cbse cce sample question paper...cosec2 0 + sec2 0 (a) - 1 (b) 1 solution. choice (c) is correct....

CBSE CCE SAMPLE QUESTION PAPER

FIRST TERM (SA-I) MATHElMATICS

(Wzth Solutions)

I CLASS X Ttme Allowed : 3 to 3% Hours1 lMaximu,n Marks : 80

General Instructions : ( i ) All questions are compulsory.

(ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 10 questions o f1 mark each, Section B comprises of 8 questions of 2 marks each, Sectioi C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each.

(iii) Question numbers 1 to 10 in Section A are multiple choice questions where you are to select one correct option out of the given four:

( i v ) There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.

( u ) Use of calculators is not permitted.

Question numbers 1 to 10 are of oiie mark each. 1. Euclid's Division Lemma states that for any two positive integers a and b,

there exist unique integers q and r such that a = bq + r, where r must satisfy. ( a ) l < r < b (b) O < r < b (c ) O < r < b ( d ) O < r < b

Solution. Choice (c) i s correct. 2. In figure, the graph of a polynomial p ( x ) is shown. The number of zeroes of

p ( x ) is Y

Y' ( a ) 4 (b) 1 (c ) 2 ( d ) 3

Solution. Choice (b) i s correct1 T h e number of zeroes i s 1 as the graph intersects the z-axis a t one point in the given figure.

( 1 )

trainer

Rectangle

3. In figure, if DE 11 BC, then x equals

(a) 6 cm (b) 8 cm (c) 10 cm . ( d ) 12.5 cm

Solution. Choice (c) is correct. In figure, since DE 11 BC, therefore

L D = L B [Corresponding Lsl In AADE and AABC, we have

L D = L B proved above1 . L A = A [Common]

So, by A criterion of similarity, we have AADE - AABC

* AD DE -= - [Corresponding sides of similar triangles are proportional] AB BC

I * 2 4 - = - [. : AB =AD +-QB = ( 2 + 3 ) cm = 5 c ~ I 5 x

- . * x = 10 cm. 4. 1f sin 30 = cos (0 - 6 9 , where (30) and (9 - 6") are both acute angles, then the

value of 0 is ( a ) 18" (b) 24" (c) 36' ( d ) 30"

Solution. Choice (b) is correct. Given, sin 30 = cos (0 - 6")

cos (90" - 30) = cos (0 - 6") [.; cos (90" -A) = sin A1 ==) 90"-30=016" a 96" = 40 =$ 0=96 '+4 * 0 = 24" -

2 2 cosec 0 - sec 0 5. Given that tan 0 = is

cosec2 0 + sec2 0

( a ) - 1 (b) 1

Solution. Choice (c) is correct. 1 Given, tan 0 = - , then J5

CBSE CCE Sample Question Paper 3

cosecz e - see" cosecz e + seez e

= (9 - 1)13 ( 6 + 9 + 1)13

- 813 -- 1613 .

- 8 - - 16 1 - - - 2

6. In figure, AD - 4 cm, BD - 3 6m and CB - 12 cm, then cot I3 equals

' . 'AD' c B . m 3

(a) - 5 4

(b) 12 4

(c) .- 12

3 (d) 7

Solution. Choice (d) is correct. In right AADB, ABZ AD^ t BD' a A B ~ = (4Y t (312 ==+ ~ ~ = 1 6 + 9 = 2 5 = ( 5 ) ~

CB In MCB, cot 8 = - AB

14' will terminate after how many places. of 7. The decimal expansion of - 120

decimal ? ( a ) 1 (b ) 2 (c) 3 (d ) wilhnot terminate

Solution. Choice (c) is correct.

147 It shows that aRer 3 places of decimal, decimal expansion - will terminate. 120

8. The pair of linear equations 32 + 2y = 5; 22 - 3y = 7 have (a ) one solution (b) two solutions (c) many solutions ( d ) no solution

Solution. Choice (a ) is. correct. Here, a l = 3 , b l = 2 , c l = - 5

a z = 2 , b Z = - 3 , c z = - 7

Clearly, al h G f % The given pair of lines have one solution.

9. If sec A = cosec B, then A + B is equal to ( a ) Zero .(b) 90" (c) < 90' ( d ) z 90"

Solution. Choice (b ) is correct. Given, sec A = cbsec B * sec A = cosec B

cosec (90" - A ) = cosec B 3 90"-A= B 3 A + B = 9 0 °

[.; cosec (90" - 0) = sec 81


10. For a given data with 70 observations the 'less than ogive' and the 'more than ogive' intersect at (20.5,35). The median of the data is

(a) 20 (b) 35 (c) 70 (dl 20.5

Solution. Choice (d) is correct. The x-coordinate of the point of intersection of the 'less than ogive' and the 'more than

ogive' is the median. Since the given intersection point is (20.5, 351, therefore the median is 20.5.

-1 Question numbers 11 to 18 carry 2 marks each. 11. Is 7 x 5 x 3 x 2 + 3 a composite number ? Justify your answer. Solution. We have 7 x 5 x 3 x 2 + 3

= 7 x 5 x 2 x 3 + 1 x 3 [Writing 3 as 1 x 31 = ( 7 ~ 5 x Z + l ) x 3 [Taking 3 as common1 = ( 7 0 + 1 ) x 3 = 7 1 x 3 = 213

Fundamental Theorem of Arithmetic states that "Every composite number can be ex- pressed (factorised) as a product of primes and their factorisation is unique, apart from the order in which the prime factors occur".

So, 7 x 5 x 3 x 2 + 3 = 71 x 3 is a composite number. 12. Can (x - 2) be the remainder on division of a polynomial p(x) by (22 + 3) ?

Justify your answer. Solution. No, (x - 2) cannot be the remainder on division of given polynomidp(x) by

(2% + 3) as the degree of remainder pblynomid and quotient polynomial is same. 13. In figure, AECD is a rectangle. Find the values of x and y.

D-x+Y-C .:-x-;rii . ,

T A-12-B

Solution. Since ABCD is a rectangle, therefore AB =DC and AD = BC

(opposite sides of a rectangle) (opposite sides of a rectangle) =1 1 2 = x + y ... (1) and x -y = 8 . . $2) Solving these equations by adding and subtracting the given equations, i.e., we have

(x+y)+(x-y)=12+8 (x+y)-(x-y).=12-8 22 = 20 2y = 4

1 14. If 7 sinz 8 + 3 cos2 0 = 4, show that tan 8 = -. J5 Solution. Given :

7 ~ i n ~ e + 3 ~ 0 ~ ~ e s 4

sin2 0 c0s2 8 =, 7- 4 +3T=-

cos2 0 cos 8 cos2 0

a 7 t a g 0 + 3 = 4 sec2 0 j 7tan2e+3=4( i+ tan2e) a 7 t a n 2 8 + 3 = 4 + 4 t a n 2 0 * 7tanz0-4tan20=4-3 * 3 t a n 2 0 = l

15 (2 + 2 sin 0)(1- sin 0) If cot 8 = -, evaluate

8 (I+ cos 8)(2 - 2cos 0) ' Solution. We have

(2 + 2 sin 0)(1- sin 8) (1 + cos 0)(2 - 2 cos 8)

- - [2(1+ sin 8)1(1- sin 0) (1 + cos 0)[2(1- cos €01

- - 2(1- sin2 0) 2(1- cos2 0)

[Dividing both sides by cos2 01

[.; 8 lies in 1st quadrant]

15 . [cot 8 = ("ven) I


FE EC , 15. In figure, DE 11 AC and DF 11 AE. Prove that - = - BF BE'

'

Solution. In ABCA, DE 11 AC BE BD . . - = - EC DA

In ABEA, DF 11 AE

. . BF BD - FE DA

From (1) and (2), we conclude that

1 16. In figure, AD I BC and BD - -CD. Prove that

3

1 BD = -CD 3

=j 3BD = CD a 3BD+BD=CD+BD =+ 4BD = BC In right AAE3D;right-angled at D, we have

A B ~ AD^ + B D ~ - M B 2 = 2AD2 + ~BD' --

In right AACD, right-angled at D, we have C A ~ = + C D ~

==j = 2AD2 + ~CD'

.. .(l)'By Thales Theorem]

. . .(2) m y Thales Theoreml

[Adding BD to both sides1

[Pythagoras Theoreml . . . (1)

[Pythagoras Theoreml ... (2).

Subtracting (1) from (21, we get ~CA' - 2AB2 = (2110' + ~CD') - ( 2 . 4 ~ ~ + 2BD2)

- 3 2CA2 - 2AB2 = 2 [ 9 B ~ ~ - BD2] a 2CA2 - 2AB2 = 2(8BD2)

2CA2 - ~ A B ~ = 16BD2 a 2CA2 - 2AB2 = ( ~ B D ) ~

2CA2 - 2AB2 = BC2 [.; BC = 4BD (proved above)] a ~ C A ~ = 2AB2 + BC'. 17. The following distribution gives the dailv income of 50 workers of a factory : - - Daily income (in 7) 1 100 - 120 1 120 - 140 1 140 - 160 1 160 - 180 1 180 - 200

Solution. Since the maximum number of students is 32, therefore, the modal class is 30 - 40. Thus, the lower limit ( I ) of the modal class = 30.

. . fi = 32, fo = 12, f2 = 20, h = 10. Using the formula :

Number of Workers 1 12

18. Find the mode of the following distribution of marks obtained by 80 students :

Marks obtained 1 0-10 / 10-20 1 20-30 1 30-40 1 40-50

14

Number of students I 6

Write the above distribution as less than type cumulative frequency distribution. Solution. We calculate 'less than type" cumulative frequency distribution by adding

frequencies from top to bottom. Less Than TyGe Cumulative Frequency Distribution

32

8

20 10

Cumulative Frequency

12 26 34 40 50

Daily income (in 7 )

100 - 120 120 - 140 140 - 160 160 - 180 180 - 200

12

6 10

Number of Workers

, 12 14 8 6

10

Daily income Less than

120 140 160 180 200

CBSE CCE Sample Question Paper ~ 9

25 =.30 + - 4

= 30 + 6.25 .

= 36.25 So, the maximum number of students have 36.25 marks.

~uestion numbers 19 to 28 carry 3 marks each. 19. Show that any positive odd integer is of the form 49 + 1 or 4q + 3, where q is a

positive integer. Solution. Let a be anypositive integer, when a is divided by 4. Then, by Euclid's division

lemna, we get a = 4q + r, 0 5 r c 4, where q and r are whole numbers.

So, r = O , 1 , 2 , 3 . When r = 0 or 2, then a becomes an even integer, i.e., a = 49 or 4q + 2. When r - 1 or 3, then a becomes positive odd integer, i.e., a = 4q + 1 or 4q + 3. Hence any positive odd integer is of the form 4 i ~ + 1 or 4q + 3, where q is a positive integer.

I 2& . .

20. Prove that - is irrational. 5

I 2 h . Solution. Let us assume, to the contrary, that - is rational. 5

2h - , wherep and q are coprime, i.e., their ~ C F is 1 5 q

=> & = 5 p 2q

5 ~ . Since is a rational number, therefore, -- 1s also rational number. q 2 q

=> h is rational number

But this contradicts the fact that h is irrational.

I 2 h . This contradiction has arisen because of our incorrect assumption that - is rationd.

5

2h is irrational. So, we conclude that - 5

or Prove that (5 - &) is irrational.

Solution. Let us assume, to the contrary, that (5 - A) is rational, i.e., we can find c o p d e a and b (b * 0 ) such that

5 - A = E b

a * 5 - - = A b

I Since a.and b are integers, we get 5 - a is rational. . b

+ is rational

But this contradicts the fact that & is irrational.

This contradiction has arisen because of our incorrect assumption that 5 - f i is rational.

So, we conclude that ( 5 - A) is irrational. 21. Aperson rowing a boat at the rate of 5 kmhour in still water, takes thrice as

much time in going 40 inn upstream as in going 40 km downstream. Find the speed of the stream.

Solution. Let the speed of the stream be x kmh and the speed of the boat in still water be 5 kmh

. Relative speed of the boat downstream = (5 + x) kmlh and relative speed of the boat upstream = (5 - x) km/h

I . Time taken by the boat downstream = - 40 h 5 + x

I and time taken by the boat upstream = .% h 5 - x

According to the given condition, a boat takes thrice as much time in going 40 krn upstream as in going 40 k m downstream, we have

40 - = 3[40) . 5 - x 5 + x

==. 1 . - 3 5 - x 5 + x

==. 5 + x = 1 5 - 3 x + 3 x + x = 1 5 - 5 ==t. 42 = 10

x = 2.5 Hence the speed of the stream is 2.5 km/h.

Or In a competitive examination, one mark is awarded for each correct answer

1 while - mark is deducted for each wrong answer. Jayanti answered 120 questions

2 - and got 90 marks. How many questions did she answer correctly ?

Solution. Let the number of correctly answered questions be x and the number of wrongly answered questions bey.

It is given that "the total number questions answered is 120". . . x + y = 1 2 0 ... ( 1 )

1 Also, it is given that "one mark is awarded for each correct answer and - mark is deducted

2 for each wrong answer and Jayanti has scored 90 marks".

CBSE CCE SampleQuestion Paper 11

. . 1 1.x-- .y=90

2 + 2 2 - y = 1 8 0 ... (2) Adding (1) and (21, we get

( x + y ) + ( & - y ) = 1 2 0 + 180 - a 32 = 300 * x = 100 Substituting x = 100 in (11, we get : y 120 - x = 120 100 = 20. Hence, the number of correctly answered questions by Jayanti are 100. Alternative Method : Total number of questions = 120. Let the number of correctly answered questions be x , then the number of wrongly answered

questions = 120 L x .

I 1 It is given that "one mark is awarded for each correct answer and - mark is deducted for , 2

l each wrong answer". Thus, marks obtained for correctly answered question = 1.x = x.

I 1 and, marks obtained for wrongly answered question = 4 1 2 0 - x )

2

. . 1 Total marks obtained = x - 4 1 2 0 - x)

2 But the total marks obtained by Jayanti = 90 (given)

- Hence, the number of correctly answered questions by Jayanti are 100. 22. If a, 8 are zeroes of the polynomial 2 - % - 15, then form a quadratic

polynomial whose zeroes are (2a) and (28). Solution. Since a and P are zeroes of the polynomial, x2 - 22 - 15, therefore

Let S and P denote respectively the sum and product of zeroes of the required polynomial, then

S = 2 a + 28 = 2(a + $1 = 2(2) = 4

12 U-Like Mathematics-X

and P = (2n)(2P) = 4(ap) = 4(- 15) = - 60 Hence, the required polynomial p(x) is given by

p(x) = (x2 - SX + P) * p (x )= .%-42 -6~ . .

1 , 23. Prove that (cosec 0 - sin g)(sed 0 - cos 0).- 1 t a n 0 + c o t e '

Solution. We have L.H.S. = (cosec 8 -sin 0)(sec - cos 8)

cosZ 8 sin2 8 =-.- sin 8 cos 0

- - sin e cos 0 1

- - sin 0 cos 0 sin2 0 + c0sz0

= sin 8 cos 0lsin 0 cos.0

[Dividing numerator and denominator by.sin 0 cos 81 sin2 e cos2 0 +

sinecose S ~ ~ O C O S O

- - 1 sin 0 cos 0 . -+- cos 0 sin 0

- - 1 tane+cote

= R.H.S. 24. If cos 0 + sin 0 - & cos 0, show that cos 0 - sin 0 = & sin 0.

Solution. Given, cos 8 + sin 8 = ficos 8 a sin 8 = & cos 0- cos e =) sin e = (& - 1) cos o

sin 0 - = cos e .l- 1

* (JZ.+ 1) sine , = cos 0

(JZ - I)(& + 1)


* sin 0 + sin 0 = cos 0 a cose-sine= sin0 25. In figure, AB I BC, FG I BC and DE I AC. Prove that AADE - AGCF.

A

Solution.. Since AB I BC and GF I BC, thereforem 11 GF. . . L1= L3 [Corresponding Lsl ... (1) In M E , we have A

L1+ L2 = 90° L3+L2=90° ... (2) [using (I)]

In AGCF, we have L3 + L4 = 90' ... (3)

From (2) and (3), we have

In M E and AGCF, we have L1= L3 [Proved in (111

and ~2 = L4 [Proved in (411 So, by AA smilarity of criterion, we have

AADE - AGCF Alternative Method : Since AB I BC and GF I BC, therefore AB 11 GF . . L1= L3 ... (1) [Corresponding Lsl In M E and AGCF, we have

L1= L3 Proved in (1)l and LAED = LGFC [Each = 907 So, b y AA similarity criterion, we have

AADE - AGCF 0

26. AABC and hDBC are on the same base BC and on opposite sides of BC and 0 is the point of intersection of AD and BC.

Prove that area(AABC) A 0 =-

area (ADBC) DO '


Solution. In the figure AABC and ADBC are on t h e same base BC, AD and BC intersect a t 0. Draw AE I BC and DF I BC.

In h4EO and ADFO, we have LAEO = LDFO [Each= 90°1 B C

and LAOE=LDOF [Vertically opposite Lsl So, by AA similarity criterion, we have

AAEO - ADFO

AE A0 - = - DF DO ... (1) [In similar triangles corresponding % sides are proportional]

1 - x B C x A E

Now, area (AABC) - 2 area (ADBC) - 1 BC DF

2

[using (111 -

Hence ~roved. A

27. Find mean of the following frequency distribution, using step-deviation method :

Classes 1 0-10 1 10-20 1 20-30 1 30-40 1 40-50 Frequency 7 12 13 10 8

Solution. Let the assumed mean be a = 25 and h = 10. Calculation of Mean

Classes Frequency Class-mark x, - 25 cf,)

d, = x, - 25 U , = - (r,) 10 f, u'

0-10 7 5 - 20 - 2 - 14 10 - 20 12 15 - 10 - 1 - 12 20 - 30 13 25 0 0 0 30 - 40 10 35 10 1 10 40 - 50 8 45 20 2 16 Total - n =Xf, = 50

*

X&uz = 0

From the table, n = Zfi = 50, Zfiui = 0. '

Using the formula :

CBSE CCE Sample Question Paper . 15

Or The mean of the following fi-equency 'distribution is 25. Find the value ofp.

Classes 1 0 -10 1 10-20 1 20-30 1 30-40 1.40-50

From the table, n = Zfi = 13 +p , .Zfixi = 285 + 45p Using the formula :

20 - 30 30 - 40 40 - 50 Total

Zfifixi Mean = - Zf i

P

(given) 25 = 285 + 45p

13 + p

Solution. Calculation of Mean I

3

5 3 P

n=Z&=13+p'

* 25(13 + p ) = 285 + 45p =? 325 + 25p = 285 + 45p a 45p - 25p = 325 - 285 * 20p = 40 * P - 2 28: Find the median of the following data :

5

Classes 1 0-10 ~10-20~20-30)30-40)40-50/50-60~60-70~70-80)80-90/90-100 Frequency 1 5 3 4 3 3 4 7 1 9 1 7 8

3 Frequency

25 35 45

2

125 105 4 5 ~

Zf?, = 285 + 45p

Solution. - Calculation of Median

Cumulative Freqeuncy (c f ) 5 8

12 15

Classes 0 -10

10 - 20 20 - 30 30 - 40

Frequency (f) 5 3 4 3


n 53 Here, n = 53 and - = - = 26.5, .

2 2 n

NOW, 66- 70 is the class whose cumulative frequency is 29 is greater than = 26.5.

:. 60 - 70 is the median class. From the table, f = 7 , cf = 22, h = 10 Using the formula :

I Question numbers 29 to 34 carry 4 marks each. 29. Find other zeroes of the polynomialp(z) - 22" + 7 2 - 1922 - 14s + 30 if two of its

I zeroes are & and -&. Solution. Since two zeroes are f i and - f i , therefore ( x - &)(x + &) = (x2 - 2) is a

factor of the given polynomial. - Now, we divide the given polynomial by (2 - 2).

2x2 + 7x - 15

x2-2 2x4+7x3-19x2-14x+30 , - 224 -4.9 ,

2~~ + First term of quotient is - - 2x

7x3 - 15x2 - 14x + 30 x2 -

- 7x3 - 14x + Second term of quotient is '" =

7.1 -15x2 + 30 x -15x2 + 30 + - Third term of quotient is = -

0 x

so, 2x4 + 7x3 - 19x2 - 14x + 30 = (x2 - 2)(2x2 + 72 - 15)

= ( x - f i ) ( x + h ) [ 2 x 2 + lox-3x-151


3 So, the zeroes of 2x2 + 72 - 15 = (x + 5)(2x - 3) are.given by - 5 and -. 2 3 Hence all the zeroes of the given polynomial are f i, - f i, - 5 and -. 2

30. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Solution. Given : AABC and APQR such that AABC - APQR.

To prove : ar(AABC) AB2 BC' CA' =-- -=- ar (APQR) P Q ~ - Q R ~ RP'

Construction : Draw AD I BC and PS I QR. P

I -xBCxAD ar (.AABC) 2 Proof :, -

ar (PQR) - 1 QR BS ,. i:

%. ar (AABC) BC x AD - - ar(APQR) QRx PS

Now, in AADB and @SQ, we have

1 [Area of A = -(base) 2 x height]

LB=LQ [AS AABC - APQR] LADB = LPSQ [Each = 9Oo1

3rd LBAD = 3rd LQPS Thus, AADB and P S Q are equiangular and hence, they are similar.

AD AB Consequently - = - PS PQ

[If A's are similar, the ratio of their corresponding sides is same]

But

Now, from (1) and (31, we get

[.: AABC - APQR]

. . .(3) [using (211


As M C - APQR, therefore

Hence, ar(h4BC) - AB' B C ~ CA' --=-=- ~ ~ ( A P Q R ) PQ' . QR' w2 [From (4) and (511

Or Prove that in a triangle, if the square of one side is equal to the sum of the

squares of the other two sides, then the angle opposite to the first side is a right angle.

Solution. Given : A triangle ABC such that : AC2 = AB2 + B C ~

To prove : AABC is a right-angled at B, i.e., L B = 90". Construction : Construct a APQR such that LQ = 90' and PQ = AB and QR = BC.

Proof : In APQR, as LQ = 90°, we have P R ~ = P Q ~ + Q R ~ [By Pythagoras Theorem1 P R ~ = A B ~ + I)c2 ( 1 ) [AsPQ=ABandQR=BCI Ac2 = AB2 + B C ~ ... (2)

From (1) and (2), we have P R ~ = A C ~

=+ PR = AC ... (3) Now in AABC and APQR, we have

AB=PQ BC = QR

and AC = PR [using (311 . . AABC s APQR [SSS congruency] * . L B = L Q = 9 0 ° [CPCTI Hence, L B - 90". . 31. Wove that

s e c e + t a n e - 1 cose - - tane:sece+l 1-s in8

Solution. We have secd+tane-1

L.H.S. = tane-sece+1


- - sec8+tan0.- 1 2 tan 8 - sec 8 + (sec2 0 - tan 8)

- - secO+tanO-1 (tan 8 - sec 8) + (sec 8 - tan 8)(sec 8 + tan 8)

- - sec8+tan8-1 -(set 8 - tan 0) + (sec 8 - tan 0)(sec 8 + tan 8)

sec8+tan8-1 - - (set 0 - tan €0- 1'+ sec 0 + tan 01

- 1 - 1 sin 8

cos 8 cos!

- - cos 8 1-sine

= R.H.S. Or

sec 0 cosec (90" - 8) - tan 0 cot (90" - 0) + sin2 55" + sin2 35" Evaluate :

tan 10" tan 20" tan 60" tan 70" tan 80" Solution. We have

sec 0 cosec (90" - 0) - tan 8 cot (90" - 8) + sin2 55" + sin2 35" tan 10" tan 20" tan 60" tan 70" tan 80"

- - see 8 sec 8 - tan 8 tan 8 + sin2 (90" - 35") + sin2 35" tan 10" tab 20" tan 60" tan (90" - 20") tan (90" - 10")

['.' cosec (90" - 8) = sec 0, cot (90" - 0) =tan 81

= see2 0 - tan2 0 + ~os"5~ + sin2 35". p,. sin (900 - 8) = cos 0, tan (90" - 0) = cot 01 tan 10" tan 20" (&I cot 20" cot 10"

- - 1+ 1 (tan 10". cot lO0)(tan 20". cot 20°)(&)

[.: sec2 0 -tad 8 = 1, sin2 0 + cos2 8 = 1, tan (90" >0) = cot 01

[:. tan 0 .,cot9 = 11

p2-1 32. If see 0 + tan 0 = p, prove that sin 0 = -

p 2 + 1 ' Solution. We have

p2 - 1 (see 0 + t a n 0)' - 1 -= p2 + 1 (sec e + tan el2 + 1

- - (sec2 0 - 1) + tan2 0 + 2 sec 0 tan 0. sec2 0 + ( I + tan2 0) + 2 sec 0 tan 0

- - tan2 0 + t a n 2 e + 2 s e c 0 t a n 0 [.: sec 2 0 - 1 = tan2 0 and 1 + tanz 0 = sec2 01 sec2 0 + set" +'2 sec 0 tan 0

- - 2tan2 0 + 2 s e c 0 t a n 0 2 sec2 0 + 2 sec 0 tan 0

2 tan @(tan 0 + sec 0) - - 2 sec 0(sec 0 + tan 0)

tan 0 -- - set 0

- - sin 0 cos 0.sec 0

= sig 0. 33. Draw th6 graphs of following equations :

2 2 - y = l , x + 2 y - 1 3 (i) Find the solution of the equation from the graph.

(ii) Shade the triangular region formed by the lines and the y-axis. '

Splution. We have & l y = 1 1 and x +2y = 13 ... (2)

13 - x * y = 2 x - 1 * y = - 2

13 - x Table of y = 2c - 1 Table of y = -

2

Plot the pointsA(0, - l), B -, 0 , C 0, - , D(13,O) and E(3,5) on graph paper and join the [: j ( 3 points to form the lines 1 and m as shown in figure.

-

(i) From the graph, clearly lines (1) and (2) intersect each other at the point E(3,5). So, x - 3, y = 5 is the required solution of the given equations. (ii) Shaded triangdar region ACE is shown in the figure. 34. The following table gives the production yield p e r hectare of wheat of 100

farms of a village :

50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 , 75 - 80 in kglhectare Number of farms

Change the above distribution to more than type distribution and draw its ogive. Solution. First we change the distribution to a more than type distribution and draw its

Here 50, 55, 60, 65, 70, 75 are the lower limits of the respective class intervals more than 50-55,55-60,60-65,65-70,70-75and75-80.

To represent the data in the table graphically,,we mark the lower limits of class intervals on thehorizontal axis (x-axis) and their coirespondig cumulative frequencies on the vertical axis Cy-axis), choosing convenient scale.

Now plot the point corresponding to the orderedpairs given by (lowerlimit, corresponding cumulative frequency)

wive (of more than type) (see figure).

I Lower limit -

Cumulative Frequency

(more than type) 100

55 - 60 60 - 65 65 - 70 70 - 75 75 - 80

Yield in kghectare

Number of farms

(Frequency) 2

Classes

50 - 55

Production field

(in kglhectare) more than or equal to 50 more than or equal to 55 more than or equal to 60 more than or equal to 65 more than or equal to 70 more than or equal to 75

8 12 24

- 38 16

98 90 78 54 16

cbse cce sample question paper...cosec2 0 + sec2 0 (a) - 1 (b) 1 solution. choice (c) is correct....

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