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Ex 5.8

Ex 5.8, 1
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Ex 5.8, 2 (ii) Important Deleted for CBSE Board 2023 Exams

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Ex 5.8, 6 Deleted for CBSE Board 2023 Exams You are here

Last updated at March 16, 2023 by Teachoo

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) =[๐ฅ] 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) =[๐ฅ] 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) = ๐ฅ2 โ 1 ๐(๐ฅ) is a polynomial & Every polynomial function is continuous โ ๐(๐ฅ) is continuous at ๐ฅโ[1, 2]Conditions of Mean value theorem ๐(๐ฅ) is continuous at (๐ , ๐) ๐(๐ฅ) is derivable at (๐ , ๐) If both conditions satisfied, then there exist some c in (๐ , ๐) such that ๐โฒ(๐) = (๐(๐) โ ๐(๐))/(๐ โ ๐) ๐(๐)" = " ๐(2) = (2)^2โ1 = 4โ1 = 3 By Mean Value Theorem ๐^โฒ (๐) = (๐(๐) โ ๐(๐))/(๐ โ ๐) 2๐ = (3 โ (0))/(2 โ 1) 2๐ = 3/1 ๐ = 3/2 ๐ = ๐/๐ lies between 1 & 2 i.e., ๐ = 3/2โ [1, 2] Hence Mean Value Theorem is satisfied. 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.