`y=f(x)` satisfies the relation `int_(2)^(x)f(t)dt=(x^(2))/2+int_(x)^(2)t^(2)f(t)dt`
The range of `y=f(x)` is

Options

A. `[0,oo)`
B. `R`
C. `(-oo,0]`
D. `[-1/2,1/2]`

Answer

D

Explanation

`int_(2)^(x)f(t)dt=(x^(2))/2+int_(x)^(2)t^(2)f(t)dt`
Differentiating w.r.t `x` we get

`f(x)=x+(-x^(2)f(x))`
or `f(x)[1+x^(2)]=x`
Let `y=f(x)=x/(1+x^(2))`
or `yx^(2)-x+y=0`
Since `x` is real `Dge0`
or `1-4y^(2)ge0`
or `y epsilon [-1/2,1/2]`
Also `f(x)` is an odd funtion. Hence `int_(-2)^(2)f(x)dx=0`
`f'(x)=(1+x^(2)-2x^(2))/(1+x^(2))-(1-x^(2))/(1+x^(2))ge0`
or `x^(2)-1le0`
or `x epsilon[-1,1]`