Question:
Let \(f(x)=\begin{vmatrix}\sin^2x&-2+\cos^2x&\cos2x\\2+\sin^2x&\cos^2x&\cos2x\\\sin^2x&\cos^2x&1+\cos2x\end{vmatrix},\ x\in[0,\pi]\). Then the maximum value of \(f(x)\) is equal to ______________.
Let \(f(x)=\begin{vmatrix}\sin^2x&-2+\cos^2x&\cos2x\\2+\sin^2x&\cos^2x&\cos2x\\\sin^2x&\cos^2x&1+\cos2x\end{vmatrix},\ x\in[0,\pi]\). Then the maximum value of \(f(x)\) is equal to ______________.
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Apply R2 -> R2 - R1 and R3 -> R3 - R1 to get simple rows (2,2,0) and (0,2,1), expand to get f(x) = 4 + 2cos 2x, then take its maximum.
Updated On: Nov 4, 2026
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Correct Answer: 6
Solution and Explanation
Let \(s=\sin^2x,\ c=\cos^2x,\ k=\cos2x=c-s\).
Rows: \(R_1=(s,\ c-2,\ k)\), \(R_2=(s+2,\ c,\ k)\), \(R_3=(s,\ c,\ 1+k)\).
\(R_2\to R_2-R_1=(2,2,0)\) and \(R_3\to R_3-R_1=(0,2,1)\) (determinant unchanged).
Expanding along \(R_1\): \(f=s(2\cdot1-0\cdot2)-(c-2)(2\cdot1-0\cdot0)+k(2\cdot2-2\cdot0)=2s-2(c-2)+4k\).
\(=2s-2c+4+4k=4-2k+4k=4+2k\) (using \(c-s=k\)).
So \(f(x)=4+2\cos2x\), whose maximum is \(4+2=6\) (at \(x=0,\pi\)).
Answer: 6
Rows: \(R_1=(s,\ c-2,\ k)\), \(R_2=(s+2,\ c,\ k)\), \(R_3=(s,\ c,\ 1+k)\).
\(R_2\to R_2-R_1=(2,2,0)\) and \(R_3\to R_3-R_1=(0,2,1)\) (determinant unchanged).
Expanding along \(R_1\): \(f=s(2\cdot1-0\cdot2)-(c-2)(2\cdot1-0\cdot0)+k(2\cdot2-2\cdot0)=2s-2(c-2)+4k\).
\(=2s-2c+4+4k=4-2k+4k=4+2k\) (using \(c-s=k\)).
So \(f(x)=4+2\cos2x\), whose maximum is \(4+2=6\) (at \(x=0,\pi\)).
Answer: 6
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