Question

If a differentiable function ' $f$ ' satisfies the relation $\underset{0}{\overset{t^2}{\int }}{f}^2(x)dx+\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}{t}^6{\sin }^{3}xdx=\underset{0}{\overset{t^2}{\int }}2xf(x)dx\forall t\in R$, then find the number of solution (s) of the equation $f(x)=\sqrt[3]{x}$

Answer 1

Answer: 2

$\underset{0}{\overset{t^2}{\int }}{f}^2(x)dx+\frac{1}{2}{t}^6\underset{0}{\overset{\pi /2}{\int }}{\sin }^{3}xdx=\underset{0}{\overset{t^2}{\int }}2xf(x)dx$
$\underset{0}{\overset{t^2}{\int }}{f}^2(x)dx+\frac{t^6}{2}\cdot \frac{2}{3}=\underset{0}{\overset{t^2}{\int }}2xf(x)dx$
$f^2\left(t^2\right)\cdot 2t+2t^5=2t^{2}f\left(t^2\right)\cdot 2t$
$f^2\left(t^2\right)+t^4-2t^{2}f\left(t^2\right)=0\Rightarrow {\left(f\left(t^2\right)-t^2\right)}^2=0$
$\Rightarrow f\left(t^2\right)=t^2$
$\Rightarrow f(x)=x\forall x\ge 0$
From the graph, it is clear that $x=0$ and $x=1$ are only two solutions.