Let $f:[0,2] \in R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int\limits_0^{x^2} f(\sqrt{t}) d t$ for $x \in[0,2]$. If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$, then $F(2)$ equals
Answer & explanation
Correct answer: option 2
We have, $F(x)=\int\limits_0^{x^2} f(\sqrt{t}) d t$
∴ $F'(x)=2 x f(x)$
$\Rightarrow f'(x)=2 x f(x)$ $\left[∵ F'(x)=f'(x)\right]$
$\Rightarrow \frac{f'(x)}{f(x)}=2 x$
$\Rightarrow \log (f(x))=x^2+\log C$
$\Rightarrow f(x)=C e^{x^2}$
$\Rightarrow f(0)=C e^0=C$
$\Rightarrow 1=C$ [∵ f(0) = 1]
∴ $f(x)=e^{x^2}$
Now, $F(x)=\int\limits_0^{x^2} f(\sqrt{t}) d t$ and $f(x)=e^{x^2}$
$\Rightarrow F(x)=\int\limits_0^{x^2} e^t d t=e^{x^2}-1$
$\Rightarrow F(2)=e^4-1$