6

5) =

–4

5 (𝑥–

6

5)

⇒ 20x + 25y – 54 = 0

S26. Ans.(a)

Sol.

Distance of a point from the plane is

|𝑎𝑥1+𝑏𝑦1+𝑐𝑧1+𝑑

√𝑎2+𝑏2+𝑐2|

i.e. |3×2–6×3+2×4+11

√32+(–6)2+(2)2|

= |6–18+8+11

√9+36+4|

= |7

7|

= 1

S27. Ans.(c)

S28. Ans.(d)

Sol.

The line 𝑥−1

2=

𝑦−2

3=

𝑧−3

4 can be written

As 𝑥−1

3−1=

𝑦−2

5−2=

𝑧−3

7−3(𝑖. 𝑒.

𝑥−𝑥1

𝑥2−𝑥1=

𝑦−𝑦1

𝑦2−𝑦1=

𝑧−𝑧1

𝑧2−𝑧1)

This gives (𝑥1, 𝑦1, 𝑧1) = (1, 2, 3) 𝑎𝑛𝑑 (𝑥2, 𝑦2, 𝑧2) = (3, 5, 7)

As line is passing through (𝑥1, 𝑦1, 𝑧1) & (𝑥2, 𝑦2, 𝑧2)

∴ These two points must satisfy the equation of line.

Taking point (𝑥2, 𝑦2, 𝑧2) = (3, 5, 7)

Consider

L.H.S

3x + 2y – 3z

Put (x, y, z) = (3, 5, 7)

9 + 10 – 21 = – 2 = R. H. S

Again,

L.H.S

3x – 6y + 3z

Put (x, y, z) = (3, 5, 7)

9 – 30 + 21

= 0 = R. H. S.

S29. Ans.(c)

Sol.

I. cos 𝜃 = ±𝑎1𝑎2+𝑏1𝑏2+𝑐1𝑐2

√𝑎12+𝑏1

2+𝑐12√𝑎2

2+𝑏22+𝑐2

2

= ± (2–1+2

√(4+1+1)√1+1+4)

= ± (3

√6√6)

= 3

6=

1

2

∴ θ = 𝜋

3

II. 6x – 3y + 6z + 2 = 0

⇒ 2x – y + 2z + ⅔ = 0

Distance between the planes 2x – y + 2z + ⅔ = 0

And 2x – y + 2z + 4 = 0 is

|d1–d2

√a2+b2+c2|

= |2

3 – 4

√4+1+4|

= |2–12

3√9|

= |–10

9|

= 10

9

S30. Ans.(d)

Sol.

I.

Line OP = y – 0 = 𝑛−0

𝑚−0(𝑥 − 0)

𝑦 =𝑛

𝑚𝑥

nx – my = 0

and OQ = y – 0 = 𝑠−0

𝑟−0(𝑥 − 0)

𝑦 =𝑠

𝑟𝑥

sx - ry = 0

we know that if 𝑎1𝑥 + 𝑏1𝑦 + 𝑐1 = 0 𝑎𝑛𝑑 𝑎2𝑥 + 𝑏2𝑦 + 𝑐2 = 0 are two lines, the angle between them can be

calculated using the following formula

cos θ = 𝑎1𝑏2+𝑏1𝑏2

√(𝑎12+𝑏1

2)√𝑎22+𝑏2

2

∴ cosθ = 𝑛𝑠+𝑚𝑟

√(𝑛2+𝑚2√𝑠2+𝑟2

II. By the cosine rule,

cos A = 𝑏2+𝑐2−𝑎2

2𝑏𝑐

⇒ 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 𝑐𝑜𝑠𝐴

S31. Ans.(c)

Sol.

log7 log7√7√7√7

log7 log7√7√71+

1

2

log7 log7√7.7

3

2.1

2

log7 log7√71+

3

4

log7 log7 77

4.1

2

log7 log7 77

8

log7 (7

8log7 7)

= log7 (7

8)

= log7 7 – log7 8

= 1– log7 23

= 1– 3 log7 2

S32. Ans.(c)

Sol.

Sum of an infinite G.P. is

S =a

1–r, |r| < 1

i.e. 5 =x

1–r, |r| < 1.

⇒ 5 (1 – r) = x, |r| < 1

Given |r| < 1

i.e. –1 < r < 1

when r < 1

– 1 < – r

0 < 1 – r

⇒ 0 < 5 (1 – r) = x

⇒ x > 0 …(i)

When r > –1

1 > –r

1 + 1 > 1 – r

2 > 1 – r

⇒ 10 > 5 (1 – r) = x

i.e. x < 10 …(ii)

from (i) and (ii)

0 < x < 10.

S33. Ans.(a)

Sol.

As per the definition of Rational expressions

1, 4 & 5 are the rational functions

S34. Ans.(b)

Sol.

A square matrix A is called orthogonal if

ATA = I.

i.e. AT = A–1.

S35. Ans.(c)

Sol.

Let U = {1, 2, 3, 4, 5, 6, 7, 8}

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

C = {2, 3, 7, 8}

L.H.S.

A′ ∪ (B ∪ C) = (5, 6, 7, 8) ∪ [ 2, 3, 4, 5, 6, 7, 8]

= {2, 3, 4, 5, 6, 7, 8}

R.H.S. (C′ ∩ B)′ ∩ A′= [(1, 4, 5, 6) ∩ (3, 4, 5, 6)]′ ∩ (5, 6, 7, 8)

= (4, 5, 6)′ ∩ (5, 6, 7, 8)

= (1, 2, 3, 7, 8) ∩ (5, 6, 7, 8)

= (7, 8)

L.H.S. ≠ R.H.S.

S36. Ans.(b)

Sol.

Total numbers between 2999 to 8001= 5001

We find the numbers in which number doesn’t repeat than subtract it from total numbers

Numbers without repetition

5×9×8×7 = 2520

Required Answer = 5001 – 2520

= 2481

S37. Ans.(c)

Sol.

a = 3

r =–1

3

sum of infinite terms of G.P. is a

1 – r

= 3

1+1

3

= 34

3

= 3

1×

3

4 =

9

4

S38. Ans.(a) Sol.

Badminton = 125 Volleyball = 145 Tennis = 90 (B.+V.) + (V. +T.) + (T. + B.) = 32 Atq, 300 = B. + V. + T. – [common area] – 24 300 = 360 – 32 – 2x x = 14

S39. Ans.(c) Sol. Show figure in solution 3 Required value B + V + T – 2(32) – 3(14) = 360 – 64 – 42 = 254

S40. Ans.(d)

Sol.

Given that α & β are the roots of the equation x² + αx – β = 0

⇒ α + β = – α & αβ = –β

⇒ β = – 2α ⇒ α = – 1

⇒ β = 2

∴ we can write quadratic expression – x2 + αx + β as f(x) =– x2– x + 2

f ′(x) = – 2x– 1 = 0

𝑥 = –1

2

𝑓′′(𝑥) = – 2 < 0

⇒ f(x) is an increasing function,

∴ Greatest value of f(x) is f (–1

2)

i.e. 9

4

S41. Ans.(d)

Sol.

Middle term of (2 + 3𝑥)4 = (4

2+ 1)

𝑡ℎ

𝑡𝑒𝑟𝑚 = (3)𝑟𝑑𝑡𝑒𝑟𝑚

𝑇3+1 = 𝐶3(2)4−3(3𝑥)3. 4

= 𝐶3 (2)33 𝑥3

4

=4!

3!1!× 2 × 27 × 𝑥3

=4×3!

3!× 2 × 27 × 𝑥3

= 4 × 2 × 27 × 𝑥3

= 216

S42. Ans.(a)

Sol.

(𝜆𝐴)−1 =𝑎𝑑𝑗(𝜆𝐴)

|𝜆𝐴|=

𝜆𝑛−1 𝑎𝑑𝑗 𝐴

𝜆𝑛|𝐴|

=𝜆𝑛(𝐴)−1

𝜆 𝜆𝑛

=𝐴−1

𝜆

S43. Ans.(a)

Sol.

|

𝑥 𝑦 3

𝑥2 5𝑦3 9

𝑥3 10𝑦5 27

|

𝐶1 → 𝐶1 − 𝐶3.

|

𝑥 − 3 𝑦 3

(𝑥2 − 9) 5𝑦3 9

(𝑥3 − 27) 10𝑦5 27

|

= |

𝑥 − 3 𝑦 3

(𝑥 + 3)(𝑥 − 3) 5𝑦3 9

(𝑥 − 3)(𝑥2 + 9 + 3𝑥) 10𝑦5 27

|

= (𝑥 − 3) |

1 𝑦 3

𝑥 + 3 5𝑦3 9

𝑥2 + 3𝑥 + 9 10𝑦5 27

|

S44. Ans.(a)

Sol.

𝐴 = (cos(−𝜃) − sin(−𝜃)

− sin(−𝜃) cos(−𝜃))

A = (𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

)

Adjoint A = [𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃

−𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃]

S45. Ans.(b)

Sol.

We know that

1 + 𝜔𝑟 + 𝜔2𝑟 = {0, 𝑖𝑓 𝑟 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3

3, 𝑖𝑓 𝑟 𝑖𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3.

& 𝜔 =−1+𝑖√3

2, 𝜔2 =

−1−𝑖√3

2

Given (𝜔)3𝑛 + (𝜔2)3𝑛

⇒ 1 + 𝜔3𝑛 + 𝜔2(3𝑛) − 1

⇒ 3 – 1 = 2.

S46. Ans.(c)

Sol.

Consider (0.2)𝑥 = 2

log10(0.2)𝑥 = log10 2.

𝑥 log10(0.2) = log10 2.

𝑥 log10 (2

10) = log10 2.

𝑥[log10 2 − log10 10] = log10 2.

𝑥 =log10 2

log10 2−log10 10

=0.3010

0.3010−1

=0.3010

−0.6990

= −0.4

S47. Ans.(c)

Sol.

There are 9 choices for the first digits, since 0 can’t be used. For the second digit, you can use any of the

remaining 9 digits. For the third digit you can use any of the 8 digits not already used. For the next digit,

there are 7 choices. And for the final digit there are 6 choices left. Multiplying the values together gives

the stated answer.

9 × 9 × 8 × 7 × 6 = 27216

S48. Ans.(d)

Sol.

(𝑥 𝑦 𝑦 + 𝑧𝑧 𝑥 𝑧 + 𝑥𝑦 𝑧 𝑥 + 𝑦

)

𝑅1 → 𝑅1 + 𝑅2 + 𝑅3

(𝑥 + 𝑦 + 𝑧 𝑥 + 𝑦 + 𝑧 2(𝑥 + 𝑦 + 𝑧)

𝑧 𝑥 𝑧 + 𝑥𝑦 𝑧 𝑥 + 𝑦

)

= (𝑥 + 𝑦 + 𝑧) (1 1 1𝑧 𝑥 𝑧 + 𝑥𝑦 𝑧 𝑥 + 𝑦

)

= (𝑥 + 𝑦 + 𝑧)(𝑧 − 𝑥)2

S49. Ans.(b)

Sol.

|1 1 1

1 + 𝑆𝑖𝑛𝐴 1 + 𝑆𝑖𝑛𝐵 1 + 𝑆𝑖𝑛𝐶𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐴 𝑆𝑖𝑛𝐵 + 𝑆𝑖𝑛2𝐵 𝑆𝑖𝑛𝐶 + 𝑆𝑖𝑛2𝐶

|

𝐶2 → 𝐶2 − 𝐶1 𝑎𝑛𝑑 𝐶3 → 𝐶3 − 𝐶1

|1 0 0

1 + 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 − 𝑠𝑖𝑛𝐴 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐴 𝑆𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐵 − sin2 𝐴 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴 + sin2 𝐶 − 𝑆𝑖𝑛2𝐴

|

= (𝑠𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴)(𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴)

|1 0 0

1 + 𝑆𝑖𝑛𝐴 1 1𝑆𝑖𝑛𝐴 + 𝑆𝑖𝑛2𝐴 1 + 𝑠𝑖𝑛𝐵 + 𝑆𝑖𝑛𝐴 1 + 𝑆𝑖𝑛𝐶 + 𝑆𝑖𝑛𝐴

|

= (𝑆𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴)(𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴)(𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐵) = 0 ⟹ 𝑆𝑖𝑛𝐵 − 𝑆𝑖𝑛𝐴 = 0 , 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐴 = 0 , 𝑆𝑖𝑛𝐶 − 𝑆𝑖𝑛𝐵 = 0

𝑆𝑖𝑛𝐵 = 𝑆𝑖𝑛𝐴 𝑆𝑖𝑛𝐶 = 𝑆𝑖𝑛𝐴 𝑆𝑖𝑛𝐶 = 𝑆𝑖𝑛𝐵𝐴 = 𝐵 𝐶 = 𝐴 𝐶 = 𝐵

⇒ The triangle ABC is equilateral.

S50. Ans.(b)

S51. Ans.(c)

Sol.

=2 tan 𝜃

1+tan2 𝜃

=2 tan 𝜃

sec2 𝜃

=2 sin 𝜃

cos 𝜃cos2 𝜃

= 2 sin 𝜃 cos 𝜃 = sin 2 𝜃.

S52.Ans (a)

Sol.

2sec𝜃 =sec(𝜃 + 𝛼)+sec(𝜃 − 𝛼)

2

𝑐𝑜𝑠𝜃=

cos(𝜃+𝛼)+cos (𝜃−𝛼)

cos(𝜃+𝛼)cos (𝜃−𝛼)

2

𝑐𝑜𝑠𝜃=

2cos(𝜃)+cos (𝛼)

cos(𝜃+𝛼)cos (𝜃−𝛼)

𝑐𝑜𝑠2𝜃𝑐𝑜𝑠𝛼 = 𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛2𝛼

𝑠𝑖𝑛2𝛼 = 𝑐𝑜𝑠2𝜃(1 − 𝑐𝑜𝑠𝛼)

1-𝑠𝑖𝑛2𝜃 = 1 + 𝑐𝑜𝑠𝛼

𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠𝛼 = 0

53. Ans.(d)

Sol.

sin 2𝐴 − sin 2𝐵 − sin 2𝐶.

sin (2(180 − (𝐵 + 𝐶)) − sin 2𝐵 − sin 2𝐶.

sin (360 − (2𝐵 + 2𝐶) − sin 2𝐵 − sin 2𝐶.

− sin(2𝐵 + 2𝐶) − sin 2𝐵 − sin 2𝐶.

− sin 2𝐵 cos 2𝐶 − cos 2𝐵 sin 2𝐶 − sin 2𝐵 − sin 2𝐶

− sin 2𝐵(cos 2𝐶 + 1) − sin 2𝐶(cos 2𝐵 + 1)

− sin 2𝐵(2 cos2 𝐶 − 1 + 1) − sin 2C(2 cos2 𝐵 − 1 + 1)

−2 cos2 𝐶 sin 2𝐵 − 2 cos2 𝐵 sin 2𝐶

−4 cos2 𝐶 sin 𝐵 cos 𝐵 − 4 cos2 𝐵 sin 𝐶 cos 𝐶

−4 cos 𝐶 cos 𝐵[cos 𝐶 sin 𝐵 + sin 𝐶 cos 𝐵)

−4 𝑐𝑜𝑠𝐶 𝑐𝑜𝑠𝐵 sin (𝐵 + 𝐶)

−4 𝑐𝑜𝑠𝐶 𝑐𝑜𝑠𝐵 sin (180 − 𝐴)

−4 𝑐𝑜𝑠𝐶 𝑐𝑜𝑠𝐵 𝑠𝑖𝑛𝐴

S54. Ans.(b)

Sol.

tan 48° = 122

𝑥

𝑥 =122

tan 48°

=122

cot 42°

= 122 tan 42°

S55. Ans.(a)

Sol.

By going through the options

Option (A)

300° = (360° – 60°)

∵ the angle lies in 4th quadrant

∴ 300° = – 60°

Putting – 60° in option A

3[3 − tan2(−60°) − cot (−60°)]2

= 3 [3 − 3 +1

√3]

2

= 3×1

3= 1 𝐴𝑛𝑠.

S56. Ans.(b)

Sol.

In ∆ ABC.

tan 30° = 𝑦

𝑥

1

√3=

𝑦

𝑥

𝑥 = √3 𝑦 __________(1)

In ∆ ADE

tan 60° = 𝑦+ℎ

𝑥

√3 =𝑦+ℎ

𝑥

√3 𝑥 = 𝑦 + ℎ

√3 × √3𝑦 = 𝑦 + ℎ

3𝑦 = 𝑦 + ℎ

2y = h

y = ℎ

2

∴ height of hill = y + h = ℎ

2+ ℎ =

3ℎ

2

S57. Ans.(b)

Sol.

𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥 = √3 1

sin 𝑥+

cos 𝑥

sin 𝑥= √3

1+cos 𝑥

sin 𝑥= √3

1+2 cos2𝑥

2−1

2 sin𝑥

2cos

𝑥

2

= √3

cot𝑥

2= cot

𝜋

6

𝑥

2=

𝜋

6

𝑥 =𝜋

3

S58. Ans.(b)

Sol. (2𝑐𝑜𝑠𝜃 + 1)10(2 cos 2𝜃 − 1)10(2 cos 𝜃 − 1)10(2 cos 4𝜃 − 1)10

[(2 cos 𝜃 + 1)(2 cos 𝜃 − 1)(2 cos 2 𝜃 − 1)(2 cos 4𝜃 − 1)]10

[(4 cos2 𝜃 − 1)(2 cos 2𝜃 − 1)(2 cos 4𝜃 − 1)]10

[(2 (2 𝑐𝑜𝑠2 𝜃 − 1) + 1)(2 cos 2 𝜃 − 1)(2 cos 4𝜃 − 1)]10

[(2 cos 2𝜃 + 1)(2 cos 2𝜃 − 1)(2 cos 4 𝜃 − 1)]10

[(4 cos2 2 𝜃 − 1)(2 cos 4 𝜃 − 1)]10

[(2(2 cos2 2𝜃 − 1) + 1)(2 cos 4𝜃 − 1)]10

[(2 cos 4𝜃 + 1)(2 cos 4 𝜃 − 1)]10 [4 cos2 4𝜃 − 1]10

𝑎𝑡 𝜃 =𝜋

8

[4 cos2 𝜋

2− 1]

10

(−1)10 = 1

S59. Ans.(a)

Sol.

Given that

𝑐𝑜𝑠𝛼 + 𝑐𝑜𝑠𝛽 = 0 ⇒ 𝑐𝑜𝑠𝛼 = −𝑐𝑜𝑠𝛽 _______(1)

& 𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛽 =−3

4

⇒ − cos2 𝛽 =−3

4 [𝑓𝑟𝑜𝑚 (1)]

⇒ cos2 𝛽 =3

4

Consider

𝑠𝑒𝑐𝛼 × 𝑠𝑒𝑐𝛽 1

𝑐𝑜𝑠𝛼×

1

𝑐𝑜𝑠𝛽

−1

cos2 𝛽= −

4

3

S60. Ans.(a)

Sol.

tan−1(2𝑥) + tan−1 3𝑥 =𝜋

4

tan−1 (2𝑥+3𝑥

1−6𝑥2) =𝜋

4

5𝑥

1−6𝑥2 = 1

5𝑥 = 1 − 6𝑥2

6𝑥2 + 5𝑥 − 1 = 0

6𝑥2 + 6𝑥 − 𝑥 − 1 = 0

6𝑥(𝑥 + 1) − 1(𝑥 + 1) = 0 (6𝑥 − 1)(𝑥 + 1) = 0

⇒ 𝑥 =1

6𝑜𝑟 𝑥 = −1.

S61. Ans.(a)

Sol.

lim𝑥→1

𝑓(𝑥)−𝑓(1)

𝑥−1

= lim𝑥→1

√25−𝑥2−√24

𝑥−1= (

0

0) 𝑓𝑜𝑟𝑚

Applying L’Hospital rule

= lim𝑥→1

1 (−2𝑥)

2√25−𝑥2= lim

𝑥→1

(−𝑥)

√25−𝑥2=

−1

√24

S62. Ans.(a)

Sol.

y = tan−1 (5−2 tan √𝑥

2+5 tan √𝑥)

𝑙𝑒𝑡 𝑢 =5−2 tan √𝑥

2+5 tan √𝑥

y = tan−1 𝑢 𝑑𝑦

𝑑𝑢=

1

1+𝑢2 _______(1)

u = 5−2 tan √𝑥

2+5 tan √𝑥

𝑑𝑢

𝑑𝑥=

(2+5 tan √𝑥)𝑑

𝑑𝑥(5−2 tan √𝑥)−(5−2 tan √𝑥)

𝑑

𝑑𝑥(2+5 tan √𝑥)

(2+5 tan √𝑥)2

=(2+5 tan √𝑥) (−2 sec2 √𝑥.

1

2√𝑥)−(5−2 tan √𝑥).5 sec2 √𝑥 .

1

2√𝑥

(2+5 tan √𝑥)2

=

−2 sec2 √𝑥

√𝑥−

5 𝑡𝑎𝑛√𝑥 sec2 √𝑥

√𝑥−

25

2

sec2 √𝑥

√𝑥+

5 tan √𝑥 sec2 √𝑥

√𝑥

(2+5 tan √𝑥)2

𝑑𝑢

𝑑𝑥=

−29

2√𝑥

sec2 √𝑥

(2+5 tan √𝑥)2 ______(2)

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢×

𝑑𝑢

𝑑𝑥

=1

1+(5−2 tan √𝑥

2+5 tan √𝑥)

2(−29)

2√𝑥

sec2 √𝑥

(2+5 tan √𝑥)2

=1

(2+5 tan √𝑥)2+(5−2 𝑡𝑎𝑛√𝑥)2

(2+5 tan √𝑥)2

.(−29)

2√𝑥.

sec2 √𝑥

(2+5 tan √𝑥)2

=1

29 (1+tan2 √𝑥)

(−29)

2√𝑥 . sec2 √𝑥.

=1

29 sec2 √𝑥

(−29) sec2 √𝑥

2√𝑥

=−1

2√𝑥

S63. Ans.(a)

Sol.

𝑓(𝑥) = 𝑥 sin 𝑥 + cos 𝑥 +1

2cos2 𝑥

⇒ 𝐹′(𝑥) = 𝑥 cos 𝑥 + sin 𝑥 − sin 𝑥 − sin 𝑥 cos 𝑥

cos 𝑥(𝑥 − sin 𝑥) > 0 𝑖𝑛 [0,𝜋

2]

S64. Ans.(c)

Sol.

= lim𝜃→0

√1−cos 𝜃

𝜃

= lim𝜃→0

√1−1+2 sin2𝜃

2

𝜃

= lim𝜃→0

√2 sin𝜃

2

𝜃= (

0

0) 𝑓𝑜𝑟𝑚

Applying L’Hospital rule

lim𝜃→0

√2 cos𝜃

2.

1

2=

√2

2=

1

√2

S65. Ans.(c)

Sol.

𝑓(𝑥) = 𝑥2 − 4𝑥 + 5

The coordinates x and y of the vertex of the graph of f are given by

x = – b/2a

= 4/2 = 2.

& y = f(2) = 4 – 8 + 5 = 1

The leading coefficient a = 1 is positive & therefore the graph of f has a minimum point at (x, y) = (2, 1)

the range of f is given by the interval [1, ∞)

But A = (1, 4)

at x = 4, y = 5

Hence the range of function is [1, 5]

S66. Ans.(c)

Sol.

∫ [𝑥]𝑑𝑥 + ∫ [−𝑥]𝑑𝑥.𝑏

𝑎

𝑏

𝑎

∫ [𝑥]𝑑𝑥 + ∫ −[𝑥]𝑑𝑥𝑏

𝑎

𝑏

𝑎 [∵ [−𝑥] = {

−[𝑥], 𝑖𝑓 𝑥 𝜖 𝐼

−[𝑥] − 1, 𝑖𝑓 𝑥 ∉ 𝐼

𝑎 ∫ 𝑑𝑥 − 𝑎 ∫ 𝑑𝑥𝑏

𝑎

𝑏

𝑎

= 0

Q67. Ans.(d)

Sol.

∫ |𝑥 − 5|𝑑𝑥8

2

− ∫ (𝑥 − 5)𝑑𝑥 + ∫ (𝑥 − 5)𝑑𝑥8

5

5

2

− [𝑥2

2− 5𝑥]

2

5

+ [𝑥2

2− 5𝑥]

5

8

− [25

2− 25 − 2 + 10] + [

64

2− 40 −

25

2+ 25]

= 9

2+

9

2

= 9

S68. Ans.(d)

Sol.

∫ 𝑆𝑖𝑛3𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥

∫ sin2 𝑥 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥

∫(1 − cos2 𝑥)𝑐𝑜𝑠𝑥 𝑆𝑖𝑛𝑥 𝑑𝑥

Let cos x = t

–Sin x dx = dt

Sin x dx = – dt

− ∫(1 − 𝑡2)𝑡 𝑑𝑡

∫ 𝑡3 − 𝑡 𝑑𝑡 𝑡4

4−

𝑡2

2+ 𝐶′

𝑡4−2𝑡2

4+ 𝐶′

cos4 𝑥−2 cos2 𝑥+1

4−

1

4+ 𝐶′.

(1−cos2 𝑥)2

4+ 𝐶

Where C = 𝐶′ −1

4

S69. Ans.(b)

Sol.

∫ 𝑒𝐿𝑛(tan 𝑥)𝑑𝑥.

∫ tan 𝑥 𝑑𝑥.

𝑙𝑛 | sec 𝑥| + 𝑐

S70. Ans.(d)

Sol.

= ∫𝑑

𝑑𝑥(tan−1 1

𝑥)𝑑𝑥1

−1

= 2 ∫𝑑

𝑑𝑥(tan−1 1

𝑥) 𝑑𝑥

1

0

= 2 [tan−1 1

𝑥]

0

1

= 2[tan−1 1 − tan−1 0]

= 2 [𝜋

4− 0] =

𝜋

2.

S71. Ans.(a)

Sol.

𝑓(𝑥) = 𝑥2 − 5𝑥 + 6

𝑓′(𝑥) = 2𝑥 − 5

As function is decreasing

∴ 𝑓′(𝑥) < 0

⇒ 2x – 5 < 0

⇒ 𝑥 <5

2

𝑖. 𝑒. 𝑥 < 2.5

Hence f(x) is decreasing in the interval (−∞, 2]

S72. Ans.(d)

Sol.

y = pcos (ax) + q sin (ax) 𝑑𝑦

𝑑𝑥= −𝑎𝑝 sin 𝑎𝑥 + 𝑎 𝑞 cos 𝑎𝑥

𝑑2𝑦

𝑑𝑥2 = −𝑎2𝑝 cos 𝑎 𝑥 − 𝑎2𝑞 sin 𝑎 𝑥.

𝑑2𝑦

𝑑𝑥2 = −𝑎2(𝑝 cos 𝑎 𝑥 + 𝑞 sin 𝑎𝑥)

𝑑2𝑦

𝑑𝑥2 = −𝑎2𝑦

𝑑2𝑦

𝑑𝑥2 + 𝑎2𝑦 = 0

Q73. Ans.(b)

Sol.

Given 𝑑𝑦

𝑑𝑥= −𝑥2 −

1

𝑥3

Integrate both sides w.r.t x.

∫𝑑𝑦

𝑑𝑥= ∫ (−𝑥2 −

1

𝑥3) 𝑑𝑥

= y = −𝑥3

3+

1

2𝑥2 + 𝐶

At (–1, –2)

−2 =−(−1)3

3+

1

2(−1)2 + 𝐶

−2 =1

3+

1

2+ 𝐶

−2 −5

6= 𝐶

−17

6= 𝐶

∴ 𝑦 = −𝑥3

3+

1

2𝑥2 −17

6

6𝑥2𝑦 = −2𝑥5 + 3 − 17𝑥2

6𝑥2𝑦 + 17𝑥2 + 2𝑥5 − 3 = 0

S74. Ans.(d)

Sol.

y = a cos x + b sin x + 𝑐𝑒−𝑥 + 𝑑

Order = 4.

As the order of differential equation is equal to the number of arbitrary constants in the given relation.

S75. Ans.(d)

Sol.

𝑙𝑛 (𝑑𝑦

𝑑𝑥) = 𝑎𝑥 + 𝑏𝑦

𝑑𝑦

𝑑𝑥= 𝑒𝑎𝑥+𝑏𝑦

𝑑𝑦

𝑑𝑥= 𝑒𝑎𝑥 . 𝑒𝑏𝑦

𝑒−𝑏𝑦𝑑𝑦 = 𝑒𝑎𝑥𝑑𝑥.

𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠

∫ 𝑒−𝑏𝑦𝑑𝑦 = ∫ 𝑒𝑎𝑥𝑑𝑥

𝐶 +𝑒−𝑏𝑦

−𝑏=

𝑒𝑎𝑥

𝑎

𝐶 =𝑒𝑎𝑥

𝑎+

𝑒−𝑏𝑦

𝑏

S76. Ans.(b)

Sol.

Let 𝐸1, 𝐸2, 𝐸3 & 𝐴 be the events defined as follows:

𝐸1 = 𝐵𝑜𝑙𝑡 is manufactured by machine X.

𝐸2 = 𝐵𝑜𝑙𝑡 is manufactured by machine Y.

𝐸3 = Bolt is manufactured by machine Z.

A = Bolt is defective.

𝑃(𝐸1) =25

100

𝑃(𝐸2) =35

100

𝑃(𝐸3) =40

100

𝑃 (𝐴

𝐸1) = Probability that the bolt is defective given that it is manufactured by machine X =

2

100

𝑃 (𝐴

𝐸2) =

4

100

𝑃 (𝐴

𝐸3) =

5

100

Required Probability = Probability that the bolt is manufactured by machine X given that the bolt is

defective.

= 𝑃(𝐸2

𝐴)

=𝑃(𝐸2)𝑃(

𝐴

𝐸2)

𝑃(𝐸1)𝑃(𝐴

𝐸1)+𝑃(𝐸2)𝑃(

𝐴

𝐸2)+𝑃(𝐸3)𝑃(

𝐴

𝐸3)

=35

100×

4

40025

100×

2

100+

35

100×

4

100+

40

100×

5

100

=35×4

50+140+200

=35×4

390

=14

39

S77. Ans.(c)

Sol.

Let P denote the probability of getting head in a single toss of a coin. Then

P = 1

2 𝑎𝑛𝑑 𝑠𝑜, 𝑞 =

1

2

Let X denote the number of heads in a toss of 8 coins. Then X is a binomial variate with parameters n = 8

and p = 1

2 such that

𝑃(𝑋 = 𝑟) = 𝐶𝑟 8 (

1

2)

8−𝑟

(1

2)

𝑟

= 𝐶𝑟 8 (

1

2)

8

Where r = 0, 1, 2, ….8.

Probability of at least 6 heads = 𝑃(𝑋 ≥ 6)

= 𝑝(𝑋 = 6) + 𝑝(𝑋 = 7) + 𝑝(𝑋 = 8)

= 𝐶6 (1

2)

8

+ 𝐶7 8 (

1

2)

8

+ 𝐶8 8 (

1

2)

8

8

= (1

2)

8

[ 𝐶6 + 𝐶7 + 𝐶8 8 ]

8

8

= (1

2)

8[28 + 8 + 1]

=37

256

S78. Ans.(a)

Sol.

One girl and 2 boys can be selected in the following mutually exclusive ways:

Group 1 Group2 Group3

(I) Girl Boy Boy

(II) Boy Girl Boy

(III) Boy Boy Girl

Thus if we define 𝐺1, 𝐺2, 𝐺3 as the events of selecting a girl from first, second & third group respectively

and 𝐵1, 𝐵2, 𝐵3, as the events of selecting a boy from first, second and third group respectively. Then

𝐵1, 𝐵2, 𝐵3, 𝐺1, 𝐺2, 𝐺3 are independent events such that

𝑃(𝐺1) =3

4, 𝑃(𝐺2) =

2

4, 𝑃(𝐺3) =

1

4

𝑃(𝐵1) =1

4, 𝑃(𝐵2) =

2

4, 𝑃(𝐵3) =

3

4

Required Probability = P (selecting 1 girl and 2 boys)

= (I or II or III)

= P(I ∪ II ∪ III)

= 𝑃[(𝐺1 ∩ 𝐵2 ∩ 𝐵3) ∪ (𝐵1 ∩ 𝐺2 ∩ 𝐵3) ∪ (𝐵1 ∩ 𝐵2 ∩ 𝐺3)]

= 𝑃(𝐺1 ∩ 𝐵2 ∩ 𝐵3) + 𝑃(𝐵1 ∩ 𝐺2 ∩ 𝐵3) + 𝑃(𝐵1 ∩ 𝐵2 ∩ 𝐺3)

= 𝑃(𝐺1) 𝑃(𝐵2) 𝑃(𝐵3) + 𝑃((𝐵1)𝑃(𝐺2)𝑃(𝐵3) + 𝑃(𝐵1)𝑃(𝐵2)𝑃(𝐺3)

=3

4×

2

4×

3

4+

1

4×

2

4×

3

4+

1

4×

2

4×

1

4

=9

32+

3

32+

1

32=

13

32

S79. Ans.(c)

Sol.

1 and 3 only.

S80. Ans.(b)

Sol.

𝜎2 =1

𝑛Σ(𝑥𝑖 − 𝑥)2

Given that 𝜎2 = 4. 𝑎𝑛𝑑 𝑛 = 25

⇒ 4 = 1

25Σ(𝑥𝑖 − 𝑥)2 _________(1)

If 2 is added to each observation

Then the variance of 25 observation is

𝜎2 =1

25Σ(𝑥𝑖 + 2 − (𝑥 + 2))

2

=1

25Σ(𝑥𝑖 + 2 − 𝑥 − 2)2

1

25Σ(𝑥𝑖 − 𝑥)2 = 4 [form(1)]

S81. Ans.(a)

Sol.

Given that

P = (𝑥1. 𝑥2 … . 𝑥𝑛)1/𝑛

𝐿𝑜𝑔 𝑃 =1

𝑛 𝐿𝑜𝑔(𝑥1. 𝑥2, … . . 𝑥𝑛)

𝑛 𝐿𝑜𝑔 𝑃 = log 𝑥1 + log 𝑥2 + ⋯ + log 𝑥𝑛 _______(1)

Also,

Q = (𝑦1. 𝑦2. … … 𝑦𝑛)1/𝑛.

Log Q = 1

𝑛log(𝑦1. 𝑦2. … . . 𝑦𝑛)

𝑛 log 𝑄 = log 𝑦1 + log 𝑦2 + ⋯ + log 𝑦𝑛 ________(2)

Now,

G. M of 𝑥

4= (

𝑥1

𝑦1,

𝑥2

𝑦2. … .

𝑥𝑛

𝑦𝑛)

1/𝑛

Log (𝐺. 𝑀) =1

𝑛 𝐿𝑜𝑔 (

𝑥1

𝑦1.

𝑥2

𝑦2. … .

𝑥𝑛

𝑦𝑛).

n log (G. M) = 𝐿𝑜𝑔𝑥1

𝑦1+ log

𝑥2

𝑦2+ ⋯ + log

𝑥𝑛

𝑦𝑛

n log (G. M) = 𝐿𝑜𝑔 𝑥1 + 𝐿𝑜𝑔 𝑥2 + ⋯ + 𝐿𝑜𝑔 𝑥𝑛 − (log 𝑦1 + log 𝑦2 + ⋯ + log 𝑦𝑛).

𝑛 (𝐿𝑜𝑔 (𝐺. 𝑀) = 𝑛 𝐿𝑜𝑔 𝑃 − 𝑛 𝐿𝑜𝑔 𝑄

𝑛 log(𝐺. 𝑀. ) = 𝑛(𝐿𝑜𝑔 𝑝 − log 𝜃)

𝐿𝑜𝑔 𝐺. 𝑀 = 𝐿𝑜𝑔𝑃

𝑄

G. M = 𝑃

𝑄

S82. Ans.(d)

Sol.

I. Since P(exactly one of A, B occurs) = q (given), we get

𝑃(𝐴 ∪ 𝐵) − 𝑃(𝐴 ∩ 𝐵) = 𝑞

p – 𝑃(𝐴 ∩ 𝐵) = 𝑞

= 𝑃(𝐴 ∩ 𝐵) = 𝑝 − 𝑞

1 − 𝑃(𝐴 ∪ 𝐵) = 𝑝 − 𝑞

𝑃(𝐴 ∪ 𝐵) = 1 − 𝑝 + 𝑞

𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 1 − 𝑝 + 𝑞

𝑃(𝐴) + 𝑃(𝐵) = (1 − 𝑝 + 𝑞) + 𝑃(𝐴 ∩ 𝐵)

= (1 − 𝑝 + 𝑞) + (1 − 𝑃(𝐴 ∪ 𝐵))

= (1 − 𝑝 + 𝑞) + (1 − 𝑝)

= 2 − 2𝑝 + 𝑞

II. P(𝐴 ∩ 𝐵) = 1 − 𝑃(𝐴 ∪ 𝐵)

= 1 – p

S83. Ans.(b)

Sol.

Correlation coefficient = √𝑟𝑥𝑦 + 𝑟𝑦𝑥

−1

2= √𝑟𝑥𝑦 × −6

1

4= 𝑟𝑥𝑦 × −6

−1

24= 𝑟𝑥𝑦

S84. Ans (d) by definition.

S85. Ans.(c)

Sol.

No. of pairs of numbers from the set {0, 1, ---10}

{

(0,0), (0,1) … . . (0, 10)(1, 0), (1, 1) … . . (1, 10)

(10, 0) (10, 1) … (10, 10)

}

= 121

Favorable Outcomes = {(6, 0), (7, 0), (8, 0),(9, 0), (10, 0), (10, 1), (9, 1), (8, 1), (7, 1), (10, 2),

(9, 2), (8, 2), (10, 3), (9, 3), (10, 4), (0,6), (0, 7), (0, 8), (0, 9), (0, 10), (1, 10), (1, 9), (1, 8), (1, 7),

(2, 10), (2, 9), (2, 8), (3, 10), (3, 9), (4, 10)}

∴ Probability = 30

121

S86. Ans.(d)

Sol.

Average = 500×1860+600×1750

1100= 1800

Combined variance

=500(81+3600)+600(100+2500)

1100

=(5×3681)+(6×2600)

1100

≈ 3092

S87. Ans.(d)

Sol.

I II III

⑥, ⑤, ④, ③ ← 2 ← 1

⑥, ⑤, ④ ← 3

⑥, ⑤ ← 4

⑥ ← 5

I II III

⑥, ⑤, ④ ← 3 ←2

⑥, ⑤ ← 4

⑥ ← 5

I II III

⑥, ⑤ ← 4 ← 3

⑥ ←5

I II III

⑥ ← 5 ←4

Number of possible outcome =20

S88. Ans.(b)

S89. Ans.(b)

Sol.

For discrete series

σ = √1

𝑁Σ 𝑓𝑖(𝑥𝑖 − 𝑥)2

i.e. y = √𝑥

𝑦2 = 𝑥

⇒ 𝑥 ≥ 𝑦

S 90.Ans (c)

S91. Ans.(d)

Sol.

(𝑥1,1

𝑥1) , (𝑥2,

1

𝑥2) , (𝑥3,

1

𝑥3) are the vertices of the triangle.

Area of the triangle

=1

2|𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3(𝑦1 − 𝑦2)|

=1

2|𝑥1 (

1

𝑥2−

1

𝑥3) + 𝑥2 (

1

𝑥3−

1

𝑥1) + 𝑥3 (

1

𝑥1−

1

𝑥2)|

=1

2|𝑥1

(𝑥3−𝑥2)

𝑥2 𝑥3+

𝑥2(𝑥1−𝑥3)

𝑥1 𝑥3+

𝑥3(𝑥2−𝑥1)

𝑥1 𝑥2|

1

2|

𝑥12(𝑥3−𝑥2)+𝑥2

2(𝑥1−𝑥3)+𝑥32(𝑥2−𝑥1)

𝑥1𝑥2𝑥3|

=1

2|

𝑥12𝑥3−𝑥1

2𝑥2+𝑥1𝑥22−𝑥2

2𝑥3+𝑥2𝑥32−𝑥1𝑥3

2

𝑥1𝑥2𝑥3|

=1

2|

𝑥1𝑥2𝑥3+𝑥12𝑥3−𝑥1

2𝑥2+𝑥1𝑥22−𝑥2

2𝑥3+𝑥2𝑥32−𝑥1𝑥3

2−𝑥1𝑥2𝑥3

𝑥1𝑥2𝑥3|

=1

2|

(𝑥1𝑥2−𝑥1𝑥3−𝑥22+𝑥2𝑥3)(𝑥3−𝑥1)

𝑥1𝑥2𝑥3|

=1

2|

(𝑥1−𝑥2)(𝑥2−𝑥3)(𝑥3−𝑥1)

𝑥1𝑥2𝑥3|

S92. Ans.(a)

Sol.

Let the circle touches the Y-axis and has its centre C(h, k).

Then the equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = ℎ2

Or 𝑥2 + 𝑦2 − 2ℎ𝑥 − 2𝑘𝑦 + 𝑘2 = 0 _____(1)

Given equation is 𝑥2 + 𝑦2 + 𝑔𝑥 + 𝑓𝑦 +𝑐

4= 0 _____(2)

By comparison of (1) & (2), we get

–2h = g and –2k =f

h = −𝑔

2 k = –

𝑓

2

i.e. (−𝑔

2,

−𝑓

2)

S93. Ans.(d)

Sol.

(�⃗� + �⃗⃗�). (�⃗� + �⃗⃗�) = �⃗�. �⃗� + �⃗�. �⃗⃗� + �⃗⃗�. �⃗� + �⃗⃗� . �⃗⃗�

= |�⃗�|2 + �⃗�. �⃗⃗� + �⃗⃗�. �⃗� + |�⃗⃗�|2

We can conclude that (�⃗� + �⃗⃗�). (�⃗� + �⃗⃗�) = |�⃗�|2 + |�⃗⃗�|2

If and only if �⃗�. �⃗⃗� = −�⃗⃗�. �⃗�

i.e. �⃗� 𝑎𝑛𝑑 �⃗⃗� are anti parallel.

S94. Ans.(d)

Sol.

Given that 𝑟 = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧�̂�

Consider

�̂�. (𝑖̂ + 𝑗̂ + �̂�).

(𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧�̂�)(𝑖̂ + 𝑗̂ + �̂�)

= (𝑥 + 𝑦 + 𝑧) [

∵ 𝑖̂. 𝑖̂ = 1 𝑖̂. 𝑗̂ = 0

𝑗̂. 𝑗̂ = 1 𝑎𝑛𝑑 𝑗̂. �̂� = 0

�̂�. �̂� = 1 �̂�. 𝑖̂ = 0

]

S95. Ans.(a)

Sol.

Let unit vector �̂� = 𝑥𝑖̂+y𝑗̂ + 𝑧�̂�

Let A = 2𝑖̂ − 𝑗̂ + �̂�

B = 3𝑖̂ − 4𝑗̂ − �̂�

Given that �̂�. 𝐴 = 0

⇒ 2𝑥 − 𝑦 + 𝑧 = 0 _______(1)

And �̂�. 𝐵 = 0

⇒ 3x – 4y – z = 0 _________(2)

From (1) & (2), we get

x = y = – z.

Hence required unit vector

=𝑥

√𝑥2+𝑦2+𝑧2𝑖̂ +

𝑦

√𝑥2+𝑦2+𝑧2𝑗̂ −

𝑧

√𝑥2+𝑦2+𝑧2�̂�

=1

√3𝑖̂ +

1

√3𝑗̂ −

1

√3�̂�

S96. Ans.(d)

Sol.

Given that,

|�⃗� − �⃗⃗�| = 5

|�⃗� − �⃗⃗�|2

= 25

|�⃗�|2 + |�⃗⃗�|2

− 2(�⃗�. �⃗⃗�) = 25

9 + 16 − 2(�⃗�. �⃗⃗�) = 25

�⃗�. �⃗⃗� = 0 _____(1)

Consider

|�⃗� + �⃗⃗�|2

= |�⃗�|2 + |�⃗⃗�|2

+ 2(�⃗�. �⃗⃗�)

= 9 + 16 + 0

= 25

⇒ |�⃗� + �⃗⃗�| = 5

S97. Ans.(c)

Sol.

Given that �⃗�, �⃗⃗�, 𝑐 are three mutually perpendicular vectors.

∴ �⃗�. �⃗� = �⃗⃗�. �⃗⃗� = 𝑐. 𝑐 = 1

And �⃗�. �⃗⃗� = �⃗⃗�. 𝑐 = 𝑐. �⃗� = 0

Also given that each of unit magnitude

∴ |�⃗�| = 1 = |�⃗⃗�| = |𝑐|

|𝐴| = √�⃗�2 + �⃗⃗�2 + 𝑐2 = √3

|�⃗⃗�| = √�⃗�2 + (−�⃗⃗�)2

+(𝑐)2 = √3

|𝑐| = √�⃗�2 + (−�⃗⃗�)2

+ (−𝑐)2 = √3

Hence, |𝐴| = |�⃗⃗�| = |𝐶|

S98. Ans.(c)

Sol.

(�⃗� − �⃗⃗�) × (�⃗� + �⃗⃗�)

|𝑖̂ 𝑗̂ �̂�

𝑎1 − 𝑏1 𝑎2 − 𝑏2 𝑎3 − 𝑏3

𝑎1 + 𝑏1 𝑎2 + 𝑏2 𝑎3 + 𝑏3

|

= 𝑖̂[(𝑎2 − 𝑏2)(𝑎3 + 𝑏3) − (𝑎2 + 𝑏2)(𝑎3 − 𝑏3)] − 𝑗̂[(𝑎1 − 𝑏1)(𝑎3 + 𝑏3) − (𝑎1 + 𝑏1)(𝑎3 − 𝑏3)]

+�̂�[(𝑎1 − 𝑏1)(𝑎2 + 𝑏2) − (𝑎1 + 𝑏1)(𝑎2 − 𝑏2)]

= 𝑖̂[�⃗�2 𝑎⃗⃗⃗ ⃗3 + �⃗�2 �⃗⃗�3 − �⃗�3 �⃗⃗�2 − �⃗⃗�2 �⃗⃗�2 − �⃗�2 �⃗�3 + �⃗�2 �⃗⃗�3 − �⃗⃗�2 �⃗�3 + �⃗⃗�2 �⃗⃗�3]

−𝑗̂[�⃗�1�⃗�3 + �⃗�1�⃗⃗�3 − �⃗⃗�1 �⃗�3 − �⃗⃗�1 �⃗⃗�3 − �⃗�1 �⃗�3 + �⃗�1 �⃗⃗�3 − �⃗⃗�1 �⃗�3 + �⃗⃗�1 �⃗⃗�3]

+�̂�[�⃗�1�⃗�2 + �⃗�1 �⃗⃗�2 − �⃗⃗�1 �⃗�2 − �⃗⃗�1 �⃗⃗�2 − �⃗�1 �⃗�2 + �⃗�1 �⃗⃗�2 − �⃗⃗�1 �⃗�2 + �⃗⃗�1 �⃗⃗�2]

= 𝑖̂(2�⃗�2�⃗⃗�3 − 2�⃗⃗�2�⃗�3) − 𝑗̂(2�⃗�1 �⃗⃗�3 − 2�⃗⃗�1 �⃗�3) + �⃗⃗� (2�⃗�1 �⃗⃗�2 − 2�⃗⃗�1 �⃗�2)

= 2[𝑖(�⃗�2 �⃗⃗�3 − �⃗⃗�2�⃗�3) − 𝑗(�⃗�1�⃗⃗�3 − �⃗⃗�1�⃗�3) + �̂�(�⃗�1�⃗⃗�2 − �̂�1�̂�2)

= 2(�⃗� × �⃗⃗�)

S99. Ans.(b)

Sol.

Moment = r × F, where r be the position vector.

𝑀 = |𝑖̂ 𝑗̂ �̂�1 2 30 0 𝜆

|

= 𝑖(2𝜆) − 𝑗(𝜆)

|𝑀| = √(2𝜆)2 + (−𝜆)2 = √5 𝜆

S100. Ans.(a)

Sol.

𝐴𝐶⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗ ⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗⃗ (Triangle law)

−𝐶𝐴⃗⃗⃗⃗⃗⃗ = 𝐴𝐵⃗⃗⃗⃗ ⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗⃗

𝐴𝐵⃗⃗⃗⃗ ⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗⃗ + 𝐶𝐴⃗⃗⃗⃗⃗⃗ = �⃗⃗�

S101. Ans(b)

Sol.

𝑦 = cos−1(sin 𝑥) 𝑑𝑦

𝑑𝑥=

−1

√1−sin2 𝑥. 𝑐𝑜𝑠𝑥

=−1 .cos 𝑥

√cos2 𝑥

= −− cos 𝑥

cos 𝑥

= −1

Given that 𝑑𝑦

𝑑𝑥= 𝑡𝑎𝑛𝜃

⇒ tan 𝜃 = −1

⇒ 𝜃 =3𝜋

4

S102. Ans.(d)

Sol.

For 𝑓(𝑥) 𝑡𝑜 𝑏𝑒 𝑑𝑖𝑓𝑖𝑛𝑒𝑑,

𝑥 − 1 ≥ 0 ⇒ 𝑥 ≥ 1.

And 𝑥 − 4 > 0 ⇒ 𝑥 > 4

∴ 𝑥 𝜖 [1, 4) ∪ (4, ∞)

S103. Ans.(a)

Sol.

𝑓(𝑥) = {

𝑠𝑖𝑛2𝑥

5𝑥𝑖𝑓 𝑥 ≠ 0

2

15𝑖𝑓 𝑥 = 0

lim𝑥→0+

𝑓(𝑥) = lim𝑥→0+

sin 2𝑥

5𝑥

= limℎ→0

sin 2(0+ℎ)

5(0+ℎ)

= limℎ→0

sin 2 ℎ

5ℎ

= limℎ→0

2 cos 2 ℎ

5

=2

5

And 𝑓(0) =2

15

𝑓(0) ≠ lim𝑥→0+

𝑓(𝑥)

⇒ 𝑓(𝑥) is not continuous at x = 0

S104. Ans.(b)

Sol.

𝑓(𝑥) = |𝑥 − 3|

= {𝑥 − 3 𝑖𝑓 𝑥 ≥ 3

−(𝑥 − 3) 𝑖𝑓 𝑥 < 3

lim𝑥→3+

𝑓(𝑥)

= lim𝑥→3+

(𝑥 − 3)

= limℎ→0

(3 + ℎ − 3)

= limℎ→0

ℎ = 0

And

lim𝑥→3−

𝑓(𝑥)

lim𝑥→3−

−(𝑥 − 3)

limℎ→0

−(3 − ℎ − 3)

limℎ→0

ℎ

= 0

Hence, lim𝑥→3+

𝑓(𝑥) = lim𝑥→3−

𝑓(𝑥)

⇒ f is continuous at x = 3.

S105. Ans.(b)

Sol.

∵ f is continuous at each point in its domain

⇒ 𝑓(0) = lim𝑥→0

𝑓(𝑥)

= lim𝑥→0

2𝑥−sin−1 𝑥

2𝑥+tan−1 𝑥

= (0

0) form

Applying L’ Hospital rule

lim𝑥→0

2−1

√1−𝑥2

2+1

1+𝑥2

=1

3

S106. Ans.(a)

Sol. 𝑢 𝑑𝑢

𝑑𝑥+

𝑣 𝑑𝑣

𝑑𝑥

𝑒𝑎𝑥 sin 𝑏𝑥[𝑏. 𝑒𝑎𝑥 cos 𝑏𝑥 + sin 𝑏𝑥. 𝑎. 𝑒𝑎𝑥] + 𝑒𝑎𝑥 cos 𝑏𝑥[−𝑏𝑒𝑎𝑥 sin 𝑏𝑥 + 𝑎 cos 𝑏𝑥. 𝑒𝑎𝑥]

= 𝑏𝑒2𝑎𝑥 sin 𝑏𝑥 cos 𝑏𝑥 + 𝑎. 𝑒2𝑎𝑥 sin2 𝑏𝑥 − 𝑏𝑒2𝑎𝑥 sin 𝑏𝑥 𝑐𝑜𝑠𝑥 + 𝑎𝑒2𝑎𝑥 cos2 𝑥.

= 𝑎 𝑒2𝑎𝑥

S107. Ans.(c)

Sol.

𝑦 = sin (𝑙𝑛𝑥) 𝑑𝑦

𝑑𝑥=

cos (𝐿𝑛𝑥)

𝑥

𝑑2𝑦

𝑑𝑥2 =−𝑥 sin(𝐿𝑛𝑥).

1

𝑥−cos (𝐿𝑛𝑥)

𝑥2

=− sin(𝐿𝑛𝑥)−cos (𝐿𝑛𝑥)

𝑥2

𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟,

𝑥2 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑦

𝑥2 − (− sin(𝑙𝑛𝑥)−cos (𝑙𝑛𝑥)

𝑥2 ) +𝑥.cos (𝐿𝑛𝑥)

𝑥+ sin 𝐿𝑛𝑥

− sin(𝐿𝑛𝑥) − cos(𝐿𝑛𝑥) + cos(𝐿𝑛𝑥) + sin (𝐿𝑛𝑥)

= 0

S108. Ans.(c)

Sol.

Length of wire = arc length + r + r

40 = (𝜃

360× 2𝜋𝑟) + 2𝑟

40−2𝑟

2𝜋𝑟=

𝜃

360

20−𝑟

𝜋𝑟=

𝜃

360 ______(1)

Now the area of sector

A = 𝜃

360× 𝜋𝑟2

=(20−𝑟)

𝜋𝑟× 𝜋𝑟2[𝑓𝑟𝑜𝑚 (1)]

= (20 − 𝑟)𝑟

= 20 r – r²

Area is greatest

∴ A’ = 0

20 – 2r = 0

⇒ r = 10

S109. Ans.(a)

Sol.

𝑓(𝑥) = [𝑥(𝑥 − 1) + 1)1

3

𝑓′(𝑥) =1

3[𝑥(𝑥 − 1) + 1]−

2

3[𝑥 + 𝑥 − 1]

𝑓′(𝑥) =1

3

(2𝑥−1)

[𝑥(𝑥−1)+1]2/3

𝑃𝑢𝑡 𝑓′(𝑥) = 0

0 =1

3

(2𝑥−1)

(𝑥(𝑥−1)+1]23

⇒ 𝑥 =1

2

Put the value, x = 1

2 in f(x)

𝑓 (1

2) = [

1

2(

1

2− 1) + 1]

1/3

= [1

2×

−1

2+ 1]

1/3

= [−1

4+ 1]

1/3

= [3

4]

1/3

S110. Ans.(c)

Sol.

𝑦 = |𝑠𝑖𝑛𝑥||𝑥|

𝐿𝑛𝑦 = |𝑥|𝐿𝑛|𝑠𝑖𝑛𝑥| 1

𝑦

𝑑𝑦

𝑑𝑥= [|𝑥|

1

|𝑠𝑖𝑛𝑥||𝑐𝑜𝑠𝑥| + log | sin 𝑥|]

𝑑𝑦

𝑑𝑥= |𝑠𝑖𝑛𝑥||𝑥| [|𝑥|

1

|sinx|| cos 𝑥| + log | sin 𝑥|]

𝐴𝑡 𝑥 = −𝜋

6

𝑑𝑦

𝑑𝑥= |sin (

−𝜋

6)|

|−𝜋

6|

[|−𝜋

6| .

|cos(−𝜋

6)|

|sin(−𝜋

6)|

+ |𝐿𝑛 (−𝜋

6)|]

= |− sin𝜋

6|

𝜋

6[

𝜋

6

|cos𝜋

6|

|− sin𝜋

6|+ 𝐿𝑛 |− sin

𝜋

6|]

= |−1

2|

𝜋

6[

𝜋

6.

√3

21

2

− 𝐿𝑛 (1

2)]

= (1

2)

𝜋

6[

𝜋√3

6+ 𝐿𝑛(2)]

(2)−𝜋/6 [√3𝜋+6𝐿𝑛(2)

6]

S111. Ans.(b)

Sol. 𝑑(√1−𝑠𝑖𝑛2𝑥)

𝑑𝑥

=𝑑

𝑑𝑥(√𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 − 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥)

=𝑑

𝑑𝑥√(cos 𝑥 − 𝑠𝑖𝑛𝑥)2

=𝑑

𝑑𝑥(cos x-sinx)

=-sinx -cosx = - (cos x+sinx)

S112. Ans.(a)

Sol.

= ∫𝑑𝑥

𝑎2 sin2 𝑥+𝑏2 cos2 𝑥

= ∫sec2 𝑥

𝑎2 tan2 𝑥+𝑏2 𝑑𝑥

Putting tan x = t and sec2 𝑥 𝑑𝑥 = 𝑑𝑡, we get

I = ∫𝑑𝑡

𝑎2𝑡2+𝑏2 =1

𝑎2 ∫𝑑𝑡

𝑡2+(𝑏

𝑎)

2 =1

𝑎2 ×1

𝑏/𝑎tan−1 (

𝑡

𝑏/𝑎) + 𝐶

=1

𝑎𝑏tan−1 (

𝑎𝑡

𝑏) + 𝐶 =

1

𝑎𝑏tan−1 (

𝑎 𝑡𝑎𝑛𝑥

𝑏) + 𝐶

S113. Ans.(d)

Sol.

𝑓′(𝑥) = limℎ→0

[𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ]

= limℎ→0

[𝑓(𝑥)𝑓(ℎ)−𝑓(𝑥)

ℎ] [∵ 𝑓 (𝑥 + 𝑦) = 𝑓(𝑥)𝑓(𝑦)]

= limℎ→0

𝑓(𝑥) [𝑓(ℎ)−1

ℎ]

= 𝑓(𝑥) limℎ→0

[𝑓(ℎ)−1

ℎ]

= 𝑓(𝑥) limℎ→0

[1+ℎ 𝑔 (ℎ) 𝜙 (ℎ)−1

ℎ]

= 𝑓(𝑥) limℎ→0

𝑔(ℎ) 𝜙 (ℎ)

= 𝑓(𝑥) (limℎ→0

𝑔(ℎ)) (limℎ→0

𝜙(ℎ))

= 𝑓(𝑥). 𝑎. 𝑏

= 𝑎𝑏 𝑓(𝑥)

S114. Ans.(c)

Sol. 𝑑𝑥

𝑑𝑦=

𝑥+𝑦+1

𝑥+𝑦−1

⇒𝑑𝑦

𝑑𝑥=

𝑥+𝑦−1

𝑥+𝑦+1

Let x + y = v and 𝑑𝑦

𝑑𝑥=

𝑑𝑣

𝑑𝑥− 1

∴𝑑𝑣

𝑑𝑥− 1 =

𝑣−1

𝑣+1⇒

𝑑𝑣

𝑑𝑥=

𝑣−1+𝑣+1

𝑣+1

⇒ 𝑣+1

2𝑣𝑑𝑣 = 𝑑𝑥 ⇒

1

2∫ 1 𝑑𝑣 +

1

2∫

1

𝑣𝑑𝑣 = ∫ 1 𝑑𝑥

⇒1

2𝑣 +

1

2log 𝑣 = 𝑥 + 𝑐1

⇒ 𝑥 + 𝑦 + log(𝑥 + 𝑦) = 2𝑥 + 𝑐 [∵ 𝑐 = 2𝑐1]

∴ (𝑦 − 𝑥) + log(𝑥 + 𝑦) = 𝑐

S115. Ans.(d)

Sol.

lim𝑥→

𝜋

6

2 sin2 𝑥+𝑠𝑖𝑛𝑥−1

2 sin2 𝑥−3𝑠𝑖𝑛𝑥+1

= (0

0) 𝑓𝑜𝑟𝑚

Applying L’ Hospital Rule

lim𝑥→

𝜋

6

4 sin 𝑥 cos 𝑥+cos 𝑥

4 sin 𝑥 cos 𝑥−3 cos 𝑥

= lim𝑥→

𝜋

6

4 sin 𝑥+1

4 sin 𝑥−3

=4×

1

2+1

4×1

2−3

=2+1

2−3= −3.

S116. Ans.(c)

Sol.

Sample space = {(1, 5), (2, 5), 3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}

Favorable outcome = {(5, 5), (5, 6), (6, 5)}

∴ Probability = 3

11

S117.Ans(b)

S118. Ans.(d)

Sol.

Given that average of group of women = 21 years

⇒𝑥1+⋯+𝑥𝑛

𝑛= 21

𝑥1 + ⋯ + 𝑥𝑛 = 21𝑛 ________(1)

And average of group of men = 26 years. 𝑦1+⋯+𝑦𝑚

𝑚= 26

𝑦1 + ⋯ + 𝑦𝑚 = 26𝑚 ________(2)

and average of combined group = 25 𝑥1+⋯+𝑥𝑛+𝑦1+⋯+𝑦𝑛

𝑚+𝑛= 25

𝑥1 + ⋯ + 𝑥𝑛 + 𝑦1 + ⋯ 𝑦𝑛 = 25(𝑚 + 𝑛)

21n + 26m = 25 (m + n) [from (1) & (2)]

⇒ 𝑚 = 4𝑛

Percentage of men in group = 𝑚

𝑚+𝑛× 100

=4𝑛

4𝑛+𝑛× 100

=4𝑛

5𝑛× 100

= 80%

∴ women = 20%

S119. Ans.(c)

Sol.

Given that sinβ is the harmonic mean of sinα and cosα.

⇒ 1

𝑠𝑖𝑛𝛽 is the arithmetic mean of

1

𝑠𝑖𝑛𝛼 and

1

𝑐𝑜𝑠𝛼

⇒ 1

𝑠𝑖𝑛𝛽=

1

𝑠𝑖𝑛𝛼+

1

𝑐𝑜𝑠𝛼

2

1

sin 𝛽=

cos 𝛼+sin 𝛼

2 sin 𝛼 cos 𝛼

sin 2𝛼

sin 𝛽= cos 𝛼 + sin 𝛼 ______(1)

Consider

I. L.H.S √2 sin (𝛼 +𝜋

4) sin 𝛽

√2 (sin 𝛼 cos𝜋

4+ cos 𝛼 sin

𝜋

4) sin 𝛽

√2 (sin 𝛼1

√2+ cos 𝛼

1

√2) sin 𝛽

(𝑠𝑖𝑛𝛼 + 𝑐𝑜𝑠𝛼) sin 𝛽 sin 2𝛼

sin 𝛽sin 𝛽 [𝑓𝑟𝑜𝑚 (1)]

= sin 2 α

= R. H. S.

∵ Given that sin θ is the arithmetic mean of sin α and cos α.

∴ sin θ = sin 𝛼+cos 𝛼

2 ______(2)

Consider,

II. R.H.S

cos (𝛼 −𝜋

4)

= cos 𝛼 cos𝜋

4+ sin 𝛼 sin

𝜋

4

= cos 𝛼1

√2+ sin 𝛼

1

√2

=1

√2(cos 𝛼 + sin 𝛼)

=2

√2(

cos 𝛼+sin 𝛼

2)

= √2 sin 𝜃

= L.H.S.

S120. Ans.(b)

Sol.

Given that

= 𝑃(𝐵) = 1.5 𝑃(𝐴) 𝑃(𝐵)

𝑃(𝐴)= 1.5 =

3

2

𝑃(𝑐) = 0.5 𝑃(𝐵) 𝑃(𝐶)

𝑃(𝐵)=

1

2

𝑃(𝐴): 𝑃(𝐵): 𝑃(𝐶) = 4 ∶ 6 ∶ 3

∴ P(A) = 4

4+6+3=

4

13