Question

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

• A. $f\left(\frac{\pi}{3}\right)=1$
• B. $f^{\prime}\left(\frac{\pi}{3}\right)=-\sqrt{3}$
• C. $f\left(\frac{\pi}{2}\right)=-1$
• D. none of these

Answer 1

Answer: (B)

Expanding along $C_1$, we get
$f(x) =2 \cos x\begin{vmatrix}2 \cos x & 1 \\1 & 2 \cos x\end{vmatrix}-\begin{vmatrix}1 & 0 \\1 & 2 \cos x\end{vmatrix}$
$ =8 \cos ^3 x-4 \cos x=4 \cos x \cos 2 x$
$ \Rightarrow f\left(\frac{\pi}{3}\right)=-1, f\left(\frac{\pi}{2}\right)=0 $
$f^{\prime}(x) =-4[\sin x \cos 2 x+2 \cos x \sin 2 x] $
$ =-4[\sin 3 x+\cos x \sin 2 x]$
$\Rightarrow f^{\prime}\left(\frac{\pi}{3}\right)=-4\left[\sin \pi+\cos \left(\frac{\pi}{3}\right) \sin \left(\frac{2 \pi}{3}\right)\right] $
$ =-\sqrt{3}$