Question

Let $f(x)=\left|\begin{array}{ccc}2\cos x& 1& 0\\ 1& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$ 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
$f(x)=2\cos x\left|\begin{array}{cc}2\cos x& 1\\ 1& 2\cos x\end{array}\right|-\left|\begin{array}{cc}1& 0\\ 1& 2\cos x\end{array}\right|$
$=8{\cos }^{3}x-4\cos x=4\cos x\cos 2x$
$\Rightarrow f\left(\frac{\pi }{3}\right)=-1,f\left(\frac{\pi }{2}\right)=0$
$f^{\prime }(x)=-4[\sin x\cos 2x+2\cos x\sin 2x]$
$=-4[\sin 3x+\cos x\sin 2x]$
$\Rightarrow f^{\prime }\left(\frac{\pi }{3}\right)=-4\left[\sin \pi +\cos \left(\frac{\pi }{3}\right)\sin \left(\frac{2\pi }{3}\right)\right]$
$=-\sqrt{3}$