Trigo Question Paper

8/11/2019 Trigo Question Paper

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CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW/ NASHIK / GOA

ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI

IIT JEE 2015 TW TEST MARKS: 90

TIME: 1HR TOPIC: TRIGO I II DATE: 29/09/13

SECTION-I

This section contains 30 multiple choice questions. Each question has 4 choices (A), (B), (C)and (D) for its answer, out which ONLY ONE is correct.

(+3, - 1)

1. The value of 6 6 4 46(sin cos ) 9(sin cos ) 4 equals to(a) 3 (b) 0 (c) 1 (d) 3

2.o o o

o o o

cos12 sin12 sin147cos12 sin12 cos147

(a) o2tan33 (b) 1 (c) 1 (d) 0

3. If tan cot a and sin cos b , then 2 2 2(b 1) (a 4) (a) 2 (b) 4 (c) 4 (d) 4

4. 22sin 4cos( ) sin sin cos 2( ) equal to(a) sin2 (b) cos2 (c) t anA/2 (d) sin 2

5. If 1sin 2 sin 22

and 3cos 2 cos 22

, then 2cos ( ) equal to

(a) 38

(b) 58

(c)

34

(d)

54

6. The value of 5cos 3cos( ) 33

lies between

(a) 4 and 4 (b) 4 and 6 (c) 4 and 8 (d) 4 and 10

7. sin /14.sin 3 /14.sin 5 /14.sin 7 /14.sin9 /14.sin11 /14.sin13 /14 is equal to

(a) 18

(b)

116

(c)

132

(d)

164


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8. sin 2A sin 2B sin 2CsinA sin B sinC

equal to

(a)

cosAcosBsinCA B C

sin sin cos2 2 2

(b)

sin AsinB cosCA B C

cos cos sin2 2 2

(c)

cosAcosBsinCA B C

sin sin cos2 2 2

(d)

sin AsinB cos CA B C

cos cos sin2 2 2

9. If oA B C 180 , then sin 2A sin 2B sin 2CcosA cosB cosC 1

equal to

(a)

A B C8sin sin sin

2 2 2 (b)

A B C

8cos cos cos2 2 2

(c)

A B C8sin cos cos

2 2 2 (d)

A B C

8cos sin sin2 2 2

10. If oA B C 180 , then the value of (cotB cot C)(cotC cot A)(cotA cotB) will be

(a) secAsecBsecC (b) cosecA cosecB cosecC

(c) tanAtanBtanC (d) 1

11. The general solution of the equation ( 3 1) sin ( 3 1) cos 2 is

(a) 2n4 12

(b) nn ( 1)4 12

(c) 2n4 12

(d) nn ( 1)4 12

12. If 1(cos x sin x) 2 tan x 2 0cos x

then x

(a) 2n

3 (b)

n

3 (c)

2n

6 (d) None of these

13. If 2 2f sin cos ec cos sec , then minimum value of f is(a) 7 (b) 8 (c) 9 (d) None of these

14. If 1f ,2 3cos then range of f is

(a) 1, 1,5

(b) 11,

5

(c) 1, 1 ,5

(d) None of these


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15. If asin x x R ~ 0 ,x

then

(a) 1a4

(b) 1a2

(c) 1a4

(d) 1a2

16. If 4 2f sin cos , then range of f is

(a)1

,12 (b)

1 3,2 4

(c)

3,14

(d) None of these

17. If 4 4f sin cos 1 , then range of f is

(a) 3 , 22

(b) 31,2

(c) [1, 2] (d) None of these

18. If 6 6f x sin x cos x, then range of f (x) is

(a) 1 ,14

(b) 1 3,

4 4

(c) 3 ,1

4

(d) None of these

19. If A + B + C = 0 then value of the expression sin 2 A + cos C (cos A.cos B cos C) + cos B (cos A.cos C cos B) is equal to(a) 1 (b) 2 (c) 0 (d) None of these

20. The value of0 0

0 0

tan 205 tan115tan 245 tan335

is equal to

(a) 0sec25 (b) 0cos25 (c) 0sec50 (d) 0cos50

21. If 1 2sin sin a and 1 2cos cos b , then(a) 2 2a b 4 (b) 2 2a b 4 (c) 2 2a b 3 (d) 2 2a b 2

22. If 3cos x cos y cos x y ,2

then

(a) x y 2n (b) x 2y 2n (c) x y 2n (d) 2x = y + 2n

23. The numerical value of 0 03 cosec20 sec20 is equal to

(a) 2 (b) 4 (c) 6 (d) None of these

24. The value of 0 0tan 40 2 tan10 is equal to

(a) 0cot 50 (b) 0cot 40 (c) cot 10 0 (d) None of these

25. The numerical value of 0 0 0sin12 .sin 48 .sin 54 is equal to


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(a) 12

(b) 14

(c) 116

(d) 18

26. The numerical value of 0sin6 , sin 42 0, sin 66 0 , sin 78 0 is equal to

(a) 14

(b) 18

(c) 132

(d) 116

27. The numerical value 3 5cos cos cos7 7 7

is equal to

(a) 12

(b) 32

(c) 32

(d) 12

28. The numerical value of 0 0 0tan 20 .tan80 .cot 50 is equal to

(a) 3 (b) 13

(c) 2 3 (d) 12 3

29. If A B3

where A,B R, then the minimum value of sin A + sin B is equal to

(a) 23

(b) 43

(c) 2 3 (d) None of these

30. If A B2

where A, B R , then maximum value of sin A + sin B is equal to

(a) 1 (b) 3 (c) 2 (d) None of these


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ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI

TRIGO I, II (ANSWER KEY)

1. (C) 2. (D) 3. (D) 4. (C) 5. (B)

6. (D) 7. (D) 8. (A) 9. (B) 10. (B)

11. (A) 12. (A) 13. (C) 14. (C) 15. (C)

16. (C) 17. (A) 18. (A) 19. (C) 20. (C)

21. (B) 22. (C) 23. (B) 24. (B) 25. (D)

26. (D) 27. (D) 28. (A) 29. (B) 30. (C)


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TRIGO (SOLUTION)

1. (C) Solution: = 4)cos(sin9)cos(sin6 4466

= 4]cos.sin2)cos[(sin9)]cos(sincossin3)cos[(sin6 222222222322 = 4]cossin21[9]cossin31[6 2222 = 1496 .

2. (D)

Solution: = oo

o

147tan12tan1

12tan1 = 0)33tan(33tan)33180tan()1245tan( ooooo .

3. (D) Solution: Given that a cottan .(i) and b cossin .(ii)

Now, )4()1( 222 ab }4)cot{tan}1)cos{(sin 222

4cos

1

sin

1cossin4)seccosec(2sin]42cot[tan]12sin1[

2222222222

4. (C) Solution: Since ,1)(cos2)cos(2 2 2cos1sin2 2 )]cos(sinsin2)[cos(22cos

)cos().cos(22cos 2cos2cos2cos2cos .

5. (B)

Solution: Given,21

2sin2sin .......(i) and23

2cos2cos .......(ii)

Squaring and adding,49

41

]2cos.2cos2sin.2[sin2)2cos2(sin)2cos2(sin 2222

41

2sin.2sin2cos.2cos 41

)22cos( 85

)(cos 2 .

6. (D)

Solution: 3)3

cos(3cos5 = 3]3

sin.sin3

cos[cos3cos5

= 3]sin2

33cos

23

cos5[ = 3sin2

33cos

213

2222

233

213

sin2

33cos

213

233

213

7sin2

33cos

213

7

373sin2

33cos

213

37

103sin

233

cos2

134

So, the value lies between 4 and 10.


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7. (D)

Solution: 14

13sin.

1411

sin.149

sin.147

sin.14

5sin.

143

sin.14

sin

=

14sin.

143

sin.145

sin1145

sin.143

sin.14

sin

=

641

147

sin.145

sin.143

sin.14

sin2

.

8. (A)

Solution: C B A

C B A

sin)sin(sin

2sin)2sin2(sin =C B A B A

C B A B A

sin2

cos2

sin2

2sin)cos()sin(2

=

2cos

2sin2

2cos

2sin2

cossin2)cos(sin2C C B AC

C C B AC

=

22cos

22cos

2cos2

]cos)[cos(sin2 B A B AC

C B AC

2)(

cos22

sin2/sin

2cos

2sin2sin

B A B AC

C C C

=

22cos

22cos

2cos2

)]cos()[cos(sin2 B A B AC

B A B AC =

2sin

2sin2

2cos2

]coscos2[sin2 B AC

B AC =

2cos

2sin

2sin

sincoscosC B A

C B A .

9. (B)Solution:

12

sin212

cos2

cos2

cossin2)cos(.)sin(21coscoscos

2sin2sin2sin2 C B A B A

C C B A B A

C B A

C B A =

2sin2

2cos

2sin2

sincos2)cos(sin2

2 C B AC

C C B AC

= 2

)(cos

2)(

cos2

sin2

)]cos()[cos(sin2 B A B AC

B A B AC =

2sin

2sin

2sin4

sinsinsin4C B A

C B A

2cos

2cos

2cos8

2sin

2sin

2sin4

2cos

2sin2

2cos

2sin2

2cos

2sin24 C B A

C B A

C C B B A A

.

10. (B)

Solution: C B

C B BC C B

sin.sincossincossin

cotcot =

C B

A

C B

A

C B

C B o

sin.sinsin

sin.sin)180sin(

sin.sin)sin(

Similarly, AC

B AC

sin.sinsin

cotcot and B A

C B A

sinsinsin

cotcot

Therefore, )cot)(cotcot)(cotcot(cot B A AC C B

= C B A B A

C

AC

B

C B

Acosec.cosec.cosec

sinsinsin

.sin.sin

sin.

sin.sinsin .

11. (A)Solution: 2cos)13(sin)13(

Divided by 221313 22 in both sides, We get,

22

2cos

22

)13(sin

22

)13(

2

115coscos15sinsin

4cos

12cos.cos

12sin.sin


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4cos

12cos

42

12

n 124

2

n .

12. (A)

Solution: Let ,2

tan x

t and using the formula. We get,

2tan1

2tan2

2tan1

2tan1

22

2

x

x

x

x

02

2tan1

2tan1

2tan1

2tan4

2

2

2 x

x

x

x

22

2

1

2

1

1t

t

t

t 021

1

1

42

2

2

t

t

t

t 0)1()1(

3286322

234

t t

t t t t

Its roots are;3

11 t and .

3

12 t

Thus the solution of the equation reduces to that of two elementary equations,

3

12

tan,3

12

tan x x

62

n x ,

32

n x is required solution.

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Trigo Question Paper