8/17/2019 Stp - 2 (Function) - s

1/2

IIT - ian’s PACE ; ANDHERI / DADAR / CHEMBUR / THANE ; Tel : 26245223 / 09 ; .www.iitianspace.com

 

216 - 217 ,2nd floor , Shopper’s point , S. V. Road. Andheri (West) Mumbai-400058 . Tel: 26245223 / 09

B A T C H : 2 01 7 STP - 2 FUN CT IO N ) T IM E : 15 M in .

1. (C)

We must have 0 < 2 + e3{x}  1 or e2  2 + e3{x}  e3

 – 2 < e3{x}  – 1 or e2 – 2  e3{x}  e3 – 2

  x    or )2e(n3

1}x{)2e(n

3

1 32   l l 

Hence

)2e(3

1n),2e(n

3

1nUx 32

Inl   Ans. ]

2. (B)

f (x) = | Ax + c | + d

g (x) = – | Ax + u | + v

figure is parallelogram and diagonals bisect each other 

   

  

 A

u +

 

  

 A

c = 3 + 1;  

A

cu  = – 4 Ans. ]

3.   (C)

log10

 (–x) = |x|log10

log10 (–x) = )x(log10     x < 0

let y = log10

 (–x), then y – y  = 0 i.e y  ( y  – 1) = 0

i.e y  = 0 or   y  – 1 = 0

i.e y = 0 or   y  = 1

i.e log10

 (–x) = 0 or log10

 (–x) = 1

i.e –x = 1 or –x = 10

i.e x = –1 or x = –10

Thus there are two solutions.

8/17/2019 Stp - 2 (Function) - s

2/2

IIT - ian’s PACE ; ANDHERI / DADAR / CHEMBUR / THANE ; Tel : 26245223 / 09 ; .www.iitianspace.com

4.99

(A)

For domain of f(x) = )x4(cos3 1 , we must have

cos –1 4x  3

   4x  

2

1 x 

8

1  .......(1)

Also –1  4x  1  4

1x

4

1

 ........ (2)

 From (1) and (2), we get x  

8

1,

4

1]

5.   A )

We must have [x] =

  x x x x

2

 [x]{x} + [x] ( [x] + {x}) = 2{x} ( [x] + {x} )

 [x]2 = 2{x}2  {x}2 = 2

][ 2 x

 0 < 2

][ 2 x