Question

# Let $$\displaystyle \begin{vmatrix}2\cos ^{2}x & \sin(2x) &-\sin x \\ \sin(2x)& 2\sin^{2}x & \cos x \\ \sin x & -\cos x & 0 \end{vmatrix}$$ Find $$\displaystyle \frac{1}{\pi}\int_{0}^{\pi/2}\left [ f(x)+f{}'(x) \right ]dx$$

Solution

## We have, $$f\left( x \right) \quad =\quad \begin{vmatrix} 2\cos ^{ 2 } x & \quad \sin (2x) & -\sin x \\ \sin (2x) & 2\sin ^{ 2 } x & \quad \cos x \\ \sin x & -\cos x & 0 \end{vmatrix}$$$$applying\quad C_{ 1 }\rightarrow C_{ 1 }-2\sin xC_{ 3 }\quad and\quad C_{ 2 }\rightarrow C_{ 2 }+2\cos xC_{ 3 }\quad we\quad get\quad \\ f\left( x \right) =\begin{vmatrix} 2 & 0 & -\sin x \\ 0 & 2 & \cos x \\ \sin x & \quad -\cos x & 0 \end{vmatrix}\\ \quad \quad \quad$$$$f(x)=\quad 2\sin ^{ 2 } x+2\cos ^{ 2 } x\quad =\quad 2\quad \Rightarrow \quad f'\left( x \right) \quad =\quad 0$$$$\Rightarrow \displaystyle \frac { 1 }{ \pi } \int _{ 0 }^{ \pi /2 } \left[ f(x)+f'(x) \right] dx =\displaystyle \frac { 1 }{ \pi } \int _{ 0 }^{ \pi /2 } 2dx$$$$f(x)=\displaystyle \frac { 1 }{ \pi } 2\left(\displaystyle \frac { \pi }{ 2 } \right) =1$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More