Question

Let $F\left(x\right)=$$\left|\left|\right|\begin{array}{ccc}1& 1+sinx& 1+sinx+cosx\\ 2& 3+2sinx& 4+3sinx+2cosx\\ 3& 6+3sinx& 10+6sinx+3cosx\end{array}\right|$ then $F^{\prime }\left(\frac{\pi }{2}\right)$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

$f\left(x\right)=\left|\begin{array}{ccc}1& 1+\sin x& 1+\sin x+\cos x\\ 2& 3+2\sin x& 4+3\sin x+2\cos x\\ 3& 6+3\sin x& 10+6\sin x+3\cos x\end{array}\right|$

$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}0& \cos x& \cos x-\sin x\\ 2& 3+2\sin x& 4+3\sin x+2\cos x\\ 3& 6+3\sin x& 10+6\sin x+\cos x\end{array}\right|+\left|\begin{array}{ccc}\left(1\right)& (1+\sin x)& 1+\sin x+\cos x\\ 0& 2\cos x& 3\cos x-2\sin x\\ 3& 6+3\sin x& 10+6\sin x+3\cos \end{array}\right|+\left|\begin{array}{ccc}1& 1+\sin x& 1+\sin x+\cos x\\ 2& 3+2\sin x& 4+3\sin x+2\cos x\\ 0& 3\cos x& 6\cos -3\sin x\end{array}\right|$ (differentiating row $mse$ )

$f^{\prime }\left(\frac{\pi }{2}\right)=\left|\begin{array}{ccc}0& 0& -1\\ 2& 5& 7\\ 3& 9& 16\end{array}\right|+\left|\begin{array}{ccc}1& 2& 2\\ 0& 0& -2\\ 3& 9& 16\end{array}\right|+\left|\begin{array}{ccc}1& 2& 2\\ 2& 5& 7\\ 0& 0& -3\end{array}\right|$

$-(18-15)+2(9-6)-3(5-4)$

$=-3+6-3=0$

$\therefore$ Option $B$ is correct.