iit trigno

7/30/2019 IIT TRIGNO

1/17

L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721

PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH

SOLUTIONS ON www.pioneermathematics.com

PAPER B

IITJEE(2013)(Trigonometry and Algebra)

TOWARDS IIT JEE IS NOT A JOURNEY,ITS A BATTLE, ONLY THE TOUGHEST WILL SURVIVE

TIME: 60 MINS MAX. MARKS: 76

MARKING SCHEME

In Section I (Total Marks: 24), for each question you will be awarded 3 marks if you darken ONLYthe bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all

other cases, minus one ( 1) mark will be awarded.

In Section II (Total Marks: 16), for each question you will be awarded 4 marks if you darkenALLthe bubble(S) corresponding to the correct answer(s) ONLYand zeromarks otherwise. There are

no negative marks in this section.

In Section III (Total Marks: 24), for each question you will be awarded 4 marks if you darken ONLthe bubble corresponding to the correct answer and zero marks otherwise. There are no negative

marks in this section.

In Section IV (Total Marks: 12), for each question you will be awarded 2 marks for each row inwhich you have darkenedALL the bubble(s) corresponding to the correct answer(s) ONLYand zer

marks otherwise. Thus, each question in this section carries a maximum of 6 marks. There are no

negative marks in this section.

NAME OF THE CANDIDATE CONTACT NUMBER

L.K. Gupta (Mathematics Classes)FOR SOLUTIONS KINDLY VISIT

www.pioneermathematics.com(In latest Updates)


2/17




This section contains 8 multiple choice questions. Each question has four choices (a), (b)

(c), (d) out of which ONLY ONE is correct.

1. The equation sin (cos x) = cos (sin x) has(a) only one real solution (b) infinitely many solution

(c) no real solution (d) none of the above

Ans. (c)

Sol :

sin (cos x) = cos (sin x)

cos (sin x) sin (cos x) = 0

cos sin x cos cosx 0

2

cosx sin x cosx sin x2cos . cos 0

4 2 4 2

cosx sin xIf cos 0

4 2

cos x sin x n

4 2 2

2 sin x 4n 1 , n I4 2

4n 1sin x

4 2 2

sin x [ 1,1],

4

which does not certify the original solution,

cos x sin xcos 04 2

cos x sin xThen, cos 0

4 2

cosx sin x

2r 14 2 2


3/17




2 cos x 4r 14 2

cos x 4r 1 , r I4 2 2

cos x [ 1, 1]4

cos x sin xAlso, cos 0

4 2

LHS 0

and RHS 0

Hence, no real solution.

2. If 2x x 2 is factors of 4 2x x , then 2 2( ) equals(a) 1 (b) 3 (c) 5 (d) 7

Sol: (b)

2x x 2 0

x 2, 1

Then x 2, 1 are the roots of 4 2x x 0

then 4 2(2) (2) 0

4 16

and 4 2( 4) ( 1) 0

1

we get 5 and 4

2 2( ) 25 16 9 3

3. The number of solutions of 5r 1

cos r x = 5 in the interval 0, 2 is

(a) 0 (b) 1 (c) 5 (d) 10

Ans. (b)


4/17




Sol :

5

r 1

cos r x 5

cos x + cos 2x + cos 3x + cos 4x + cos 5x = 5 which is possible only, when

cos x = cos 2x = cos 3x = cos 4x = cos 5x = 1and is satisfied by x = 0 only.

Hence number of solution = 1.

4. Set a, b , be such that cos (a b) = 1 and cos (a + b) = 1 .e

The number of pair

of a, b satisfying the above system of equation is

(a) 0 (b) 1 (c) 2 (d) 4

Ans. (d)

Sol :

cos a b cos 0

a b 2n, n I

a b 2, 0, 2

a b 2 2a, 2a, 2a 2

1

cos a b cos 2a, cos 2a, cos 2ae

1y cos 2a

e

Hence, number of solutions is 4.


5/17




5. The number of values of x for which sin 2x + cos 4x = 2 is(a) 0 (b) 1 (c) 2 (d) infinite

Ans. (a)

Sol :

sin 2x + cos 4x = 2

It is possible only, when

sin 2x = 1 and cos 4x = 12x 2n and 2x 2m

2

x n and x m m, n I

4

Then, solution n , n I m, m I4

= 6. If 1, 2, are the three cube roots of unity, then for

2 2

2

, , , R, the expression is

(a) 1 (b) (c) (d) 1

Ans. (b)

Sol :

2 2

2

2 2

3 2 2

3 1 . 7. The number of positive integral solutions of

3 42

5 6

x 3x 4 x 20 is

x 5 2x 7

(a) four (b) three (c) two (d) only one

Ans. (b)

Sol :

3 42

5 6

x 3x 4 x 2Since, 0

x 5 2x 7


6/17




4 7

x , 7/2, 53 2

x 2, 3, 4.

8. If tan cos cot sin , then the value of cos is4

(a)1

2(b)

1

2 (c)

1

2 2(d) none of these

Ans. (c)

Sol :

tan cos cot sin

tan cos tan sin 2

cos sin

2

1cos sin 2

1 1 1cos sin

2 2 2 2

1cos

4 2 2

Section II (Total Marks: 16)

(Multiple Correct Answer (s) Type)

This section contains 4 multiple choice questions. Each question has four choices (a), (b),

(c), and (d) out of which ONE or MORE may be correct.


7/17




9. If tan and tan are the roots of the equation 2x px q 0 p 0 , then (a) 2 2sin p sin cos qcos q

(b) tan p/ q 1 (c) cos 1 q

(d) sin p Ans. (a, b)

Sol :

tan tan p, tan tan q q 0

tan tan p

tan 1 tan tan 1 q

p[Alternate.(b)]

q 1

Alternate. (a) :

2 2LHS cos tan p tan q

2

2

tan p tan q

1 tan

2 2

2 22 2

2 2 2

2

p pq

q 1q 1 p q q 1 q q 1

p q 1 p1q 1

22

22

q p q 1

p q 1

= q

Alternate. (c) : p

tan q 1

22

q 1cos

p q 1

22

pAlternate. (d) : sin

p q 1


8/17




10. The argument and the principal argument of the complex number

22 i

, where i 1 are4i 1 i

(a) 1tan 2 (b) 1tan 2 (c) 11

tan2

(d) 11

tan2

Ans. (a, b)

Sol :

2 2

2 i 2 i 2 i 2 i

4i 1 i 2i 4i 1 1 2i 6i4i 1 i

1 1i

6 3

1 1 1

1

3 tan tan 2 tan 21

6

1 1imaginary part

Argument tan tan 2real part

1and principal arg ument tan 2

11. The roots of the equation, (x2 + 1)2 = x(3x2 + 4x + 3), are given by(a) 2 3 (b) 1 i 3 /2, i 1

(c) 2 3 (d) 1 i 3 /2, i 1

Ans. (a, b, c, d)

Sol :

Given equation is

2

2 2x 1 x 3x 4x 3 4 3 2x 3x 2x 3x 1 0

2 2

2

3 1x x 3x 2 0

x x

x 0

2

2

1 1x 3 x 2 0

x x

21 1

x 3 x 4 0x x


9/17




1 1x 4 x 1 0

x x

2 2or x 4x 1 x x 1 0

2

1 3or x 2 h 3 x 0

2 4

1 i 3

x 2 3 , .2

12. In a triangletan A + tan B + tan C = 6 and tan A tan B = 2, then the values of tan A, tan B and tan C are

(a) 1, 2, 3 (b) 2, 1, 3 (c) 1, 2, 0 (d) none of these

Ans. (a, b)

Sol :

In a triangle

tan A + tan B + tan C = tan A tan B tan C

or 6 = 2 tan C

tan C = 3 tan A + tan B = 3, tan A tan B = 2 tan A = 1 or 2and tan B = 2 or 1. (i)

Section III (Total Marks: 24)

(Integer Answer Type)

This section contains 6 questions. The answer to each of the questions is a single-digit

integer, ranging from 0 to 9.The bubble corresponding to the correct answer is to be

darkened in the Answer sheet.

13. 4 20 0

1 1If , then the value of 9 81 97 782 must be

cos 290 3 sin250

Ans. 3

Sol :

Here, cos 2900 = cos (2700 + 200) = sin 200 and sin 2500 = sin (2700 200) = cos 200

0 0

1 1The given expression

sin 20 3 cos 20


10/17




0

0 0 0

1 cos 60

sin 20 sin60 cos 20

0 0 0 0

0 0 0

sin 60 cos20 cos 60 sin20

sin 20 cos 20 sin60

0 0

0sin 60 20 sin40 3

2 2

4

3

2 163

4 2 256 16Then, 9 81 97 9 81 979 3

= 256 + 432 + 97-782

= 3

14. The least degree of a polynomial with integer coefficient whose one of the roots may bcos 120 is

Ans. 4

Sol :

0 012 5 60

0let 12

05 60

0

3 2 60

0cos 3 2 cos 60

1cos 3 cos 2 sin 3 sin 2

2

3 2 31

4 cos 3 cos 2cos 1 3 sin 4sin 2 sin cos 2


11/17




Let cot x

3 2 2 2 14x 3x 2x 1 2x 3 4 1 x 1 x2

5 3 3 21

8x 10x 3x 2x 2x 4x 12

5 3 3 216x 20x 6x 4x 4x 4x 1 1 0

5 332x 40x 10x 1 0

4 3 21

x 32x 16x 32x 16x 2 02

4 3 21but x , 16x 8x 16x 8x 1 02

Degree is 4.

15. The three angles of a quadrilateral are 600, 60g and 5 ,6

if fourth angle is then 0,

the value of -90 must be

Ans. 6

Sol :

First angle =600,second angle=60g=

0

0

0

0

9060 54

100

5 180Third angle 150

6

Fourth angle=3600-(600+540+1500)

=960= 0-90=6

16. The number of solutions of 3sec 5 4 tan in [0,4 ] must beSol. 8

The given equation can be written as


12/17




3 4sin5

cos cos

4 sin 5 sin 3,cos 0

4 5 3sin cos

41 41 41 (i)

Let5 4

cos , then sin41 41

Now, from Eq. (i),

3

cos cos41

32n , where cos

41

or 2n

or 2n

Now, in [0,2 ], 2n gives two solutions.

So, in total in [0,4 ]

we have , 2 + 2 + 2 + 2 = 8 solutions

17. The sum of the roots of equation cos 4x + 6 = 7 cos 2x over the interval [0, 314] is , then the numerical quantity 4949 must be

Ans. 1

Sol :

On putting cos 2x = 1 we get

2t2 1 + 6 = 7t

5t 1,

2

5t (impossible)

2

t 1


13/17




cos 2x 1

2x 2n

x n , n I

The roots over [0, 314] are , 2, 3, ...., 99

100 314

Sum of roots 2 3 .... 99 4950

4950

18. The sides of a cyclic quadrilateral are in AP, the shortest is 6 and the difference of thelongest and the shortest is also 6. The square of the area of the quadrilateral is n-5756,find

the value of n.

Ans.4

Sol :

Let sides are a, a + d, a + 2d, a + 3d

Given, a = 6 and a + 3d a = 6 d 2

Sides 6, 8, 10, 12

2s = 6 + 8 + 10 + 12

= 36

s 18

Then, (Area)2 = (18 6) (18 8) (18 10) (18 12)

= 12 10 8 6

= 5760


14/17




Section IV (Total Marks: 12)

(Matrix-Match Type)

This section contains 2 questions. Each question has three statements (a, b and c) given in

Column I and five statements (p, q, r, s and t) in Column II. Any given statements in

Column I can have correct matching with ONE or MORE statements(s) given in Column II

For example, if for a given question, statement B matches with the statements given in qand r, then for the particular question, against statement B, darken the bubbles

corresponding to q and r in the ANSWER SHEET.

19. Observe the following columns :Column I Column II

(A)If maximum and minimum values of

2

2

7 6tan tan for

1 tan

all real values of

are and respectively , then2

(P) 2

(B) If maximum and minimum values of

5cos 3cos 3

3

for all real values of are and

respectively, then

(Q) 6

(C) If maximum and minimum values of

1 sin 2cos

4 4

for all real values of

are and respectively, then

(R) 6

(S) 10

(T) 14

Ans. A R, S ; B R, T ; C P, Q Sol :

2

2

7 6 tan tan A Let y

1 tan

2 27 cos 6 sin cos sin

1 cos 2 1 cos 27 3 sin 2

2 2

3 sin 2 4 cos 2 3

2 2 2 23 4 3 sin 2 4 cos 2 3 3 4 3


15/17




2 y 8 8, 2

6, 10 R, S

B Let y 5 cot 3 cos /3 3

1 35 cos 3 cos sin 3

2 2

13 3 3cos sin 3

2 2

2 22 213 3 3 13 3 3 13 3 3

3 cos sin 3 32 2 2 2 2 2

3 7 y 3 7 4 y 10

10, 4 6, 14 R, T

C Let y 1 sin 2 cos 4 4

1 cos 2 cos

2 4 4

1 cos 2 cos

4 4

1 3 cos

4

1 cos 1

4

3 3 cos 3

4

1 3 1 3 cos 1 3

4

2 y 4 4, 2

2, 6 P,Q


16/17




20. Observe the following columns :Column I Column II

(A) 1If , are the solutions of sin x in

2 0, 2 and , are

the solutions of cos x =

3in 0,2 , then

2

(P)

(B) If , are the solutions of cot x 3 in 0,2 and

, are the solutions of cosec x 2 in 0, 2 , then

(Q)

(R)

(C)

1If , are the solutions of sin x in 0, 2 and ,

2

are the solutions of tan x = 1

in 0, 2 , then3

(S) 3

(T) 2

Ans. A Q, S ; B P,T ; C R, S, T Sol :

1

A sin x2

sin

6

sin , sin 2

6 6

7 11x ,6 6

..(i)

3 and cos x cos

2 6

cos , cos

6 6

5 7x ,

6 6 ..(i)

From Eqs. (i) and (ii). It is clear that7 11 5

, , y6 6 6

3 S , Q

B cot x 3

cot6


17/17


PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S C O 320 SECTOR 40 D CHANDIGARH

cot , cot 2

6 6

5 11x ,

6 6 .(i)

and cosec x 2 cosec 6

cosec , cosec 26 6

7 11x ,

6 6 (ii)

From Eqs. (i) and (ii), It is clear that

11 5 7 , , 6 6 6

2 T , P

1

C sin x2

sin

6

sin , sin 26 6

7 11x ,

6 6 .(i)

1and tan x

3

tan

6

tan , tan 6 6

7

x ,6 6

(ii)

From Eqs. (i) and (ii), it is clear that

7 11 , ,

6 6 6

3, S , 2 T , R

iit trigno

Documents