Ex 5.8, 2 - Examine if Rolle’s theorem - Chapter 5 Class 12

Ex 5.8, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.8, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.8, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.8, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Ex 5.8, 2 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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

Transcript

Ex 5.8, 2 Examine if Rolle’s theorem is applicable to the functions. Can you say some thing about the converse of Rolle’s theorem from this function? (𝑖) 𝑓 (π‘₯) = [π‘₯] π‘“π‘œπ‘Ÿ π‘₯ ∈ [5, 9]Greatest Integer less than equal to π‘₯ 𝑓 (π‘₯) =[π‘₯] is not continuous & differentiable β‡’ Condition of Rolle’s Theorem is not satisfied. Therefore, Rolle’s Theorem is not applicable . Ex 5.8, 2 Examine if Rolle’s theorem is applicable to the functions. Can you say some thing about the converse of Rolle’s theorem from this function? (𝑖𝑖) 𝑓 (π‘₯) = [π‘₯] π‘“π‘œπ‘Ÿ π‘₯ ∈ [βˆ’2, 2]Greatest Integer less than equal to π‘₯ 𝑓 (π‘₯) =[π‘₯] is not continuous & differentiable β‡’ Condition of Rolle’s Theorem is not satisfied. Therefore, Rolle’s Theorem is not applicable . Ex 5.8, 2 Examine if Rolle’s theorem is applicable to the functions. Can you say some thing about the converse of Rolle’s theorem from this function? (𝑖𝑖𝑖) 𝑓 (π‘₯) = π‘₯2 – 1 π‘“π‘œπ‘Ÿ π‘₯ ∈ [1, 2]𝑓 (π‘₯) = π‘₯2 – 1 π‘“π‘œπ‘Ÿ π‘₯ ∈ [1 , 2] Condition 1 𝑓(π‘₯) = π‘₯2 – 1 𝑓(π‘₯) is a polynomial & Every polynomial function is continuous β‡’ 𝑓(π‘₯) is continuous at π‘₯∈[1, 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 Condition 2 𝑓(π‘₯)=π‘₯2 – 1 𝑓(π‘₯) is a polynomial & Every polynomial function is differentiable β‡’ 𝑓(π‘₯) is differentiable at π‘₯∈[1, 2] Condition 3 𝑓(π‘₯) = π‘₯2 – 1 𝑓(1) = (1)^2+(1) = 1 βˆ’ 1 = 0 & 𝑓(2) = (2)^2βˆ’1= 4βˆ’1 = 3 Since 𝒇(𝟏) β‰  𝒇(𝟐) Thus third condition of Rolle’s Theorem is not satisfied. Therefore Rolle’s theorem is not applicable 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 Converse of Rolle’s Theorem If 𝑓 [π‘Ž, 𝑏]→𝑅 for some π‘βˆˆ[π‘Ž, 𝑏] for which 𝑓^β€² (𝑐)=0 then (i) 𝑓(π‘Ž) = 𝑓(𝑏) (ii) 𝑓 is continuous at [π‘Ž, 𝑏] (iii) & Differentiable at [π‘Ž, 𝑏] Now, 𝑓(π‘₯)=π‘₯^2βˆ’1 𝑓^β€² (π‘₯)=2π‘₯ 𝑓^β€² (𝑐)=2𝑐 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 If 𝑓^β€² (𝑐)=0 2𝑐=0 𝑐=0 Since 𝑐=0 does not belong in (1, 2) i.e. c = 0 βˆ‰ (1 , 2) β‡’ There is no value of c for which 𝑓^β€² (𝑐)=0 ∴ Converse of Rolle’s Theorem is also not applicable.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.