Check sibling questions

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

(A) 3/2Β 

(B) 2/3Β 

(C) 1/2Β 

(D) 3/2

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

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Transcript

Question 12 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 𝑓(π‘₯)=π‘₯" (" π‘₯" – 2)" 𝑓(π‘₯) = π‘₯^2 – 2π‘₯ in interval [1, 2]. Checking conditions for Mean value Theorem Conditions of Mean value theorem 𝑓(π‘₯) is continuous at {π‘Ž, 𝑏} 𝑓(π‘₯) is differentiable at (π‘Ž , 𝑏) If both conditions satisfied, then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) Condition 1 We need to check if𝑓(π‘₯) = π‘₯^2 – 2π‘₯ is continuous in interval [1, 2] Since 𝑓(π‘₯) is polynomial . it is continuous ∴ 𝒇(𝒙) is continuous at [1, 2] Condition 2 We need to check if 𝑓(π‘₯) = π‘₯^2 – 2π‘₯ is differentiable in interval (1, 2) Since 𝑓(π‘₯) is a polynomial . it is Differentiable ∴ 𝒇(𝒙) is differentiable in (1, 2) Since both conditions are satisfied From Mean Value Theorem, There exists a c ∈ (1, 2) such that, 𝒇^β€² (𝒄) = (𝒇(𝟐) βˆ’ 𝒇(𝟏))/(𝟐 βˆ’ 𝟏) 2π‘βˆ’2= ((2^2 βˆ’ 2(2)) βˆ’ (1^2 βˆ’ 2(1)))/1 πŸπ’„βˆ’πŸ = ((πŸ’ βˆ’ πŸ’) βˆ’ (𝟏 βˆ’ 𝟐))/𝟏 πŸπ’„βˆ’πŸ = (𝟎 βˆ’ (βˆ’πŸ))/𝟏 2π‘βˆ’2 = 1 2𝑐 = 1+2 2𝑐 = 3 𝒄=πŸ‘/𝟐 Since c = πŸ‘/𝟐 ∈(𝟏, 𝟐) So, the correct answer is (A)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.