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1. If 4cos3cos2cos 4sin3sin2sin = tan k is an identity then the value k is equal to :(a) 2 (b) 3 (c) 4 (d) 6

Sol. (b) 3cos4cos2cos 3sin4sin2sin = 3coscos3cos2 3sincos3sin2 = 3cos3sin= tan3 = tanK K = 3

2. Three real numbers a, b, c satisfy 2b = a + c, then ccosbcosacoscsinbsinasin

(a) tan b (b) tan a (c) tan c (d) none

Sol. (a) LHS = bcos

2ca

cos2

cacos2

bsin2

cacos

2ca

sin2

=

bcos2

cacosbcos2

bsin2

cacosbsin2

(using 2b = a + c)

=

12

cacos2bcos

12

cacos2bsin

= tan b hence proved.

3. If A + B + C = & sin A C 2 = k sin C2 , then tan A2 tan B2 =(a) k

k 11 (b) kk 11 (c) kk 1 (d) k k 1

Sol. (a)

4. Let a and b are two real numbers such that, sin a + sin b = 22

and cos a + cos b = 26

.Then value of cos(a b) equals(a) 1 (b) 0 (c) 2 (d) 2 - 2

Sol. (b) squaring and adding

2 + 2 cos(a b) = 21

+ 23

= 2

cos(a b) = 0

5. The exact value of 69sin51cos39cos21sin 66sin6sin6cos24sin is(a) 0 (b) 1 (c) 1 (d) 2

IIT ASHRAMCommon Practice Test-8

CLASS - XI MATHEMATICS SOLUTION 27-07-2014

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Sol. (b) 21sin39cos39cos21sin 24cos6sin6cos24sin = 18sin18sin = 1 ]6. If tan A & tan B are the roots of the quadratic equation, a x2 + b x + c = 0 then evaluate a sin2 (A + B) + b sin

(A + B). cos (A + B) + c cos2 (A + B).(a) c (b) a (c) b (d) none

Sol. (a) tan A + tan B = ba

; tan A. tan B = c

a; tan (A + B) = baca1 = bc a

Now E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c ]

=1

1 2 2 bc a( ) a b

c a

bc a

c2

2

2

( ) = ( )( )c ab c a 2

2 2 b

c a

a

c ac

2

1 =

( )( )

c a

b c a 22 2 b cc a c2 2( ) E = c ]

7. In a triangle ABC, the value of CsinBsinAcos

+ AsinCsinBcos

+ BsinAsinCcos

(a) is prime (b) is composite(c) is rational which is not an integer (d) negative integer

Sol. (a) [(cos B cos C 1) + (cos C cosA 1) + (cos A cos B 1)] = 2

8. If tanB = Acosn1AcosAsinn

2 then tan(A + B) equals(a) Acos)n1(

Asin (b) Asin Acos)1n( (c) Acos)1n( Asin (d) Acos)1n( AsinSol. (a)tan(A + B) = BtanAtan1

BtanAtan = Acosn1AcosAsinn

Atan1Acosn1AcosAsinnAtan

2

2

= AcosAsinn)Acosn1(AcosAcosAsinn)Acosn1(Asin

22

22

= )AsinnAcosn1(Acos0Asin

22 = Acos)n1( Asin ]9.

The expression 20sin20tan 20sin20tan 22 22 simplifies to(a) a rational which is not integral (b) a surd(c) a natural which is prime (d) a natural which is not composite

Sol. (d) tan220 sin220 = tan220 (1 cos220) = tan220 sin220Hence Nr = Dr (D) ]

10. The value of tan27 + tan18 + tan27 tan18, is(a) an irrational number (b) rational which is not integer(c) integer which is prime (d) integer which is not a prime.

Sol. (d) Consider tan(27 + 18) = 18tan27tan1 18tan27tan1 = 18tan27tan1 18tan27tan

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1 tan27 tan18 = tan27 + tan18 tan27 + tan18 + tan27 tan18 = 1 ]11. If cos( + ) + sin( ) = 0 and tan = 20061 . Then tan equals

(a) 3 (b) 1 (c) 1 (d) 2Sol. (c) cos cos sin sin + sin cos cos sin = 0 [11th J-Batch (02-10-2005)]

cos (cos sin ) + sin (cos sin ) = 0(cos sin ) (cos + sin ) = 0if cos sin = 0 tan = 1 which is not possible sin + cos = 0 tan = 1 Ans. ]

12. If tanx tan2x = 1, then the value of tan4x 2tan3x tan2x + 2tanx + 1 is(a) 1 (b) 2 (c) 3 (d) 4

Sol. (D) tan4x 2 tan3 x tan2x + 2tanx + 1tan4x + tan2x 2tan3x + 2tan x 2tan2x + 1= (tan2 tanx)2 + 2(tan x tan2x) + 1= 4.

13. If tan 25 = a then the value of tan tantan tan

205 115245 335

in terms of a is(a) 2

2

a1a1 (b) 2

2

a1a1 (c) 22a1 a1 (d) none of these

Sol. (a) tan205 = tan(180 + 25) ; tan115 = tan(90 + 25) and similarly]14. If cos25 + sin 25 = k, then cos20 is equal to

(a) 2k (b) 2

k (c) 2k (d) None of theseSol. (A) cos 20= cos(45 - 25)

= (cos25 + sin25)15. If )cos(

)cos()cos()cos(

43

43

21

21 = 0, then tan 1 tan 2 tan 3 tan 4 equals to(a) 0 (b) 1 (c) - 1 (d) 2

Sol. (C) On expansion we getcos ( 1 - 2) cos ( 3 - 4) + cos ( 1 + 2) cos ( 3 + 4) = 0 cos 1 . cos 2 cos 3 cos 4 + sin 1 sin 2 sin 3 sin 4 = 0 tan 1 tan 2 tan 3 tan 4 = - 1

16. The value of tan 9

tan 92

tan 94

is equal to

(a) 1 (b) 2 (c) 3 (d) 2Sol. (C) tan20 tan40 tan80 =

tan20 tan(60 20) tan(60 + 20) = tan3(20) = tan60 = 3 .

17. How many of the arrangements of the letter of the word LOGARITHM begin with a

vowel and end with a consonant?

(a) 90720 (b) 90820 (c) 80210 (d) 80270

Sol. (a)

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18. In a telephone system four different latter P, R, S, T and the four digits 3, 5, 7, 8 are

used. Find the maximum number of telephone numbers the system can have if each

consists of a letter followed by a four-digit number in which the digit may be repeated.

(a) 408 (b) 956 (c) 1024 (d) 1120

Sol. (c)

19. Find the number of ways in which 4 boys and 3 girls can be seated in a line if the

terminal seats are occupied by the boys.

(a) 1440 (b) 1540 (c) 1210 (d) 1105

Sol. (a)

20. How many 5-digit numbers can be formed from the digit 2, 3, 4, 5 and 9 assuming no

digit is repeated? How many of them will be divisible by 5?

(a) 120, 60 (b) 120, 24 (c) 100, 22 (d) 100, 24

Sol. (b)

21. A new flag is to be designed with six vertical strips using some or all the colours

yellow, green, blue and red. Then, the number of ways this can be done such that no

two adjacent strips have the same colour is

(a) 12 81 (b) 16 192 (c) 20 125 (d) 24 216

Sol. (a)

22. The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 & 7

so that digits do not repeat and the terminal digits are even is:

(a) 144 (b) 72 (c) 288 (d) 720

Sol. (d)

23. Number of natural numbers less than 1000 and divisible by 5 can be formed with the

ten digits, each digit not occurring more than once in each number is

(a) 150 (b) 154 (c) 175 (d) 172

Sol. (b)

24. Number of numbers from 1 to 2010 having atleast one digit 4 is

(a) 541 (b) 542 (c) 543 (d) 544

Sol. (c)

25. How many numbers between 400 and 1000 (both exclusive) can be made with the

digits 2, 3, 4, 5, 6, 0 if repertition of digits not allowed.

(a) 60 (b) 70 (c) 56 (d) 72

Sol. (a)

26. How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7

when repetition of digits is not allowed?

(a) 121 (b) 200 (c) 36 (d) 21

Sol. (d) An odd number less than 1000 may be a one-digit number, two-digit number

or a three-digit number. So, required no. of numbers is

27. How many different numbers of six digits each can be formed from the digits 4, 5, 6, 7,

8, 9 when repetition of digits is not allowed?

(a) 530 (b) 567 (c) 720 (d) 610

Sol. (c) Required no. of numbers = 6 5 4 3 2 1.

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28. A coin is tossed 4 times and the outcomes are recorded. How many possible outcomes

are there?

(a) 8 (b) 16 (c) 32 (d) 64

Sol. (b) Required number of outcomes = 2 2 2

29. How many five digits telephone numbers can be constructed using the digits 0 to 9. If

each number starts with 67 and no digit appears more than once?

(a) 256 (b) 73 (c) 336 (d) 253

Sol. (c) Required number of telephone numbers = 8 7 6

30. 11880)!1n()!3n( , find n.

(a) 9 (b) 11 (c) 13 (d) 7

Sol. (a)