Question

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

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

Answer 1

Answer: (C)

$F\left(x^2\right)=x^2(1+x)$
Differentiating w.r.t to $x$, we get
$2x{F}^{{}^{\prime }}\left(x^2\right)=2x(1+x)+x^2\Rightarrow F^{{}^{\prime }}\left(x^2\right)=(1+x)+\frac{x}{2}$
Substituting $x^2=t\Rightarrow x=\sqrt{t}$, we get
$F^{{}^{\prime }}\left(t\right)=(1+\sqrt{t})+\frac{\sqrt{t}}{2}$ ...(1)
Now, in $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$
Using Leibnitz theorem,
$F^{{}^{\prime }}\left(x\right)=1\cdot f\left(x\right)+0$
Substituting value from (1),
$f\left(x\right)=(1+\sqrt{x})+\frac{\sqrt{x}}{2}f\left(4\right)=(1+\sqrt{4})+\frac{\sqrt{4}}{2}=4$