Ex 5.8, 6 - Examine the applicability of Mean Value Theorem

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

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

Transcript

Ex 5.8, 6 Examine the applicability of Mean Value Theorem in the function (๐‘–) ๐‘“ (๐‘ฅ) = [๐‘ฅ] ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [5, 9]Greatest Integer less than equal to ๐‘ฅ ๐‘“ (๐‘ฅ) =[๐‘ฅ] is not continuous & differentiable โ‡’ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Ex 5.8, 6 Examine the applicability of Mean Value Theorem in the function (๐‘–๐‘–) ๐‘“ (๐‘ฅ) = [๐‘ฅ] ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [โˆ’2, 2]Greatest Integer less than equal to ๐‘ฅ ๐‘“ (๐‘ฅ) =[๐‘ฅ] is not continuous & differentiable โ‡’ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Ex 5.8, 2 Examine the applicability of Mean Value Theorem in the function (๐‘–๐‘–๐‘–) ๐‘“ (๐‘ฅ) = ๐‘ฅ2 โ€“ 1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [1, 2]๐‘“ (๐‘ฅ) = ๐‘ฅ2 โ€“ 1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [1 , 2] Condition 1 ๐‘“(๐‘ฅ) = ๐‘ฅ2 โ€“ 1 ๐‘“(๐‘ฅ) is a polynomial & Every polynomial function is continuous โ‡’ ๐‘“(๐‘ฅ) is continuous at ๐‘ฅโˆˆ[1, 2]Conditions of Mean value theorem ๐‘“(๐‘ฅ) is continuous at (๐‘Ž , ๐‘) ๐‘“(๐‘ฅ) is derivable at (๐‘Ž , ๐‘) If both conditions satisfied, then there exist some c in (๐‘Ž , ๐‘) such that ๐‘“โ€ฒ(๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) ๐‘“(๐‘)" = " ๐‘“(2) = (2)^2โˆ’1 = 4โˆ’1 = 3 By Mean Value Theorem ๐‘“^โ€ฒ (๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) 2๐‘ = (3 โˆ’ (0))/(2 โˆ’ 1) 2๐‘ = 3/1 ๐‘ = 3/2 ๐’„ = ๐Ÿ‘/๐Ÿ lies between 1 & 2 i.e., ๐‘ = 3/2โˆˆ [1, 2] Hence Mean Value Theorem is satisfied. Ex 5.8, 6 Examine the applicability of Mean Value Theorem in the function (๐‘–) ๐‘“ (๐‘ฅ) = [๐‘ฅ] ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [5, 9] Greatest Integer less than equal to ๐‘ฅ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ๏ทฏ is not continuous & differentiable โ‡’ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Ex 5.8, 6 Examine the applicability of Mean Value Theorem in the function (๐‘–๐‘–) ๐‘“ (๐‘ฅ) = [๐‘ฅ] ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [โˆ’2, 2] Greatest Integer less than equal to ๐‘ฅ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ๏ทฏ is not continuous & differentiable โ‡’ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Ex 5.8, 2 Examine the applicability of Mean Value Theorem in the function (๐‘–๐‘–๐‘–) ๐‘“ (๐‘ฅ) = ๐‘ฅ2 โ€“ 1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [1, 2] ๐‘“ (๐‘ฅ) = ๐‘ฅ2 โ€“ 1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ [1 , 2] Condition 1 ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 โ€“ 1 ๐‘“(๐‘ฅ) is a polynomial & Every polynomial function is continuous โ‡’ ๐‘“ ๐‘ฅ๏ทฏ is continuous at ๐‘ฅโˆˆ[1, 2] Condition 2 ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 โ€“ 1 ๐‘“(๐‘ฅ) is a polynomial & Every polynomial function is differentiable โ‡’ ๐‘“ ๐‘ฅ๏ทฏ is differentiable at ๐‘ฅโˆˆ 1, 2๏ทฏ Condition 3 ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 โ€“ 1 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘ฅ๏ทฏ = 2๐‘ฅ โ€“0 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = 2๐‘ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 โ€“ 1 ๐‘“ ๐‘Ž๏ทฏ = ๐‘“ 1๏ทฏ = 1๏ทฏ๏ทฎ2๏ทฏโˆ’1 = 1 โ€“ 1 = 0 ๐‘“ ๐‘๏ทฏ = ๐‘“ 2๏ทฏ = 2๏ทฏ๏ทฎ2๏ทฏโˆ’1 = 4โˆ’1 = 3 By Mean Value Theorem ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = ๐‘“ ๐‘๏ทฏ โˆ’ ๐‘“ ๐‘Ž๏ทฏ๏ทฎ๐‘ โˆ’ ๐‘Ž๏ทฏ 2๐‘ = 3 โˆ’ 0๏ทฏ๏ทฎ2 โˆ’ 1๏ทฏ 2๐‘ = 3๏ทฎ1๏ทฏ ๐‘ = 3๏ทฎ2๏ทฏ ๐’„ = ๐Ÿ‘๏ทฎ๐Ÿ๏ทฏ lies between 1 & 2 i.e., ๐‘ = 3๏ทฎ2๏ทฏโˆˆ [1, 2] Hence Mean Value 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 10 years. He provides courses for Maths and Science at Teachoo.