# Ex 5.8, 6

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.8, 6 Examine the applicability of Mean Value Theorem in the function (๐) ๐ (๐ฅ) = [๐ฅ] ๐๐๐ ๐ฅ โ [5, 9] Greatest Integer less than equal to ๐ฅ ๐ ๐ฅ๏ทฏ = ๐ฅ๏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 (๐๐) ๐ (๐ฅ) = [๐ฅ] ๐๐๐ ๐ฅ โ [โ2, 2] Greatest Integer less than equal to ๐ฅ ๐ ๐ฅ๏ทฏ = ๐ฅ๏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 (๐๐๐) ๐ (๐ฅ) = ๐ฅ2 โ 1 ๐๐๐ ๐ฅ โ [1, 2] ๐ (๐ฅ) = ๐ฅ2 โ 1 ๐๐๐ ๐ฅ โ [1 , 2] Condition 1 ๐ ๐ฅ๏ทฏ = ๐ฅ2 โ 1 ๐(๐ฅ) is a polynomial & Every polynomial function is continuous โ ๐ ๐ฅ๏ทฏ is continuous at ๐ฅโ[1, 2] Condition 2 ๐ ๐ฅ๏ทฏ=๐ฅ2 โ 1 ๐(๐ฅ) is a polynomial & Every polynomial function is differentiable โ ๐ ๐ฅ๏ทฏ is differentiable at ๐ฅโ 1, 2๏ทฏ Condition 3 ๐ ๐ฅ๏ทฏ = ๐ฅ2 โ 1 ๐๏ทฎโฒ๏ทฏ ๐ฅ๏ทฏ = 2๐ฅ โ0 ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = 2๐ ๐ ๐ฅ๏ทฏ = ๐ฅ2 โ 1 ๐ ๐๏ทฏ = ๐ 1๏ทฏ = 1๏ทฏ๏ทฎ2๏ทฏโ1 = 1 โ 1 = 0 ๐ ๐๏ทฏ = ๐ 2๏ทฏ = 2๏ทฏ๏ทฎ2๏ทฏโ1 = 4โ1 = 3 By Mean Value Theorem ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = ๐ ๐๏ทฏ โ ๐ ๐๏ทฏ๏ทฎ๐ โ ๐๏ทฏ 2๐ = 3 โ 0๏ทฏ๏ทฎ2 โ 1๏ทฏ 2๐ = 3๏ทฎ1๏ทฏ ๐ = 3๏ทฎ2๏ทฏ ๐ = ๐๏ทฎ๐๏ทฏ lies between 1 & 2 i.e., ๐ = 3๏ทฎ2๏ทฏโ [1, 2] Hence Mean Value Theorem is satisfied.

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .