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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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If $f(x)=\begin{vmatrix}1&a&a^2\\\sin(n-1) x&\sin nx&\sin(n+1)x\\\cos(n-1)x&\cos nx&\cos(n+1)x\end{vmatrix}$ then $\int_0^{\large\frac{\pi}{2}}f(x)dx$ is equal to

$\begin{array}{1 1}(A)\;a-(1+a^2)&(B)\;1+a+a^2\\(C)\;-a+(1+a^2)&(D)\;-(1+a+a^2)\end{array}$

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1 Answer

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Apply $C_1+C_3-2(\cos x)C_2$
$f(x)=\begin{vmatrix}1+a^2-2a\cos x&a&a^2\\\sin(n-1) x+\sin(n+1)x-2\cos x\sin nx&\sin nx&\sin(n+1)x\\\cos(n-1)x+\cos(n+1)x-2\cos x\cos nx&\cos nx&\cos(n+1)x\end{vmatrix}$
$\Rightarrow \begin{vmatrix} 1+a^2-2a\cos x& a&a^2\\0& \sin nx&\sin(n+1)x\\0&\cos nx & \cos(n+1)x\end{vmatrix}$
$\Rightarrow (1+a^2-2a\cos x)[\sin nx\cos(n+1)x-\cos nx\sin (n+1)x]$
$\Rightarrow -(1+a^2-2a\cos x)\sin x$
$\Rightarrow \int_0^{\large\frac{\pi}{2}}f(x) dx=-(1+a^2)\int_0^{\large\frac{\pi}{2}}\sin xdx+\int _0^{\large\frac{\pi}{2}}\sin xdx$
$\Rightarrow a-(1+a^2)$
Hence (A) is the correct answer.
answered Apr 22, 2014 by sreemathi.v

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