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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

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