Question

If $f\left(x\right)=\left|\begin{array}{ccc}\cos x& e^{x^2}& 2x{\cos }^2\frac{x}{2}\\ x^2& \sec x& \sin x+x^3\\ 1& 2& x+\tan x\end{array}\right|$, then the value of $\underset{-\pi /2}{\overset{\pi /2}{\int }}(x^2+1)\left[f\right(x)+f^{\prime \prime }(x\left)\right]dx$

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

As,
$f\left(x\right)=\left|\begin{array}{ccc}\cos x& e^{x^2}& 2x{\cos }^2\frac{x}{2}\\ x^2& \sec x& \sin x+x^3\\ 1& 2& x+\tan x\end{array}\right|\Rightarrow f(-x)=\left|\begin{array}{ccc}\cos x& e^{x^2}& -2x{\cos }^2\frac{x}{2}\\ x^2& \sec x& -(\sin x+x^3)\\ 1& 2& -(x+\tan x)\end{array}\right|\Rightarrow f(-x)=-f\left(x\right)$
So, $f\left(x\right)$ is an odd function.
Differentiating w.r.t $x$
$\Rightarrow -f^{\prime }(-x)=-f^{\prime }\left(x\right)\Rightarrow f^{\prime }(-x)=f^{\prime }\left(x\right)$
Again Differentiating
$\Rightarrow -f^{\prime \prime }(-x)=f^{\prime \prime }\left(x\right)\Rightarrow f^{\prime \prime }(-x)=-f^{\prime \prime }\left(x\right)$
So, $f^{\prime \prime }\left(x\right)$ is an odd function
Thus, $f\left(x\right)+f^{\prime \prime }\left(x\right)$ is odd function
Let,
$\varphi \left(x\right)=(x^2+1)\cdot \left[f\right(x)+f^{\prime \prime }(x\left)\right]\Rightarrow \varphi (-x)=-\varphi \left(x\right)$
Therefore, $\varphi \left(x\right)$ is an odd function.
$\therefore \underset{-\pi /2}{\overset{\pi /2}{\int }}\varphi \left(x\right)dx=0$