JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
The maximum value of $f(x)=\begin{vmatrix}\sin^{2}x&1+\cos^{2}x&\cos2x\\1+\sin^{2}x&\cos^{2}x&\cos2x\\\sin^{2}x&\cos^{2}x&\sin2x\end{vmatrix}$, $x\in\mathbf{R}$ is :
Determinant
Solution
Expanding and simplifying with $\sin^2x+\cos^2x=1$ gives $f(x)=\cos2x-2\sin2x$ (verified symbolically). Max $=\sqrt{1^2+2^2}=\sqrt5$. Option (A).