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

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

Answer 1

The correct option is A $\pi$

$f\left(x\right)=\left|\begin{array}{ccc}2{\cos }^{2}x& \sin 2x& -\sin x\\ \sin 2x& 2{\sin }^{2}x& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$
$C_1\to C_1-2\sin x{C}_3,C_2\to C_2+2\cos x{C}_3$
$f\left(x\right)=\left|\begin{array}{ccc}2& 0& -\sin x\\ 0& 2& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$
$\Rightarrow f\left(x\right)=2{\cos }^{2}x+2{\sin }^{2}x=2$
$\Rightarrow f^{\prime }\left(x\right)=0$
So, ${\int }_0^{\pi /2}\left(f\right(x)+f^{\prime }(x\left)\right)dx={\int }_0^{\pi /2}2dx=\pi$