Question

Let $\Delta (x)=\left|\begin{array}{ccc}{\cos }^{2}x& \cos x\sin x& -\sin x\\ \cos x\sin x& {\sin }^{2}x& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$ then $\underset{0}{\overset{\pi /2}{\int }}\left[\Delta (x)+{\Delta }^{\prime }(x)\right]dx$ equals

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

Answer 1

Answer: (B)

Applying $C_1\to C_1-\sin x{C}_3$ and $C_2\to C_2+$ $\cos x{C}_3$, we get
$\Delta (x)=\left|\begin{array}{ccc}1& 0& -\sin x\\ 0& 1& \cos x\\ \sin x& -\cos x& 0<br/>\end{array}\right|$
Applying $R_3\to R_3-\sin x{R}_1+\cos x{R}_2$, we get
$\Delta (x)=\left|\begin{array}{ccc}1& 0& -\sin x\\ 0& 1& \cos x\\ 0& 0& {\cos }^{2}x+{\sin }^{2}x\end{array}\right|=1$
$\Rightarrow {\Delta }^{\prime }(x)=0$
Thus, $\underset{0}{\overset{\pi /2}{\int }}\left[\Delta (x)+{\Delta }^{\prime }(x)\right]dx=\underset{0}{\overset{\pi /2}{\int }}dx=\frac{\pi }{2}$.