Question

Let $f:\left(0,+\infty \right)\to R$ and$F\left(x\right)={\int }_0^{x}f\left(t\right)dt$. If $F\left(x^2\right)=x^2\left(1+x\right)$, then $f\left(4\right)=$

• A. $\frac{5}{4}$
• B. $7$
• C. $4$
• D. $2$

Answer 1

The correct option is C

$4$

Explanation for correct option:

Finding the value of $f\left(4\right)$

Given data,

$f:\left(0,+\infty \right)\to R$

$F\left(x\right)={\int }_0^{x}f\left(t\right)dt$

$F\left(x^2\right)=x^2\left(1+x\right)$

Consider the given data as,

$$\begin{gathered} F\left(x^2\right)={\int }_0^{x^2}f\left(t\right)dt \\ {x}^2\left(1+x\right)={\int }_0^{x^2}f\left(t\right)dt \end{gathered}$$

Differentiate both the side with respect to $x$

$\frac{d\left(x^2\left(1+x\right)\right)}{dx}=\frac{d}{dx}{\int }_0^{x^2}f\left(t\right)dt$

We know that

$\frac{d}{dx}{\int }_a^{b}f\left(t\right)dt=f\left(b\right)\times \frac{db}{dx}-f\left(x\right)\frac{da}{dx}$

Then,

$$\begin{gathered} \Rightarrow \frac{d\left(x^2\left(1+x\right)\right)}{dx}=\frac{d}{dx}{\int }_0^{x^2}f\left(t\right)dt \\ \Rightarrow 2x+3x^2=f\left(x^2\right)\times 2x-0 \\ \Rightarrow f\left(x^2\right)=\frac{2x+3x^2}{2x} \\ \Rightarrow f\left(x^2\right)=1+\frac{3x}{2} \\ \Rightarrow f\left(2^2\right)=1+\frac{3\times 2}{2} \\ \Rightarrow f\left(4\right)=1+3 \\ \Rightarrow f\left(4\right)=4 \end{gathered}$$

Hence, the correct answer is option (C).