Question

Let $\Delta \left(x\right)$ = $\left|\begin{array}{ccc}co{s}^{2}x& cosxsinx& -sinx\\ cosxsinx& si{n}^{2}x& cosx\\ sinx& -cosx& 0\end{array}\right|$ then ${\int }_0^{\pi /2}\left[\Delta \right(x)+{\Delta }^1(x\left)\right]dx$ equals

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

Answer 1

The correct option is B $\frac{\pi }{2}$

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

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

$△\left(x\right)=\left|\begin{array}{ccc}1& 0& -\sin x\\ 0& 1& \cos x\\ \sin x& -\cos x& 0\end{array}\right|$

Applying $R_3\to R_3-\sin x{R}_1+\cos x{R}_2$

$△\left(x\right)=\left|\begin{array}{ccc}1& 0& -\sin x\\ 0& 1& \cos x\\ 0& 0& 1\end{array}\right|=1(1-0)-0-\sin x(0-0)=1$

$\therefore {△}^{\prime }\left(x\right)=0$

Now, ${\int }_0^{\pi /2}(△\left(x\right)+{△}^{\prime }\left(x\right))dx={\int }_0^{\pi /2}(1+0)dx={\left[x\right]}_0^{\pi /2}=\frac{\pi }{2}$