# Ex 5.8, 4 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.8, 4 Verify Mean Value Theorem, if ๐ (๐ฅ) = ๐ฅ2 โ 4๐ฅ โ 3 in the interval [๐, ๐], where ๐= 1 ๐๐๐ ๐= 4 ๐ (๐ฅ) = ๐ฅ2 โ 4๐ฅ โ 3 ๐ฅโ[๐, ๐] where a = 1 & b = 4 Mean Value Theorem satisfied if Condition 1 ๐(๐ฅ) is continuous ๐(๐ฅ)=๐ฅ2 โ 4๐ฅ โ 3 ๐(๐ฅ) is a polynomial & Every polynomial function is continuous โ ๐(๐ฅ) is continuous at ๐ฅโ[1, 4] 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) 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 If ๐(๐ฅ) is differentiable ๐(๐ฅ) = ๐ฅ2 โ 4๐ฅ โ 3 ๐(๐ฅ) is a polynomial & Every polynomial function is differentiable โ ๐(๐ฅ) is differentiable at ๐ฅโ[1, 4] Condition 3 ๐(๐ฅ) = ๐ฅ2 โ 4๐ฅ โ 3 ๐^โฒ (๐ฅ) = 2๐ฅโ4 ๐^โฒ (๐) = 2๐โ4 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(1) = (1)^2โ4(1)โ3 = 1 โ 4 โ 3 = โ6 ๐(๐) = ๐(4) = (4)^2โ4(4)โ3 = 16 โ 16 โ 3 = โ 3 By Mean Value Theorem ๐^โฒ (๐) = (๐(๐) โ ๐(๐))/(๐ โ ๐) ๐^โฒ (๐) = (โ3 โ (โ6))/(4 โ 1) ๐^โฒ (๐) = (โ3 + 6)/3 ๐^โฒ (๐) = 3/3 ๐^โฒ (๐) = 1 2c โ 4 = 1 2c = 1 + 4 2c = 5 c = 5/2 Value of c = 5/2 which is lies between (1, 4) c = ๐/๐โ(๐, ๐) Hence Mean Value Theorem satisfied

Ex 5.8

Ex 5.8, 1
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Ex 5.8, 2 (i) Deleted for CBSE Board 2022 Exams

Ex 5.8, 2 (ii) Important Deleted for CBSE Board 2022 Exams

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.