Question:
The function \(f(x) = 2\cos x - x + 3\) is
The function \(f(x) = 2\cos x - x + 3\) is
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If \(f'(x)<0\) for all \(x\) in an interval, the function is strictly decreasing.
Updated On: Apr 24, 2026
- increasing in \((0,\pi)\)
- decreasing in \((0,\pi)\)
- increasing in \((0,\frac{\pi}{2})\) and decreasing in \((\frac{\pi}{2},\pi)\)
- decreasing in \((0,\frac{\pi}{2})\) and increasing in \((\frac{\pi}{2},\pi)\)
- increasing in \((0,\frac{\pi}{4})\) and decreasing in \((\frac{\pi}{4},\pi)\)
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The Correct Option is B
Solution and Explanation
Step 1: Concept:
• Find \(f'(x)\) and determine its sign in the interval \((0, \pi)\).
Step 2: Detailed Explanation:
• Differentiate: \[ f'(x) = -2\sin x - 1 \]
• In the interval \((0, \pi)\): \[ \sin x>0 \]
• Therefore: \[ -2\sin x - 1<-1<0 \]
• Hence: \[ f'(x)<0 \quad \forall \; x \in (0, \pi) \]
• So, the function is decreasing in the given interval.
Step 3: Final Answer:
• The function is decreasing in \((0, \pi)\).
• Find \(f'(x)\) and determine its sign in the interval \((0, \pi)\).
Step 2: Detailed Explanation:
• Differentiate: \[ f'(x) = -2\sin x - 1 \]
• In the interval \((0, \pi)\): \[ \sin x>0 \]
• Therefore: \[ -2\sin x - 1<-1<0 \]
• Hence: \[ f'(x)<0 \quad \forall \; x \in (0, \pi) \]
• So, the function is decreasing in the given interval.
Step 3: Final Answer:
• The function is decreasing in \((0, \pi)\).
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