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📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

Verify Rolle's theorem applicability for f(x) = sin x on [0, pi]. What is f'(c) = 0 satisfied at?

Answer: c = pi/2.

  • A c = 0
  • B c = pi/2
  • C c = pi/4
  • D c = pi

Correct answer: B. c = pi/2

Explanation: Rolle's theorem conditions: sin x is continuous on [0,π], differentiable on (0,π), and f(0) = sin 0 = 0 = sin π = f(π). So theorem applies. f'(x) = cos x = 0 at c = π/2 ∈ (0,π).

f(x) = |x|: Continuous but NOT Differentiable at 0x=0 (sharp "kink")LHD = -1RHD = +1LHD ≠ RHD at the kink → not differentiable, even though the curve has no gap (continuous)

f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 - since these one-sided derivatives disagree, the function is not differentiable at that point.

Concept context

When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.

Read the full Continuity and Differentiability notes →