Question

If a differentiable function ' $f$ ' satisfies the relation $\int \limits _0^{t^2} f^2(x) d x+\frac{1}{2} \int\limits _0^{\pi / 2} t^6 \sin ^3 x d x=\int\limits _0^{t^2} 2 x f(x) d x \forall t \in R$, then find the number of solution (s) of the equation $f(x)=\sqrt[3]{x}$

Answer 1

$\int\limits_0^{ t ^2} f ^2( x ) dx +\frac{1}{2} t ^6 \int\limits_0^{\pi / 2} \sin ^3 xdx =\int\limits_0^{ t ^2} 2 x f( x ) dx $
$\int\limits_0^{ t ^2} f ^2( x ) dx +\frac{ t ^6}{2} \cdot \frac{2}{3}=\int\limits_0^{ t ^2} 2 xf ( x ) dx $
$f ^2\left( t ^2\right) \cdot 2 t +2 t ^5=2 t ^2 f \left( t ^2\right) \cdot 2 t$
$f ^2\left( t ^2\right)+ t ^4-2 t ^2 f \left( t ^2\right)=0 \Rightarrow\left( f \left( t ^2\right)- t ^2\right)^2=0$
$\Rightarrow f \left( t ^2\right)= t ^2 $
$\Rightarrow f ( x )= x \forall x \geq 0$
From the graph, it is clear that $x=0$ and $x=1$ are only two solutions.