Question

Let $f: R \rightarrow R$ be a differentiable function satisfying $f(x)=x^2+3 \int\limits_0^{x^{\frac{1}{3}}} e^{-t^3} t^2 \cdot f\left(x-t^3\right) d t$.
The value of determinant $\begin{vmatrix} f (0) & ( f (3)-9) & ( f (5)-25) \\ ( f (-3)-9) & f (0) & ( f (7)-49) \\ ( f (-5)-25) & ( f (-7)-49) & f (0)\end{vmatrix}$ is equal to

• A. 0
• B. -1
• C. 2
• D. -3

Answer 1

Answer: (A)

$f(x)=x^2+3 \int\limits_0^{x^{\frac{1}{3}}} e^{-t^3} t^2 f\left(x-t^3\right) d t$
Put $ x-t^3=\lambda$ in second term of R.H.S and simplify it,
$f(x)=x^2+e^{-x} \int\limits_0^x e^\lambda f(\lambda) d \lambda$
Finally, $f(x)=\left(\frac{x^3}{3}+x^2\right)$
It is skew symmetric determinant i.e. $\begin{vmatrix}0 & 9 & \frac{125}{3} \\ -9 & 0 & \frac{7^3}{3} \\ \frac{-125}{3} & \frac{-7^3}{3} & 0\end{vmatrix}=0$