Question

If $f\left(x\right)=\left|\begin{array}{ccc}\sin x& 1& 0\\ 1& 2\sin x& 1\\ 0& 1& 2\sin x\end{array}\right|$ then ${\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx$ equals to,

• A. $0$
• B. $-1$
• C. $1$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is A $0$

$f\left(x\right)=\left|\begin{array}{ccc}\sin x& 1& 0\\ 1& 2\sin x& 1\\ 0& 1& 2\sin x\end{array}\right|$
$f\left(x\right)=$ $\sin x(4{\sin }^{2}x-1)-1\left(2\sin x\right)$

$f\left(x\right)=4{\sin }^{3}x-3\sin x$

$\therefore f(-x)=-4{\sin }^{3}x+3\sin x$

$f(-x)=-f\left(x\right)$

$\therefore f\left(x\right)$ is an odd function

$\therefore {\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx=0$