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8/11/2019 Trigo Question Paper
1/8
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
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
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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