Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Last updated at Jan. 3, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Transcript

Ex 5.8, 5 Verify Mean Value Theorem, if ๐ (๐ฅ) = ๐ฅ3 โ 5๐ฅ2 โ 3๐ฅ in the interval [a, b], where a = 1 and b = 3. Find all ๐ โ (1, 3) for which ๐ โฒ(๐) = 0. ๐ (๐ฅ) = ๐ฅ3 โ 5๐ฅ2 โ 3๐ฅ in [a, b], where a = 1 and b = 3 Condition 1 ๐ (๐ฅ) = ๐ฅ3 โ 5๐ฅ2 โ 3๐ฅ ๐(๐ฅ) is a polynomial & every polynomial function is continuous โด ๐(๐ฅ) is continuous at ๐ฅโ[1, 3] 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) Condition 2 ๐(๐ฅ) = ๐ฅ3 โ 5๐ฅ2 โ 3๐ฅ ๐(๐ฅ) is a polynomial & every polynomial function is differentiable โด ๐(๐ฅ) is differentiable at ๐ฅโ[1, 3] Now, ๐(๐ฅ)" = " ๐ฅ3 โ 5๐ฅ2 โ 3๐ฅ ๐^โฒ (๐ฅ)" = 3" ๐ฅ2 โ10๐ฅ โ 3 ๐ฅโ[1, 3] So, ๐โฒ(๐) = " 3" ๐^2โ10๐โ3 Also, ๐(๐ฅ)" = " ๐ฅ3 โ 5๐ฅ2 โ 3๐ฅ ๐(๐)" = " ๐(1) = (1)^3โ5(1)^2โ3(1) = 1โ5โ3 = โ7 ๐(๐)" = " ๐(3) = (3)^3โ5(3)^2โ3(3) = 27โ45โ9 = โ27 By Mean Value Theorem ๐^โฒ (๐) = (๐(๐) โ ๐(๐))/(๐ โ ๐) "3" ๐^2โ10๐โ3 = (โ27 โ (โ7))/(3 โ 1) "3" ๐^2โ10๐โ3 = (โ27 + 7)/2 "3" ๐^2โ10๐โ3 = (โ20)/2 "3" ๐^2โ10๐โ3 = โ10 "3" ๐^2โ10๐โ3+10 = 0 "3" ๐^2โ10๐+7 = 0 "3" ๐^2โ3๐โ7๐+7 = 0 "3" ๐(๐โ1)โ7(๐โ1) = 0 (3๐โ7)(๐โ1) = 0 So, c = 7/3 & c = 1 Since c = 7/3 lies between 1 & 3 c = ๐/๐ โ[1, 3] Thus, Mean Value Theorem is verified. From our question Find all ๐ โ (1, 3) for which ๐ โฒ(๐) = 0. We need to find cโ[1, 3] For which ๐^โฒ (๐) = 0 ๐^โฒ (๐) = 0 "3" ๐^2โ10๐โ3 = 0 The above equation is of the form ๐ด๐ฅ^2+๐ต๐ฅ+๐ถ x = (โ๐ต ยฑ โ(๐ต^2 โ4๐ด๐ถ) )/2๐ด c = (โ(โ10) ยฑ โ((โ10)^2 โ 4(โ3)(3) ) )/2๐ด c = (10 ยฑ โ(100 + 36) )/2(โ3) c = (10 ยฑ โ136 )/6 c = (10 ยฑ โ(2 ร 2 ร 34))/6 c = (10 ยฑ 2โ34)/6 c = 2(5 ยฑ โ34 )/6 c = (5 ยฑ โ34)/3 So, c = (5 + โ34)/3 c = (5 + 5.83)/3 c = 10.83/3 c = 3.61 c = (5 โ โ34)/3 c = (5 โ 5.83)/3 c = (โ0.83)/3 c = โ0.28 Thus, c = 3.61 & c = โ0.28 But both values do not lie between [1, 3] Hence, there exists no value of ๐โ[1, 3] for which ๐^โฒ (๐) = 0

Ex 5.8

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.