Let f:[0,2] R be a function which is continuous on [0,2] and is d… - Vidyadip

MCQ

Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$.

Let $F(x)=\int_0^{x^2} f(\sqrt{t}) d t$ for $x \in[0,2]$. If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$, then $F(2)$ equals

A
$e^2-1$
✓
$e^4-1$
C
$e-1$
D
$e^4$

✓

Answer

Correct option: B.

$e^4-1$

b
$f^{\prime}(x)=2 x f(x) $

$\frac{f^{\prime}(x)}{f(x)}=2 x $

$\ell(f(x))=x^2+c $

$x=0, f(0)=1 $

$c=0 $

$\therefore \quad \ell n(f(x))=x^2 $

$f(x)=e^{x^2} $

$\therefore \quad F(x)=f(x)+c $

$F( x )= e ^{ x ^2}+ C $

$F(0)=0 $

$\therefore \quad c=-1 $

$\therefore \quad f(x)=e^{x^2}-1$

$f(2)=e^4-1 $

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