Q57 / 120 · VITEEE 2014 · Mathematics
If $\Delta(x)=\left|\begin{array}{ccc}1 & \cos x & 1-\cos x \\ 1+\sin x & \cos x & 1+\sin x-\cos x \\ \sin x & \sin x & 1\end{array}\right|$,
then $\int_{0}^{\pi / 4} \Delta(x) d x$ is equal to
- A$\frac{1}{4}$
- B$\frac{1}{2}$
- C0
- Correct answerD$-\frac{1}{4}$
Solution
$\Delta(x)=\left|\begin{array}{ccc}1 & \cos x & 1-\cos x \\ 1+\sin x & \cos x & 1+\sin x-\cos x \\ \sin x & \sin x & 1\end{array}\right|$
Applying $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}-\mathrm{C}_{1}$
$\Delta(x)=\left|\begin{array}{ccc}
1 & \cos x & 0 \\
1+\sin x & \cos x & 0 \\
\sin x & \sin x & 1
\end{array}\right|$
$=\cos x-\cos x(1+\sin x)$
$\quad\left[\because\right.$ expanding along $\left.C_{3}\right]=-\cos x \cdot \sin x$
$\quad \therefore \int_{0}^{\pi / 4} \Delta(x) d x=-\frac{1}{2} \int_{0}^{\pi / 4} \sin 2 x$
$=-\frac{1}{2}\left[-\frac{\cos 2 x}{2}\right]_{0}^{\pi / 4}$
$=+\frac{1}{2 \times 2}\left[\cos \frac{\pi}{2}-\cos 0^{\circ}\right]$
$=\frac{1}{4}(0-1)=-\frac{1}{4}$
Hence, the correct answer is option: 'D'.