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

Transcript

Ex 5.8, 5 Verify Mean Value Theorem, if ๐‘“ (๐‘ฅ) = ๐‘ฅ3 โ€“ 5๐‘ฅ2 โ€“ 3๐‘ฅ in the interval [a, b], where a = 1 and b = 3. Find all ๐‘ โˆˆ (1, 3) for which ๐‘“ โ€ฒ(๐‘) = 0. ๐‘“ (๐‘ฅ) = ๐‘ฅ3 โ€“ 5๐‘ฅ2 โ€“ 3๐‘ฅ in [a, b], where a = 1 and b = 3 Condition 1 ๐‘“ (๐‘ฅ) = ๐‘ฅ3 โ€“ 5๐‘ฅ2 โ€“ 3๐‘ฅ ๐‘“(๐‘ฅ) is a polynomial & every polynomial function is continuous โˆด ๐‘“(๐‘ฅ) is continuous at ๐‘ฅโˆˆ[1, 3] 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) Condition 2 ๐‘“(๐‘ฅ) = ๐‘ฅ3 โ€“ 5๐‘ฅ2 โ€“ 3๐‘ฅ ๐‘“(๐‘ฅ) is a polynomial & every polynomial function is differentiable โˆด ๐‘“(๐‘ฅ) is differentiable at ๐‘ฅโˆˆ[1, 3] Now, ๐‘“(๐‘ฅ)" = " ๐‘ฅ3 โ€“ 5๐‘ฅ2 โ€“ 3๐‘ฅ ๐‘“^โ€ฒ (๐‘ฅ)" = 3" ๐‘ฅ2 โ€“10๐‘ฅ โ€“ 3 ๐‘ฅโˆˆ[1, 3] So, ๐‘“โ€ฒ(๐‘) = " 3" ๐‘^2โˆ’10๐‘โˆ’3 Also, ๐‘“(๐‘ฅ)" = " ๐‘ฅ3 โ€“ 5๐‘ฅ2 โ€“ 3๐‘ฅ ๐‘“(๐‘Ž)" = " ๐‘“(1) = (1)^3โˆ’5(1)^2โˆ’3(1) = 1โˆ’5โˆ’3 = โˆ’7 ๐‘“(๐‘)" = " ๐‘“(3) = (3)^3โˆ’5(3)^2โˆ’3(3) = 27โˆ’45โˆ’9 = โˆ’27 By Mean Value Theorem ๐‘“^โ€ฒ (๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) "3" ๐‘^2โˆ’10๐‘โˆ’3 = (โˆ’27 โˆ’ (โˆ’7))/(3 โˆ’ 1) "3" ๐‘^2โˆ’10๐‘โˆ’3 = (โˆ’27 + 7)/2 "3" ๐‘^2โˆ’10๐‘โˆ’3 = (โˆ’20)/2 "3" ๐‘^2โˆ’10๐‘โˆ’3 = โˆ’10 "3" ๐‘^2โˆ’10๐‘โˆ’3+10 = 0 "3" ๐‘^2โˆ’10๐‘+7 = 0 "3" ๐‘^2โˆ’3๐‘โˆ’7๐‘+7 = 0 "3" ๐‘(๐‘โˆ’1)โˆ’7(๐‘โˆ’1) = 0 (3๐‘โˆ’7)(๐‘โˆ’1) = 0 So, c = 7/3 & c = 1 Since c = 7/3 lies between 1 & 3 c = ๐Ÿ•/๐Ÿ‘ โˆˆ[1, 3] Thus, Mean Value Theorem is verified. From our question Find all ๐‘ โˆˆ (1, 3) for which ๐‘“ โ€ฒ(๐‘) = 0. We need to find cโˆˆ[1, 3] For which ๐‘“^โ€ฒ (๐‘) = 0 ๐‘“^โ€ฒ (๐‘) = 0 "3" ๐‘^2โˆ’10๐‘โˆ’3 = 0 The above equation is of the form ๐ด๐‘ฅ^2+๐ต๐‘ฅ+๐ถ x = (โˆ’๐ต ยฑ โˆš(๐ต^2 โˆ’4๐ด๐ถ) )/2๐ด c = (โˆ’(โˆ’10) ยฑ โˆš((โˆ’10)^2 โˆ’ 4(โˆ’3)(3) ) )/2๐ด c = (10 ยฑ โˆš(100 + 36) )/2(โˆ’3) c = (10 ยฑ โˆš136 )/6 c = (10 ยฑ โˆš(2 ร— 2 ร— 34))/6 c = (10 ยฑ 2โˆš34)/6 c = 2(5 ยฑ โˆš34 )/6 c = (5 ยฑ โˆš34)/3 So, c = (5 + โˆš34)/3 c = (5 + 5.83)/3 c = 10.83/3 c = 3.61 c = (5 โˆ’ โˆš34)/3 c = (5 โˆ’ 5.83)/3 c = (โˆ’0.83)/3 c = โˆ’0.28 Thus, c = 3.61 & c = โ€“0.28 But both values do not lie between [1, 3] Hence, there exists no value of ๐œโˆˆ[1, 3] for which ๐‘“^โ€ฒ (๐‘) = 0

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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.