Question

If $f\left(x\right)=$ $\left|\begin{array}{ccc}si{n}^{2}x& co{s}^{2}x& 1\\ co{s}^{2}x& si{n}^{2}x& 1\\ x-12& 12& 2\end{array}\right|$ then $f^{\prime }\left(\frac{\pi }{2}\right)=$

• A. $-1$
• B. $0$
• C. $+1$
• D. $\pm 1$

Answer 1

The correct option is A $-1$

$c_2\to c_2+c_1$
$if\left(u\right)=$ $\left|\begin{array}{ccc}si{n}^{2}x& 1& 1\\ co{s}^{2}x& 1& 1\\ x-12& x& 2\end{array}\right|$
$R_2\to R_2+R_1$
$f\left(x\right)=$ $\left|\begin{array}{ccc}si{n}^{2}x& 1& 1\\ 1& 2& 2\\ x-12& x& 2\end{array}\right|$
${\left|f^{\prime }\left(x\right)\right|}_{\pi /2}={|\left|\begin{array}{ccc}si{n}^{2}x& 1& 1\\ 1& 2& 2\\ 1& 1& 0\end{array}\right|+\left|\begin{array}{ccc}2sinxcosx& 0& 0\\ 1& 2& 2\\ x-12& x& 2\end{array}\right||}_{\pi /2}$ and $cosx=0$ first row$=0$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 2\\ 1& 1& 0\end{array}\right|$
$c_2\to c_2-c_1$
$=\left|\begin{array}{ccc}1& 0& 1\\ 1& 1& 2\\ 1& 0& 0\end{array}\right|$
$=-1$