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If $$f(x)=\begin{vmatrix} \cos { x }  & 1 & 0 \\ 1 & 2\cos { x }  & 1 \\ 0 & 1 & 2\cos { x }  \end{vmatrix}$$, then $$\displaystyle \int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ f(x) } dx$$


A
14
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B
13
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C
12
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D
1
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Solution

The correct option is B $$-\cfrac { 1 }{ 3 } $$
Solving determinant by expanding across row 1, we get

$$f(x) = \cos x\begin{vmatrix} 2\cos x & 1\\1 & 2\cos x\end{vmatrix} -1\begin{vmatrix}1 & 1\\0 & 2\cos x\end{vmatrix} +0\begin{vmatrix} 1 & 2\cos x\\0 & 1\end{vmatrix}$$

$$\Rightarrow f(x) = \cos x(4\cos ^2x-1)-1(2\cos x-0)+0(1-0)$$

$$\Rightarrow f(x) = 4\cos ^3x - 3\cos x$$

$$\Rightarrow f(x) = \cos 3x$$

Now, $$\displaystyle \int^{\dfrac{\pi}{2}}_0 f(x)dx$$

$$=\displaystyle \int^{\dfrac{\pi}{2}}_0 \cos 3xdx$$

$$=\left(\dfrac{\sin 3x}{3}\right)^{\dfrac{\pi}{2}}_0$$

$$=\dfrac{1}{3}\left(\sin \dfrac{3\pi}{2} - \sin 0\right)$$

$$=\dfrac{1}{3} ( - 1 - 0 )$$

$$=- \dfrac{1}{3}$$

Mathematics

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