Download 3sin A 2sin B 1 + = 3sin2A 2sin2B 0, - 2 π 4 π 6 π 3sin A

1.
2
2
If A and B are acute positive angles satisfying the equation 3sin A  2sin B  1 and
3sin 2A  2sin 2B  0, then A  2B 
(a) 
Soln.
(b)

2
(c)

4
(d)

.
6
3 sin2 A + 2 sin2 B = 1
3sin 2 A  1  2sin 2 B
 3 sin2A = cos2B
Also 3 sin 2 A – 2 sin 2 B = 0
 sin 2 B =
3
sin 2 A
2
Now,
cos (A + 2B) = cosA cos 2B – sinA sin2B
= cos A (3 sin2 A) – sin A (
3
sin 2 A)
2
= 3 sin2 A cos A – 3 sin2 A cos A = 0
A+2B=
2.

2
The number of solutions of cos x  1  sin x ,0  x  3, is
(a) 3
these.
Soln.
(b) 2
Ans (a)
Clearly 1 + sin x > 0
 equation becomes
 cos x  1  sin x
 cos x – sin x = 1
Clearly
x = 0 is a solution
x = 2 is a solution
x
3π
is a solution
2
No other solution is possible
(c) 4
(d) none of
3.



 1
 x  .cos   x   ,
3

3
 4
The sum of all the solutions of the equation cos x.cos 
x  0, 6
is
(a) 25 
these.
Soln.
(b) 30 
(c) 40
(d) none of
Ans (b)


We have cos x  cos 2



 1
1
 1
 cos x   (1  cos 2 x)  
 sin 2 x  
3
 4
4
 4
3
4
 cos x  cos 2 x   
 cos 3x = 1 
1
4
 cos x (4 cos2x – 3) = 1
3x = 2n 

x
2nπ
3
Where n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
 the reqd. sum 
4.
Soln.
2π
2 π 9(9  1)
(1  2  3  ...  9) =
.
 30 π .
3
3
2
If tan m  tan n , ( m  n ) then the different values of  are in
(a) A.P.
(b) H.P.
(c) G.P.
(d) no particular sequence.
tan m  tan 
 m  n  K
  m  n    K

K
where K  I
mn
These values of  are in A.P. with common difference =
Hence (a) is the correct answer

mn
5.
The set of value of x for which sin x.cos3 x  cos x.sin 3 x, 0  x   , is
(A)  0,  
Soln.
6.


4


(B)  0,
 
, 
4 
(C) 
(D) None of these
sin x cos3 x  cos x sin3 x

sin x cos x cos 2 x  sin 2 x   0

1
sin 2x cos 2x  0
2

sin 4x  0

2n  4x  2n   ;

0x

4
n I
(given in the options)
The number of values of K for which the system of equations  K  1 x  8y  4K and
Kx   K  3 y  3K  1 has infinitely many solutions is
(A) 0
Soln.
(B) 1
For infinitely many solutions,
we have
k 1
8
4k


k
k  3 3k  1
from first two
k 2  4k  3  8k
 k 2  4k  3  0
 k  1,3
from first and last, 3k 2  2k  1  4k 2
 k 2  2k  1  0
  k  1  0  k  1
2
taking the common solution we get k = 1
Hence only one solution is possible
(C) 2
(D) Infinite
7.
Soln.
The number of different permutations of all the letters of the word 'PERMUTATION' such that
any two consecutive letters in the arrangement are neither both vowels nor both identical is
(A) 63 × 6! × 5!
(B) 57 × 5! × 5!
(C) 33 × 6! × 5!
(D) 7 × 7! × 5!
The letters other than vowels are : PRMTTN
Number of permutations with no two vowels together is
6! 7
× C5 × 5!
2!
Further among these permutations the number of cases in which T's are together is
5! × 6C5 × 5!
So the required number =
8.
Soln.
6! 7
C5 × 5! - 5! × 6C5 × 5! = 57 × (5!)2
2!
The number of different words that can be formed using all the letters of the word 'SHASHANK'
such that in any word the vowels are separated by atleast two consonants, is
(A) 2700
(B) 1800
(C) 900
(D) 600
The letters other than vowels are SHSHNK which can be arranged in
6!
ways
2! 2!
Now in its each case, let the first A be placed in the rth gap then the number of ways to place
the 2nd A will be (7 - r - 1). So, the total number of ways =
=
6!
× (5 + 4 + 3 + 2 + 1) = 2700.
2! 2!
6! 5
 (6  r)
2! 2! r 1
9.
The sum of the series
9
13
17
 3
 4
    upto infinity
5 .2.1 5 .3.2 5 .4.3
2
(A) 1
(B)
9
5
1
5
(D)
2
5
(C)
(C)
Soln.
Hint : Tr 
4r  1
,r2
5 r (r  1)
r
5r  (r  1)
1
1
 r 1
 r
r
5 r (r  1) 5 (r  1) 5 r

T
r
r 2

 1

1   1
1   1
1 
   1  2    2  3    3  4      
  5 .1 5 .2   5 .2 5 .3   5 .3 5 .4 

1
5
If x15  x13  x11  x9  x7  x5  x3  x  7 then
10.
Soln.
(A) x16 is 15
(B) x16 is less than 15
(C) x16 greater than 15
(D) Nothing can be said about x16
(C)
Hint : x15  x13  x11  x9  x7  x5  x3  x
 x( x8  1)( x4  1)( x2  1)
x16  1  ( x8  1)( x4  1)( x2  1)( x2  1)
2


7 2
1 
 ( x  1)  7  x 
  2  14
x
x


 x16  15
If a, b, c, d are distinct integers in AP such that d  a2  b2  c2 then a + b + c + d is
11.
(A) 0
(B) 1
(C) 2
(D) None
(C)
Soln.
Hint : d  a2  b2  c2  a  3t  (a  t )2  a 2  (a  2t )2
5t 2  3(2a  1)t  3a 2  a  0
D  0  24a2  16a  9  0
1
70
1
70
 
a 
3
2
3
2
 a  1,0
a  0, t  0,
3
5
a  1, t  1,
4
5
 t 1
abcd  2
12.
Soln.
The least value of sec + cosec + sec2 + cosec2 in (0, /2) is
(A) 4 + 2 2
(B) 4 - 2 2
(C) 5  2 2
(D) 5 - 2 2
2
2
Since sec + cosec and sec  + cosec  has minimum value at  = /4, we have
minimum value of
sec + cosec + sec2 + cosec2
=2 2 +4
13.
Soln.
If
2 cosA = cosB + cos3B,
2 sinA = sinB - sin3B. Then |sin(A-B)| =
(A) 1/2
(B) 1/3
(C) 2/3
(D) 1/5
sin(A - B) = sinA cosB - cosA sinB ... (i)
Substituting the values of cosA and sinA
from the given equations in (i), he have
sin(A - B) = 
sin Bcos B
... (ii)
2
squaring and adding given equation we get
cos2B = 1/3  sin(A - B) =  1/3
14.
Soln.
15.
Soln.
If x + y + z + w = 5, then the least value of x2 cot9° + y2 cot27° + z2 cot63° + w2 cot81° is
(A) 5/4
(B)
5 5
4
(C) 25/4
(D)
25 5
4
x2
y2
z2
w2
25
5 5



We have


tan 9 tan 27 tan 63 tan 81 4 5
4
The number of solutions of [sin x] + |cos x| = 1, in   x  4, [where [x] = greatest integer
function], is
(A) 3
(B) 4
(C) 5
(D) 6
(C) [sin x] = - 1, 0, 1
If [sin x] = - 1 then | cos x | 2 not possible
If sin x   0 then | cos x | 1  x = , 2, 3, 4
If sin x   1 Then | cos x | 0  x 
5
2
 Number or solutions = 5 for x   , 4 .
16.
Let f(x) = 2 cosec 2x + sec x + cosec x , then minimum value of f(x), for x  (0, /2), is
1
(A)
2 1
1
(C)
Soln.
(B)
(D)
2 1
2
2 1
2
2 1
(B)
f(x) =
=
2  sin x  cos x 
2 1  sin x  cos x 
2
=

sin 2x
sin 2x
sin 2x
2 1  sin x  cos x 
 sin x  cos x 
 f (x) 
2
1
=
2
sin x  cos x  1
2  cos x  sin x 
 sin x  cos x  12
 minimum is at x =
 f(x) is decreasing for 0 < x < /4 and increase for /4 < x < /2

4
2
2 2
2



 fmin (x) = f   
2
2 1
4
1 2  2
2
17.
Soln.
 n 1
 , where [] denotes the G.I.F, is
lim 
r
n  
 r 1 2 

(A) 1
(B) 0
(C) does not exist
(D) none of these
(B)
n
Clearly
2
1
r
1
r 1

 n 1
lim  r   0
n 
 r 1 2 
1
2n
18.
The maximum value of f(x) =
| x | 2  x 2
, (xR), is
| x | 1
(A) -4
(B) -3
(C) -2
(D) -1

4 
 3 - 2 4 = -1
(| x |  1) 
Soln.
f(x) = 3 - (| x |  1) 
19.
The number of all the odd divisor of 3600 is

(A)
45
(B)
4
(C)
18
(D)
9
Solution :
(D) 3600 = 24  32  52
so number of odd divisors = (2 + 1) (2 + 1) = 9
20.
Let A be the set of 4-digit numbers a1a2a3a4 where a1> a2> a3> a4, then n(A) is equal to
(A)
126
(B)
84
(C)
210
(D)
none of these
Solution:
Any selection of four digits from the ten digits 0, 1, 2, 3, . . . , 9 gives one such
number. So, the required number of numbers = 10C4 = 210 .
Hence (C) is the correct answer.
21.
Total number of ways of selecting two numbers from the set {1, 2, 3, 4, …., 3n} so that their sum
is divisible by 3 is equal to:
(A)
2n2  n
2
(B)
3n2  n
2
(C)
2n2  n
(D)
3n2  n
Solution :
Given numbers can be rearranged as
1 4 7 …. 3n – 2  3  – 2 type
2 5 8 …. 3n – 1  3  – 1 type
 3  type
3 6 9 …. 3n
That means we must take two numbers from last row or one number each from first and second
row.
Total ways = nC2 + nC1 . nC1
=
=
22.
n  n  1
2
 n2
3n2  n
2
A polygon has 65 diagonals. The number of its sides is
(A)
8
(B)
10
(C)
11
(D)
13
Solution :
(D) Let number of sides be n, then nC2  n = 65
 n (n  3) = 130  n = 13
23.
In a chess tournament, all participants were to play one game with the other. Two players fell ill
after having played 3 games each. If total number of games played in the tournament is equal to
84, then total number of participants in the beginning was equal to:
(A)
10
(B)
15
(C)
12
(D)
14
Solution :
Let there were ‘n’ players in the beginning. Total number of games to be played was played to
n
C2 and each player would have played (n – 1) games. Thus
n
C2 – ((n – 1) + (n – 1) – 1) + 6 = 84
 n2 – 5n – 150 = 0
 n = 15
24.
If n objects are arranged in a row, then the number of ways of selecting three
objects so that no two of them are next to each other is
(A)
(C)
Solution:
n  2 n  3 n  4 
6
(B)
n-2
C3
n-3
C3 + n-3C2
(D)
none of these.
Let x0 be the number of objects to the left of the first object chosen, x 1
the number of objects between the first and the second, x 2 the number
of objects between the second and the third and x3 the number of
objects to the right of the third object. We have
x0 , x3  0 , x1, x2  1 and x0 + x1 +x2 + x3 = n –3
. . . (1)
The number of solutions of (1)
= coefficient of yn-3 in (1+ y+ y2+...)(1+ y + y2+ ...)(y + y2+ y3+ ...)(y + y2+
y3+...)
= coefficient of yn-3 in y2( 1+ y + y2 +y3+ . . .)4 = coefficient of yn-5 in (1- y)4
= coefficient of yn-5 in ( 1+ 4C1y + 5C2 y2 + 6C3y3 + . . .) = n-5 +3Cn-5 = n-2C3
=
n  2n  3n  4
6
also n-3 C3 + n-3C2 = n-2C3 .
Hence (A) , (B) and (C) are the correct answers.
Paragraph for Questions Nos. 25 to 27
In order to solve the equation of the form a cos   bsin   c..... 1
Put
a  r cos  ;
b  r sin 
r  a 2  b2 , tan  
b
equation (1) becomes
a
r cos  cos   r sin  sin   c
cos      
c
a 2  b2
If c  a 2  b2 , then the equation a cos   bsin   c has no solution
If c  a 2  b2 , then put
c
a 2  b2
 cos t
 cos       cos t
    2n  t
25.
Soln.
  2n    t
The solution of sin x  3 cos x  2 is
(A) 2n 
5

, 2n 
12
12
(B) 2n 
5

, 2n 
12
12
(C) 2n 


, 2n 
6
6
(D) None of these
1
3
2
1
sin x 
cos x 

2
2
2
2



cos  x    cos
6
4

x  2n 
 

6 4
x  2n 
5

or x  2n 
12
12
x


 2n 
6
4
26.
Soln.
k cos x  3sin x  k  1 is solvable only if
(A) k  (, 4]
(B) k  (, )
k2  9
cos x 
3
k2  9
then cos  x    

k 1
k2  9
k 1
sin x 
k 1
k2  9
1  cos  x     1 ;
;
1 
k2  9
where cos  
k 1
k2  9
2
k
k2  9
3
and sin  
k2  9
1
 1   k  1  k 2  9 ;
k4


If 0  x  2, then number of solutions of equation 3  sin x  cos x   2 sin 3 x  cos3 x  8 is
(A) 0
Soln.
(D) None of these
k cos x  3sin x  k  1
k
27.
(C) k  1,  
(B) 1
(C) 2

(D) 4

3  sin x  cos x   2  sin x  cos x   sin 2 x  cos 2 x  sin x cos x  8
  sin x  cos x  3  2 1  sin x cos x   8
  sin x  cos x 1  2sin x cos x   8
 sin x  cos x 3  8
; sin x  cos x  2


sin  x    2  1, Not possible
4

The given equation has no. solution
Paragraph for Question Nos. 28 to 40
If a sequence or series is not a direct form of an AP, GP, etc. Then its nth term can not be
determined. In such cases, we use the following steps to find the nth term Tn  of the given
sequence.
Step – I : Find the differences between the successive terms of the given sequence. If these
differences are in AP, then take Tn  an 2  bn  c , where a,b,c are constants.
Step – II : If the successive differences found in step I are in GP with common ratio r, then
take Tn  a  bn  cr n1 , where a, b, c are constants.
Step – III : If the second successive differences (Differences of the differences) in step I are in
AP, then take Tn  an3  bn2  cn  d , where a, b, c, d are constants.
Step – IV : If the second successive differences (Differences of the differences) in step I are in
GP, then take Tn  an2  bn  c  dr n1 , where a, b, c, d are constants.
Now let sequences :
A : 1, 6, 18, 40, 75, 126, ….. B : 1, 1, 6, 26, 91, 291, …. C : ln 2 ln 4, ln 32, ln 1024 …..
28.
If the nth term of the sequence A is Tn  an3  bn2  cn  d then the value 6a  2b  d is
(A) ln 2
(B) 2
(C) ln 8
(D) 4
Soln.
(D)
Hint :
Tn  an3  bn2  cn  d
T1  a  b  c  d  1
T2  8a  4b  2c  d  6
6a  2b  d  4
50
29.
For the sequence 1, 1, 6, 26, 91, 291, …….. Find the S50 where S50   Tr
r 1
(A)


5 50
3  1  3075
8
(B)


5 50
3  1  5075
8
(C)


5 50
3  1  1275
8
Soln.
(A)
Hint :
5
5n 9
Tn  3n1  
4
2 4
S50 

(D) None of these

5
5
9
1  3  ...  349  1  2  ....  50   50.
4
2
4
5  350  1  5 50.51 450
 

 .
4  2  2 2
4
30.





5 50
125.51 450
3 1 

8
2
4

5 50
3  1  3075
8
The sum of the series 1.n  2. n  1  3. n  2   ....  n.1
(A)
(C)
n  n  1 n  2 
(B)
6
n  n  1 2n  1
6
(D)
n  n  1 n  2 
n  n  1 2n  1
3
Soln.
(A)
Hint :
 r  n  r  1    n  1 r   r 2
n
n
n
r 1
r 1
r 1
  n  1  n   n2
2
n  1 n n  n  1 2n  1



2

n  n  1
6
6
 3n  3  2n  1
=
n  n  1 n  2 
6
3