Question

Let $f\left(x\right)=\left|\begin{array}{ccc}\cos x& \sin x& \cos x\\ \cos 2x& \sin 2x& 2\cos 2x\\ \cos 3x& \sin 3x& 3\cos 3x\end{array}\right|$ then $f^{\prime }\left(\frac{\pi }{2}\right)=$.

• A. $0$
• B. $-12$
• C. $4$
• D. $12$

Answer 1

Answer: (C)

$4$

Given $f\left(x\right)=\left|\begin{array}{ccc}\cos x& \sin x& \cos x\\ \cos 2x& \sin 2x& 2\cos 2x\\ \cos 3x& \sin 3x& 3\cos 3x\end{array}\right|$

differentiate on both sides wrt $x$

$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}-\sin x& \cos x& -\sin x\\ \cos 2x& \sin 2x& 2\cos 2x\\ \cos 3x& \sin 3x& 3\cos 3x\end{array}\right|+\left|\begin{array}{ccc}\cos x& \sin x& \cos x\\ -2\sin 2x& 2\cos 2x& -4\sin 2x\\ \cos 3x& \sin 3x& 3\cos 3x\end{array}\right|+\left|\begin{array}{ccc}\cos x& \sin x& \cos x\\ \cos 2x& \sin 2x& 2\cos 2x\\ -3\sin 3x& 3\cos 3x& -9\sin 3x\end{array}\right|$

$\because$ differentiation of a determinant $=$ sum of determinants with each row differentiated

$f^{\prime }\left(\frac{\pi }{2}\right)=\left|\begin{array}{ccc}-1& 0& -1\\ -1& 0& -2\\ 0& -1& 0\end{array}\right|+\left|\begin{array}{ccc}0& 1& 0\\ 0& -2& 0\\ 0& -1& 0\end{array}\right|+\left|\begin{array}{ccc}0& 1& 0\\ -1& 0& -2\\ 3& 0& 9\end{array}\right|$

$=(2-1)+0+(-1(-9+6\left)\right)$

$=1+3=4$

$\therefore f^{\prime }\left(\frac{\pi }{2}\right)=4$