Question

Let $f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{2}x& \sin 2x& -\sin x\\ \sin 2x& 2{\sin }^{2}x& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$. then the value of ${\int }_0^{\pi /2}\left[f\right(x)+f^{\prime }(x\left)\right]dx$ is

• A. $\pi$
• B. $\pi /2$
• C. $2\pi$
• D. $3\pi /2$

Answer 1

The correct option is A $\pi$

applying $C_1\to C_1-2\sin x{C}_3;C_2\to C_2+2\cos x{C}_3$

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

$\therefore f^{\prime }\left(x\right)=0\therefore {\int }_0^{\pi /2}\left[f\right(x)+f^{\prime }(x\left)\right]=dx={\int }_0^{\pi /2}2dx=\pi$