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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Example 42 Verify Rolle’s theorem for the function y = x2 + 2, a = – 2 and b = 2. y = x2 + 2, a = –2 and b = 2 Let 𝑓(π‘₯) = π‘₯^2+2 Rolle’s theorem is satisfied if Condition 1 𝑓(π‘₯) is continuous at (βˆ’2 , 2) Since 𝑓(π‘₯) is a polynomial . it is continuous Conditions of Rolle’s theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is derivable at (π‘Ž , 𝑏) 𝑓(π‘Ž) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Condition 2 𝑓(π‘₯) is differentiable at (βˆ’2 , 2) Since 𝑓(π‘₯) is a polynomial . it is differentiable Condition 3 𝑓(βˆ’2) = (βˆ’2)^2+2 = 4+2 = 6 𝑓(2) = 2^2+2 = 4+2 = 6 Hence, 𝑓(2) = 𝑓(βˆ’2) Conditions of Rolle’s theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is derivable at (π‘Ž , 𝑏) 𝑓(π‘Ž) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Now, 𝑓(π‘₯) = π‘₯^2+2 𝑓^β€² (π‘₯) = 2x So 𝑓^β€² (𝑐) = 2𝑐 Since all 3 conditions are satisfied 𝑓^β€² (𝑐) = 0 2𝑐 = 0 𝑐 = 0 Value of c i.e. 0 lies between βˆ’2 and 2. Hence c = 0 ∈ (βˆ’πŸ, 𝟐) Thus , Rolle’s theorem is satisfied.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.