Question

Assertion :Let $f\left(x\right)={\left|\begin{array}{ccc}0& \cos x& -\sin x\\ \sin x& 0& \cos x\\ \cos x& \sin x& 0\end{array}\right|}^2$
If $\sin 2x=1$ then $f\left(x\right)=2/3$ Reason: $f\left(x\right)=0$ if $\sin x=\cos x$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is incorrect but Reason is correct
• D. Assertion is correct but Reason is correct

Answer 1

The correct option is C Assertion is incorrect but Reason is correct

Multiplication of two determinants leads us
$f\left(x\right)=\left|\begin{array}{ccc}1& -y& y\\ -y& 1& y\\ y& y& 1\end{array}\right|$
where $y=\sin x\cos x$
Using $C_1\to C_1-C_2,C_2\to C_2+C_3$ we get
$f\left(x\right)={(1+y)}^2\left|\begin{array}{ccc}1& 0& y\\ -1& 1& y\\ 0& 1& 1\end{array}\right|={(1+y)}^2\left|\begin{array}{ccc}1& 0& y\\ 0& 1& 2y\\ 0& 1& 1\end{array}\right|$
${(1+y)}^2(1-2y)$
when $\sin 2x=1,y=1/2$ and $f\left(x\right)=0$
when $\sin x=\cos x,1-2y=1-2{\sin }^{2}x=\cos 2x=0$
$\therefore f\left(x\right)=0$