Q105 / 160 · AP EAMCET 2010 · Mathematics
If $f(x)=\left|\begin{array}{ccc}2 \cos x & 1 & 0 \\ x-\frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|$ then $f^{\prime}(\pi)$ is equal to
- A0
- Correct answerB2
- C$\frac{\pi}{2}$
- D$\pi-6$
Solution
$f(x)=\left|\begin{array}{ccc}2 \cos x & 1 & 0 \\ x-\frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|$
$f^{\prime}(x)=\left|\begin{array}{ccc}-2 \sin x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos x & 0 & 0 \\ x-\frac{\pi}{2} & -2 \sin x & 1 \\ 0 & 0 & 2 \cos x\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos x & 1 & 0 \\ x-\frac{\pi}{2} & 2 \cos x & 0 \\ 0 & 1 & -2 \sin x\end{array}\right|$
$f^{\prime}(\pi)=\left|\begin{array}{ccc}-2 \sin \pi & 1 & 0 \\ 1 & 2 \cos \pi & 1 \\ 0 & 1 & 2 \cos \pi\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos \pi & 0 & 0 \\ \pi-\frac{\pi}{2} & -2 \sin \pi & 1 \\ 0 & 0 & 2 \cos \pi\end{array}\right|$ $+\left|\begin{array}{ccc}2 \cos \pi & 1 & 0 \\ \pi-\frac{\pi}{2} & 2 \cos \pi & 0 \\ 0 & 1 & -2 \sin \pi\end{array}\right|$
$f^{\prime}(\pi)=\left|\begin{array}{rrr}0 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -2\end{array}\right|$ $+\left|\begin{array}{ccc}-2 & 0 & 0 \\ \pi / 2 & 0 & 1 \\ 0 & 0 & -2\end{array}\right|+\left|\begin{array}{ccc}-2 & 1 & 0 \\ \pi / 2 & -2 & 0 \\ 0 & 1 & 0\end{array}\right|$
$\left[\begin{array}{l}\because \sin \pi=0 \\ \cos \pi=-1\end{array}\right]$
$f^{\prime}(\pi)=-(-2+0)=2$
Hence, the correct answer is option: 'B'.