Question

If $\Delta (x) \begin{vmatrix}1&\cos x&1 -\cos x\\ 1+ \sin x& \cos x &1+ \sin x - \cos x\\ \sin x &\sin x&1\end{vmatrix},$ then $ \int\limits^{\pi / 4}_{0} \Delta\left(x\right)dx $ is equal to

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

Answer 1

The Correct Option is D

$\Delta (x) = \begin{vmatrix}1&\cos x&1 -\cos x\\ 1+ \sin x & \cos x &1+ \sin x - \cos x\\ \sin x &\sin x&1\end{vmatrix}, $
Applying $C_{3} \to C_{3} + C_{2} - C_{1}$
$\Delta\left(x\right) = \begin{vmatrix}1&\cos x &0\\ 1 +\sin x&\cos x &0\\ \sin x&\sin x &1\end{vmatrix}$
$ = \cos x - \cos x \left( 1 + \sin x \right) $
$[\because$ expanding along $C_{3}$]
$= - \cos x. \sin x = - \frac{1}{2} \sin \ 2x$
$\therefore \int\limits^{\pi /4}_{0} \Delta\left(x\right)dx $
$= - \frac{1}{2} \int\limits^{\pi/ 4}_{0} \sin \ 2x \ dx $
$= - \frac{1}{2}\left[- \frac{\cos2x}{2}\right]^{\pi / 4}_{0}$
$ = + \frac{1}{2 \times2} \left[ \cos \frac{\pi}{2} - \cos0^{\circ}\right]$
$ = \frac{1}{4} \left(0-1\right) = - \frac{1}{4} $