Question

If $\Delta \left(x\right)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|$ then ${\int }_0^{\pi /2}\Delta \left(x\right)dx$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $0$
• D. $-\frac{1}{2}$

Answer 1

Answer: (D)

$-\frac{1}{2}$

$\Delta \left(x\right)=\left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|$
Applying $C_1\to C_1-C_2-C_3$
$\Delta \left(x\right)=\left|\begin{array}{ccc}0& \cos x& 1-\cos x\\ 0& \cos x& 1+\sin x-\cos x\\ -1& \sin x& 1\end{array}\right|=-\cos x\sin x=-\frac{\sin 2x}{2}$
Therefore
${\int }_0^{\pi /2}\Delta \left(x\right)dx={\int }_0^{\pi /2}-\frac{\sin 2x}{2}dx=-\frac{1}{2}$
Hence, option 'D' is correct.