Let `f:[0,2]vecR` be a function which is continuous on [0,2] and is differentiable on (0,2) with `f(0)=1` `L e t :F(x)=int_0^(x^2)f(sqrt(t))dtforx in [0,2]dotIfF^(prime)(x)=f^(prime)(x)` . for all `x in (0,2),` then `F(2)` equals `e^2-1` (b) `e^4-1` `e-1` (d) `e^4`

Options

A. `e^(2)-1`
B. `e^(4)-1`
C. `e-1`
D. `e^(4)`

Answer

B

Explanation

We have, F(x)`=overset(x^(2))underset(0)int f(sqrt(t))dt`.
`:. F'(x)=2xf(x)`
`rArr f'(x)=2xf(x) " "[:' F'(x)=f'(x)]`
`rArr (f'(x))/(f(x))=2x`
`rArr log(f(x))=x^(2)+logC`
`rArr f(x)=Ce^(x^(2))`
`rArr f(0)=C e^(0)=C`
`rArr 1=C " "[:' f(0)=1]`
`:. f(x)=e^(x^(2))`
Now, `F(x)=overset(x^(2))underset(0)int f(sqrt(t))dt` and f(x)`=e^(x^(2))`
`rArr F(x)=overset(x^(2))underset(0)int e^(t) dt =e^(x^(2))-1`
`rArr F(2)=e^(4)-1`