Question

If $f(x)=\frac{1}{9}\left|\begin{array}{ccc}\cos x& 1& 0\\ 1& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$, then $\frac{d^{2}f}{d{x}^2}=$

• A. $\cos 3x$
• B. $\cos (\pi +3x)$
• C. $\sin 3x$
• D. $\sin (\pi +3x)$

Answer 1

Answer: (B)

Given,
$f(x)=\frac{1}{9}\left|\begin{array}{ccc}\cos x& 1& 0\\ 1& 2\cos x& 1\\ 0& 1& 2\cos x\end{array}\right|$,
$=\frac{1}{9}\left[\cos x\left(4{\cos }^{2}x-1\right)-1(2\cos x)\right]$
$=\frac{1}{9}\left[4{\cos }^{3}x-\cos x-2\cos x\right]$
$=\frac{1}{9}\left[4{\cos }^{3}x-3\cos x\right]$
$f(x)=\frac{1}{9}\cos 3x$
$\Rightarrow f^{{}^{\prime }}(x)=\frac{1}{9}(-\sin 3x)3$
$\Rightarrow f^{{}^{\prime }}(x)=-\frac{1}{3}\sin 3x$
$\Rightarrow f^{{}^{\prime \prime }}(x)=-\frac{1}{3}(\cos 3x)(3)=-\cos 3x$
$\Rightarrow f^{{}^{\prime \prime }}(x)=\cos (\pi +3x)$
$\Rightarrow \frac{d^{2}f}{d{v}^2}=\cos (\pi +3x)$