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Let f:[0,2]vecR be a function which is c...

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`

A

`e^(2)-1`

B

`e^(4)-1`

C

` e - 1`

D

`e^(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
PLAN New ton - Leibnitz ' s formula
`(d)/(dx) [ int_(phi(x))^(Psi(x))f (t)dt ] = f { Psi (x) }{(d)/(dx)Psi(x)}-f{phi(x)}{(d)/(dx)phi(x)}`
Given , `F(x) = int _(0)^(x^(2))f(sqrt(t))dt`
` :. F' (x) =2x f(x)`
Also , ` F' (x) = f'(x)`
`rArr 2x f(x) = f ' (x)`
`rArr (f'(x))/(f(x))=2x`
`rArr int(f'(x))/(f(x))dx = int 2x dx rArr " In" f(x) = x^(2) +c`
`rArr f(x) =e^(x^(2)+c)rArr f(x) = Ke^(x^(2))[K=e^(c)]`
Now , `f(0)=1`
`:. 1=K`
Hence , `f(x)=e^(x^(2))`
`F(2) = int _(0)^(4) e^(t) dt = [e^(t)] = e^(4)-1`
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