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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.