Question

If $f\left(x\right)=\left|\begin{array}{ccc}{x}^2-4x+6& 2x^2+4x+10& 3x^2-2x+16\\ x-2& 2x+2& 3x-1\\ 1& 2& 3\end{array}\right|$, then

• A. $\underset{-3}{\overset{3}{\int }}\frac{x^2\sin x}{1+x^6}.f\left(x\right)dx=0$
• B. $f\left(x\right)$ is a constant function
• C. $f^{\prime }\left(x\right)$ is a constant function
  
• D. $\underset{-3}{\overset{3}{\int }}\frac{x^2\sin x}{1+x^6}.f\left(x\right)dx=2$

Answer 1

Answer: (A), (B), (C)

$f^{\prime }\left(x\right)$ is a constant function


Observe that the elements of row $R_3$ are the derivatives of the elements of row $R_2$ and they, in turn, are proportional to the derivatives of the elements of row $R_1$.
Therefore,
$f^{\prime }\left(x\right)=\left|\begin{array}{c}{R}_1^{\prime }\\ R_2\\ R_3\end{array}\right|+\left|\begin{array}{c}{R}_1\\ R_2^{\prime }\\ R_3\end{array}\right|+\left|\begin{array}{c}{R}_1\\ R_2\\ R_3^{\prime }\end{array}\right|=0,\forall x\in R$
$\Rightarrow f\left(x\right)=\text{constant}$
As $f\left(0\right)=\left|\begin{array}{ccc}6& 10& 16\\ -2& 2& -1\\ 1& 2& 3\end{array}\right|=2$

$\therefore f\left(x\right)=2,\forall x\in R$

$\underset{-3}{\overset{3}{\int }}\frac{x^2\sin x}{1+x^6}.f\left(x\right)dx=2\underset{-3}{\overset{3}{\int }}\frac{x^2\sin x}{1+x^6}dx$
Let $g\left(x\right)=\frac{x^2\sin x}{1+x^6}$
$g(-x)=\frac{-x^2\sin x}{1+x^6}=-g\left(x\right)$
Hence, $g$ is an odd function.
$\therefore \underset{-3}{\overset{3}{\int }}\frac{x^2\sin x}{1+x^6}.f\left(x\right)dx=0$