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 ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\Delta \left(x\right)dx=$

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $0$
• D. $–\left(\frac{1}{4}\right)$

Answer 1

The correct option is D

$–\left(\frac{1}{4}\right)$

Finding the value of ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\Delta \left(x\right)dx$ :

$\begin{array}{rcl}\Delta \left(x\right)& =& \left|\begin{array}{ccc}1& \cos x& 1-\cos x\\ 1+\sin x& \cos x& 1+\sin x-\cos x\\ \sin x& \sin x& 1\end{array}\right|\\ Now,C_3& \to & C_3+C_2-C_1\\ \Delta \left(x\right)& =& \left|\begin{array}{ccc}1& \cos x& 0\\ 1+\sin x& \cos x& 0\\ \sin x& \sin x& 1\end{array}\right|\\ R_1& \to & R_1-R_2\\ \Delta \left(x\right)& =& \left|\begin{array}{ccc}-\sin x& 0& 0\\ 1+\sin x& \cos x& 0\\ \sin x& \sin x& 1\end{array}\right|\\ \Delta \left(x\right)& =& -\sin x·\cos x\end{array}$

$\begin{array}{rcl}\therefore {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\Delta \left(x\right)dx& =& -1{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\sin x·\cos xdx\\ & =& {\left[-\frac{{\sin }^{2}x}{2}\right]}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ & =& \frac{-1}{4}+0\\ & =& \frac{\mathbf{-}\mathbf{1}}{\mathbf{4}}\end{array}$

Hence, the correct answer is option D.