Question

If $f\left(x\right)=\left|\begin{array}{ccc}2\cos x& 1& 0\\ x-\frac{\pi }{2}& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$ then $\frac{df}{dx}$ at $x=\frac{\pi }{2}$ is

• A. $2$
• B. $\frac{\pi }{2}$
• C. $1$
• D. $8$

Answer 1

The correct option is A $2$

$f\left(x\right)=\left|\begin{array}{ccc}2\cos x& 1& 0\\ x-\frac{\pi }{2}& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$
(Expanding along first column)
$=2\cos x(4{\cos }^{2}x-1)-(x-\frac{\pi }{2})(2\cos x-0)$
$=8{\cos }^{3}x-2\cos x-2x\cos x+\pi \cos x$
$\Rightarrow f^{\prime }\left(x\right)=24{\cos }^{2}x(-\sin x)+2\sin x-2(\cos x-x\sin x)-\pi \sin x$
$\therefore f^{\prime }\left(\frac{\pi }{2}\right)=0+2+0=2$