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If f(x)=int(x^2)^(x^2+1)e^-t^2dt , then ...

If `f(x)=int_(x^2)^(x^2+1)e^-t^2dt` , then `f(x)` increases in `(0,2)` (b) no value of `x` `(0,oo)` (d) `(-oo,0)`

A

( 2 , 2 )

B

no value of x

C

`( 0 , oo)`

D

`(- oo , 0)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given , `f(x) = int_(x^(2))^(x^(2)+1)e^(-t^(2))dt`
On differentiating both sides using Newton ' s Leibnitz's formula , we get
`f'(x)=e^-(x^(2)+1){(d)/(dx)(x^(2)+1)}-e^(-(x^(2))^(2)){(d)/(dx)(x^(2))}`
`=e^(-(x^(2)+1)^(2))*2x-e^(-(x^(2))^(2))*2x`
`=2x^(-(x^(4)+2x^(2)+1))(1-e^(2x^(2)+1))`
[ where , `e^(2x^(2)+1)gt1, AA x and e^(-(x^(4)+2x^(2)+1))gt0,AAx]`
`:. f ' (x)gt 0`
which shows `2xlt0 or x lt 0 rArr x in (-oo,0)`
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