Question:
If \(m\) and \(M\) are the minimum and the maximum values of \(4+\frac{1}{2}\sin^2 2x-2\cos^4x,\ x\in\mathbf{R}\), then \(M-m\) is equal to :
If \(m\) and \(M\) are the minimum and the maximum values of \(4+\frac{1}{2}\sin^2 2x-2\cos^4x,\ x\in\mathbf{R}\), then \(M-m\) is equal to :
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Write everything in terms of \(c=\cos^2x\) and treat it as a quadratic on \([0,1]\).
Updated On: Nov 11, 2026
- \(\frac{15}{4}\)
- \(\frac{9}{4}\)
- \(\frac{7}{4}\)
- \(\frac{1}{4}\)
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The Correct Option is B
Solution and Explanation
Let \(c=\cos^2x\in[0,1]\). Then \(\frac12\sin^22x=\frac12\cdot4\sin^2x\cos^2x=2c(1-c)\) and \(2\cos^4x=2c^2\).
Expression \(=4+2c-2c^2-2c^2=4+2c-4c^2\).
The downward parabola in c has its vertex at \(c=\frac14\): maximum \(M=4+\frac12-\frac14=\frac{17}{4}\).
The minimum on \([0,1]\) occurs at an endpoint: \(c=0\Rightarrow4\); \(c=1\Rightarrow4+2-4=2\). So \(m=2\).
\(M-m=\frac{17}{4}-2=\frac94\).
Answer: (2) \(\frac{9}{4}\)
Expression \(=4+2c-2c^2-2c^2=4+2c-4c^2\).
The downward parabola in c has its vertex at \(c=\frac14\): maximum \(M=4+\frac12-\frac14=\frac{17}{4}\).
The minimum on \([0,1]\) occurs at an endpoint: \(c=0\Rightarrow4\); \(c=1\Rightarrow4+2-4=2\). So \(m=2\).
\(M-m=\frac{17}{4}-2=\frac94\).
Answer: (2) \(\frac{9}{4}\)
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