Question

If $f\left(x\right)=\left|\begin{array}{ccc}2co{s}^{2}x& Sin2x& -Sinx\\ Sin2x& 2si{n}^{2}x& Cosx\\ Sinx& -Cosx& 0\end{array}\right|$ then ${\int }_0^{\pi /2}\left[f\right(x)+f^1(x\left)\right]dx=$

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

Answer 1

Answer: (B)

$\pi$

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

$=2{\cos }^{2}x\left({\cos }^{2}x\right)-\sin 2x\left(\sin x\cos x\right)-\sin x(-\sin 2x\cos x-2{\sin }^{3}x)$

$=2({\cos }^{4}x+{\sin }^{2}x)-{\sin }^{2}2x=2$

Therefore, $f^{\prime }\left(x\right)=0$

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