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Question 115 of 148

Q.Verify Rolle's theorem for the function f(x)=∣x−2∣+∣x−5∣f(x) = |x-2| + |x-5| in [1,6][1, 6].

Puducherry TnboardTamil Nadu HSC (DGE) Board 2019Subjective· 2mImportance★★★★★
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Although f(1)=f(6)=5f(1)=f(6)=5 and ff is continuous, ff fails to be differentiable at x=2x=2 and x=5x=5 inside (1,6)(1,6), so Rolle's theorem does not apply.

  1. f(x)=∣x−2∣+∣x−5∣f(x)=|x-2|+|x-5| is a sum of absolute-value functions, each continuous on R\mathbb{R}, so ff is continuous on [1,6][1,6].
  2. Check differentiability on (1,6)(1,6): ∣x−2∣|x-2| is not differentiable at x=2x=2, and ∣x−5∣|x-5| is not differentiable at x=5x=5. Both x=2x=2 and x=5x=5 lie in the open interval (1,6)(1,6).
  3. Hence ff is not differentiable at two points of (1,6)(1,6), so the differentiability hypothesis of Rolle's theorem fails.
  4. Check the endpoint condition anyway: f(1)=∣1−2∣+∣1−5∣=1+4=5f(1)=|1-2|+|1-5|=1+4=5; f(6)=∣6−2∣+∣6−5∣=4+1=5f(6)=|6-2|+|6-5|=4+1=5; so f(1)=f(6)f(1)=f(6). …

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