Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
then $\int_{0}^{\pi / 4} \Delta(x) d x$ is equal to
Solution:
2732 Upvotes
Verified Answer
The correct answer is:
$-\frac{1}{4}$
$\Delta(x)=\left|\begin{array}{ccc}1 & \cos x & 1-\cos x \\ 1+\sin x & \cos x & 1+\sin x-\cos x \\ \sin x & \sin x & 1\end{array}\right|$
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 $\left.C_{3}\right]=-\cos x \cdot \sin x$
$\quad \therefore \int_{0}^{\pi / 4} \Delta(x) d x=-\frac{1}{2} \int_{0}^{\pi / 4} \sin 2 x$
$=-\frac{1}{2}\left[-\frac{\cos 2 x}{2}\right]_{0}^{\pi / 4}$
$=+\frac{1}{2 \times 2}\left[\cos \frac{\pi}{2}-\cos 0^{\circ}\right]$
$=\frac{1}{4}(0-1)=-\frac{1}{4}$
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 $\left.C_{3}\right]=-\cos x \cdot \sin x$
$\quad \therefore \int_{0}^{\pi / 4} \Delta(x) d x=-\frac{1}{2} \int_{0}^{\pi / 4} \sin 2 x$
$=-\frac{1}{2}\left[-\frac{\cos 2 x}{2}\right]_{0}^{\pi / 4}$
$=+\frac{1}{2 \times 2}\left[\cos \frac{\pi}{2}-\cos 0^{\circ}\right]$
$=\frac{1}{4}(0-1)=-\frac{1}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.