Question

Let $f:R\to R$ be a differentiable function satisfying $f(x)=x^2+3\underset{0}{\overset{x^{\frac{1}{3}}}{\int }}{e}^{-t^3}{t}^2\cdot f\left(x-t^3\right)dt$.
The value of determinant $\left|\begin{array}{ccc}f(0)& (f(3)-9)& (f(5)-25)\\ (f(-3)-9)& f(0)& (f(7)-49)\\ (f(-5)-25)& (f(-7)-49)& f(0)\end{array}\right|$ is equal to

• A. 0
• B. -1
• C. 2
• D. -3

Answer 1

Answer: (A)

$f(x)=x^2+3\underset{0}{\overset{x^{\frac{1}{3}}}{\int }}{e}^{-t^3}{t}^{2}f\left(x-t^3\right)dt$
Put $x-t^3=\lambda$ in second term of R.H.S and simplify it,
$f(x)=x^2+e^{-x}\underset{0}{\overset{x}{\int }}{e}^{\lambda }f(\lambda )d\lambda$
Finally, $f(x)=\left(\frac{x^3}{3}+x^2\right)$
It is skew symmetric determinant i.e. $\left|\begin{array}{ccc}0& 9& \frac{125}{3}\\ -9& 0& \frac{7^3}{3}\\ \frac{-125}{3}& \frac{-7^3}{3}& 0\end{array}\right|=0$