Question:
The sum of absolute maximum and absolute minimum values of the function
\(f(x) = |2x^2 + 3x - 2| + \sin x \cos x\)
in the interval [0, 1] is :
The sum of absolute maximum and absolute minimum values of the function
\(f(x) = |2x^2 + 3x - 2| + \sin x \cos x\)
in the interval [0, 1] is :
\(f(x) = |2x^2 + 3x - 2| + \sin x \cos x\)
in the interval [0, 1] is :
Show Hint
Factor \(2x^2+3x-2 = (2x-1)(x+2)\), split at \(x=\frac12\), and show f decreases on [0, 1/2] and increases on [1/2, 1].
Updated On: Nov 4, 2026
- \(3 + \frac{\sin(1)\cos^2\left(\frac{1}{2}\right)}{2}\)
- \(3 + \frac{1}{2}(1 + 2\cos(1))\sin(1)\)
- \(5 + \frac{1}{2}(\sin(1) + \sin(2))\)
- \(2 + \sin\left(\frac{1}{2}\right)\cos\left(\frac{1}{2}\right)\)
Show Solution
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The Correct Option is B
Solution and Explanation
\(2x^2 + 3x - 2 = (2x-1)(x+2)\) is negative on [0, 1/2) and positive on (1/2, 1].
On [0, 1/2]: \(f(x) = -2x^2 - 3x + 2 + \frac{\sin 2x}{2}\), \(f'(x) = -4x - 3 + \cos 2x \leq -2 \lt 0\), so f decreases.
On [1/2, 1]: \(f'(x) = 4x + 3 + \cos 2x \gt 0\), so f increases.
Hence the minimum is \(f\left(\frac12\right) = 0 + \sin\frac12\cos\frac12 = \frac{\sin 1}{2}\).
The maximum is the larger of \(f(0) = 2\) and \(f(1) = 3 + \sin 1\cos 1 = 3 + \frac{\sin 2}{2} \approx 3.45\), i.e. \(f(1)\).
Sum \(= 3 + \sin 1\cos 1 + \frac{\sin 1}{2} = 3 + \frac12 \sin 1\,(2\cos 1 + 1)\).
Answer: (2) \(3 + \frac{1}{2}(1 + 2\cos(1))\sin(1)\)
On [0, 1/2]: \(f(x) = -2x^2 - 3x + 2 + \frac{\sin 2x}{2}\), \(f'(x) = -4x - 3 + \cos 2x \leq -2 \lt 0\), so f decreases.
On [1/2, 1]: \(f'(x) = 4x + 3 + \cos 2x \gt 0\), so f increases.
Hence the minimum is \(f\left(\frac12\right) = 0 + \sin\frac12\cos\frac12 = \frac{\sin 1}{2}\).
The maximum is the larger of \(f(0) = 2\) and \(f(1) = 3 + \sin 1\cos 1 = 3 + \frac{\sin 2}{2} \approx 3.45\), i.e. \(f(1)\).
Sum \(= 3 + \sin 1\cos 1 + \frac{\sin 1}{2} = 3 + \frac12 \sin 1\,(2\cos 1 + 1)\).
Answer: (2) \(3 + \frac{1}{2}(1 + 2\cos(1))\sin(1)\)
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