1d 2d 3a 4a 5a 6 ac 7abcd 8ad 9abc 10abcd 11abcd 12c 13c 14c15(3) 16(2) 17(5) 18(4) 19(1) 20(3)

1)Find domain and differentiate

ALT:

1 D Let x2 = 4 cos2 + sin2

then (4 – x2) = 3 sin2 and (x2 – 1) = 3 cos2
 f(x) = 3 sin  3 cos 
 1 1 
 ymin. = 3 and ymax. = 3   = 6.
 2 2
Hence range of f(x) is  3, 6 

2D For
' x  2, L.H.S. is always non negative and R.H.S. is always – ve.
Hence for x  2 no solution.
If 1  x < 2 then (x – 2) = (x – 1) – 1 = x – 2, which is an identity  (D)
For 0  x < 1, LHS is '0' and RHS is (–) ve  No solution.
For x < 0, LHS is (+) ve, RHS is (–) ve  No solution.]

3. If sinx   2 cos x   3, x  0, 2 ([.] denotes the greatest integer function), then x belongs to

 5   5   5   5 
(A)  ,  (B)
, 4  (C)  , 2  (D)
 4 , 2 
 4    4  

3A The only posibility is [sinx] =-1 and [ 2 cos x ]=-2 .Draw Graph

ax b
a b
cx d
4. (a) : fof (x) x
ax b
c d
cx d
(ac + dc)x2 + (bc + d2 – bc – a2)x – ab – bd = 0
ac + dc = 0, a2 – d2 = 0, ab + bd = 0
So a = –d

5. If k and K are minimum and maximum values of

sin–1x + cos–1x + tan–1x respectively, then
p 3p
(a) k = , K = (b) k = 0, K = p
4 4
(c) k = p/2, K = p (d) not defined
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7.

8 AD

9.

On comparing, we can write a = 4, b = 2, c = 7

and d = 3.
a + c = 11, a + d = 7 and b + d = 5 are primes.
10. (a, b, c, d) : Given f(x) f(y) = f(x) + f(y) + f(xy) – 2
x, y ∈ R. f(x) 10 + x
16. (a) : e = , x ∈ (–10, 10)
Put x = y = 1 then f(1)2 – 3f(1) + 2 = 0 10 − x
⇒ f(1) = 1 or f(1) = 2  10 + x 
⇒ f(x) = log 
If f(1) = 1, then f(x) = 1 x ∈ R.  10 − x 
A contradiction ∵ degree of f(x) is positive.
 200 x 
\ f(1) ≠ 1. hence, f(1) = 2. 10 + 
 200 x  100 + x 2 
1 ⇒ f  = log 
Replace 'y' with then  100 + x 2  10 − 200 x 
x
 100 + x 2 
 1  1
f(x) f   = f(x) + f   (∵ f(1) = 2)
 x  x 2
10(10 + x )   10 + x 
\ f(x) must be in the form xn + 1 or –xn + 1. = log  = 2 log  = 2f(x)
10(10 − x) 
  10 − x 
∵ f(4) = 65, f(x) = x3 + 1 ⇒ f 1(x) = 3x2.
1  200 x  1
\ f(x) = f ⇒k= = 0.5.
2  100 + x 2  2
 7  7
11. (a, b, c, d) : f   = –0.5, g   = 0.5, g(3) = 0,
 2  2 Putting x = 1, f(1) = ek ⇒ 2 = ek. Hence from (i), f(x) = 2x.
f(0) = 23; f(2) = –1, g(2) = –1; f(4) = 0; g(4) = 26 This can also be obtained by putting y = 1 so that
2 F(x + 1) = f(x) f(1) = 2f(x) ⇒ f(x) = 2f(x – 1).
12, 13, 14 Putting successively x – 1, x – 2, x – 3, ...., 2 for x in the
above and multiplying them, we get f(x) = 2x.
n
Now, ∑ f (c + r ) = f(c + 1) + f(c + 2) +...+ f(c + n)
r =1
= 2c + 1 + 2c + 2 +...+ 2c + n
= 2c 2 + 2c 22 +...+ 2c 2n
= 2c 2{1 + 2 + 22 +... to n terms}
1(2n −1)
= 2c 2 = 2c 2(2n – 1)
2 −1
⇒ 16(2n – 1) = 2c + 1(2n – 1)
⇒ 2c + 1 = 16 = 24 ⇒ c + 1 = 4 \ c = 3.

15. (3) : From the given equation

f(x + y) = f(x) f(y), we have f(x) = ekx
Where k is a constant.
integral
17. 5
Sol. Suppose k  2  log2 a  3  log3 b  log6  a  b  , then we have
1 1 ab 6k
a  2k2 , b  3k3 , a  b  6k , and therefore,    k 2 k 3  22  33 .
a b ab 2 3

   x
18 (4) The period of sin  is 24
 12


   x  24     x   24     x    
as sin 
   sin    sin  2    sin  
 12   12   12   12 

   x  x  2
Similarly, the period of tan 
 3
is 3 and the period of cos  is 8.
  4 
4

Hence, the period of the given function   LCM of (24, 8, 3) = 24.

19 1

x
19. sin    1 x 2  2x  
2

x
 sin    1  x   
2

2

LHS  0 and RHS  0 .

Then solution is obtained if LHS = RHS = 0 and x   satisfies both.

20. (3) : f(1, 1) = f(0, f(1, 0)) = f(0, f(0, 1)) = f(0, 2) = 3