Question:
The maximum value of
\(f(x)=\left|\begin{matrix}\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1+\sin^2 x & \cos^2 x & \cos 2x\\ \sin^2 x & \cos^2 x & \sin 2x\end{matrix}\right|,\ x\in R\)
is :
The maximum value of
\(f(x)=\left|\begin{matrix}\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1+\sin^2 x & \cos^2 x & \cos 2x\\ \sin^2 x & \cos^2 x & \sin 2x\end{matrix}\right|,\ x\in R\)
is :
\(f(x)=\left|\begin{matrix}\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1+\sin^2 x & \cos^2 x & \cos 2x\\ \sin^2 x & \cos^2 x & \sin 2x\end{matrix}\right|,\ x\in R\)
is :
Show Hint
Use row operations R2 - R1 and R3 - R1 to simplify the determinant to a single trigonometric expression, then use the maximum of \(|A\cos 2x + B\sin 2x|\).
Updated On: Nov 4, 2026
- \(\sqrt{5}\)
- \(\frac{3}{4}\)
- \(5\)
- \(\sqrt{7}\)
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The Correct Option is A
Solution and Explanation
Apply \(R_2\to R_2-R_1\) and \(R_3\to R_3-R_1\):
\(f(x)=\left|\begin{matrix}\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1 & -1 & 0\\ 0 & -1 & \sin 2x-\cos 2x\end{matrix}\right|\).
Expand along the first row:
\(f(x)=\sin^2 x\,[-(\sin 2x-\cos 2x)]-(1+\cos^2 x)(\sin 2x-\cos 2x)+\cos 2x\,(-1)\)
\(=-(1+\sin^2 x+\cos^2 x)(\sin 2x-\cos 2x)-\cos 2x=-2(\sin 2x-\cos 2x)-\cos 2x\)
\(=\cos 2x-2\sin 2x\).
So \(f(x)=|\cos 2x-2\sin 2x|\), whose maximum is \(\sqrt{1^2+2^2}=\sqrt{5}\).
Answer: (1) \(\sqrt{5}\)
\(f(x)=\left|\begin{matrix}\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1 & -1 & 0\\ 0 & -1 & \sin 2x-\cos 2x\end{matrix}\right|\).
Expand along the first row:
\(f(x)=\sin^2 x\,[-(\sin 2x-\cos 2x)]-(1+\cos^2 x)(\sin 2x-\cos 2x)+\cos 2x\,(-1)\)
\(=-(1+\sin^2 x+\cos^2 x)(\sin 2x-\cos 2x)-\cos 2x=-2(\sin 2x-\cos 2x)-\cos 2x\)
\(=\cos 2x-2\sin 2x\).
So \(f(x)=|\cos 2x-2\sin 2x|\), whose maximum is \(\sqrt{1^2+2^2}=\sqrt{5}\).
Answer: (1) \(\sqrt{5}\)
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