Question:
The value of \(C\) of Rolle's theorem for the function \(f(x)=\sin x\) in \([0,\pi]\) is
The value of \(C\) of Rolle's theorem for the function \(f(x)=\sin x\) in \([0,\pi]\) is
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For Rolle's theorem, first check \(f(a)=f(b)\), then solve \(f'(c)=0\) in \((a,b)\).
Updated On: Jul 3, 2026
- \(\dfrac{\pi}{3}\)
- \(\dfrac{\pi}{6}\)
- \(\dfrac{\pi}{2}\)
- \(\dfrac{\pi}{4}\)
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The Correct Option is C
Solution and Explanation
Concept:
Rolle's theorem says that if \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \[ f(a)=f(b) \] then there exists at least one \(c\in(a,b)\) such that \[ f'(c)=0 \]
Step 1: Verify endpoint values. Given, \[ f(x)=\sin x \] on \[ [0,\pi] \] Now, \[ f(0)=\sin0=0 \] and \[ f(\pi)=\sin\pi=0 \] Thus, \[ f(0)=f(\pi) \]
Step 2: Differentiate the function. \[ f'(x)=\cos x \] By Rolle's theorem, \[ f'(c)=0 \] So, \[ \cos c=0 \]
Step 3: Solve in \((0,\pi)\). In the interval \[ (0,\pi) \] we have \[ \cos c=0 \] at \[ c=\frac{\pi}{2} \]
Step 4: Final answer. \[ \boxed{\frac{\pi}{2}} \]
Rolle's theorem says that if \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \[ f(a)=f(b) \] then there exists at least one \(c\in(a,b)\) such that \[ f'(c)=0 \]
Step 1: Verify endpoint values. Given, \[ f(x)=\sin x \] on \[ [0,\pi] \] Now, \[ f(0)=\sin0=0 \] and \[ f(\pi)=\sin\pi=0 \] Thus, \[ f(0)=f(\pi) \]
Step 2: Differentiate the function. \[ f'(x)=\cos x \] By Rolle's theorem, \[ f'(c)=0 \] So, \[ \cos c=0 \]
Step 3: Solve in \((0,\pi)\). In the interval \[ (0,\pi) \] we have \[ \cos c=0 \] at \[ c=\frac{\pi}{2} \]
Step 4: Final answer. \[ \boxed{\frac{\pi}{2}} \]
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