The value of c in Mean value theorem for the function f (x) = x (x – 2), x ∈ [1, 2] is

(A) 3/2 

(B) 2/3 

(C) 1/2 

(D) 3/2

This question is similar to Example 43 - Chapter 5 Class 12 - Continuity and Differentiability

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

Transcript

Question 12 The value of c in Mean value theorem for the function f (x) = x (x โ€“ 2), x โˆˆ [1, 2] is (A) 3/2 (B) 2/3 (C) 1/2 (D) 3/2 ๐‘“(๐‘ฅ)=๐‘ฅ" (" ๐‘ฅ" โ€“ 2)" ๐‘“(๐‘ฅ) = ๐‘ฅ^2 โ€“ 2๐‘ฅ in interval [1, 2]. Checking conditions for Mean value Theorem Conditions of Mean value theorem ๐‘“(๐‘ฅ) is continuous at {๐‘Ž, ๐‘} ๐‘“(๐‘ฅ) is differentiable 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 1 We need to check if๐‘“(๐‘ฅ) = ๐‘ฅ^2 โ€“ 2๐‘ฅ is continuous in interval [1, 2] Since ๐‘“(๐‘ฅ) is polynomial . it is continuous โˆด ๐’‡(๐’™) is continuous at [1, 2] Condition 2 We need to check if ๐‘“(๐‘ฅ) = ๐‘ฅ^2 โ€“ 2๐‘ฅ is differentiable in interval (1, 2) Since ๐‘“(๐‘ฅ) is a polynomial . it is Differentiable โˆด ๐’‡(๐’™) is differentiable in (1, 2) Since both conditions are satisfied From Mean Value Theorem, There exists a c โˆˆ (1, 2) such that, ๐’‡^โ€ฒ (๐’„) = (๐’‡(๐Ÿ) โˆ’ ๐’‡(๐Ÿ))/(๐Ÿ โˆ’ ๐Ÿ) 2๐‘โˆ’2= ((2^2 โˆ’ 2(2)) โˆ’ (1^2 โˆ’ 2(1)))/1 ๐Ÿ๐’„โˆ’๐Ÿ = ((๐Ÿ’ โˆ’ ๐Ÿ’) โˆ’ (๐Ÿ โˆ’ ๐Ÿ))/๐Ÿ ๐Ÿ๐’„โˆ’๐Ÿ = (๐ŸŽ โˆ’ (โˆ’๐Ÿ))/๐Ÿ 2๐‘โˆ’2 = 1 2๐‘ = 1+2 2๐‘ = 3 ๐’„=๐Ÿ‘/๐Ÿ Since c = ๐Ÿ‘/๐Ÿ โˆˆ(๐Ÿ, ๐Ÿ) So, the correct answer is (A)

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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.