Question

If $f\left(x\right)=\left|\begin{array}{ccc}2& \sin 2x& -2\\ 3& x^2+3x& \pi +3\\ 5& 5x+\cos x& 4\end{array}\right|,$ then which of the following(s) is a solution of $f^{\prime }\left(x\right)=0$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A), (B)

$\frac{\pi }{2}$

$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2& 2\cos 2x& -2\\ 3& 2x+3& \pi +3\\ 5& 5-\sin x& 4\end{array}\right|$
Clearly we can observe, for $x=0$ we have
$f^{\prime }\left(0\right)=\left|\begin{array}{ccc}2& 2& -2\\ 3& 3& \pi +3\\ 5& 5& 4\end{array}\right|\cdots \left(i\right)$
and for $x=\frac{\pi }{2}$$f^{\prime }\left(\frac{\pi }{2}\right)=\left|\begin{array}{ccc}2& -2& -2\\ 3& \pi +3& \pi +3\\ 5& 4& 4\end{array}\right|\cdots \left(ii\right)$
Clearly two columns are identical in $\left(i\right)$ and $\left(ii\right)$. Hence, the values of both determinant are zero.
So, $x=0,\frac{\pi }{2}$ are solutions of $f^{\prime }\left(x\right)=0$
Also $f^{\prime }\left(\pi \right)\ne 0$ and $f^{\prime }\left(\frac{\pi }{4}\right)\ne 0$