NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

(D) 3/2

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

Get live Maths 1-on-1 Classs - Class 6 to 12

Transcript

Question 12 The value of c in Mean value theorem for the function f (x) = x (x β 2), x β [1, 2] is (A) 3/2 (B) 2/3 (C) 1/2 (D) 3/2 π(π₯)=π₯" (" π₯" β 2)" π(π₯) = π₯^2 β 2π₯ in interval [1, 2]. Checking conditions for Mean value Theorem Conditions of Mean value theorem π(π₯) is continuous at {π, π} π(π₯) is differentiable at (π , π) If both conditions satisfied, then there exist some c in (π , π) such that πβ²(π) = (π(π) β π(π))/(π β π) Condition 1 We need to check ifπ(π₯) = π₯^2 β 2π₯ is continuous in interval [1, 2] Since π(π₯) is polynomial . it is continuous β΄ π(π) is continuous at [1, 2] Condition 2 We need to check if π(π₯) = π₯^2 β 2π₯ is differentiable in interval (1, 2) Since π(π₯) is a polynomial . it is Differentiable β΄ π(π) is differentiable in (1, 2) Since both conditions are satisfied From Mean Value Theorem, There exists a c β (1, 2) such that, π^β² (π) = (π(π) β π(π))/(π β π) 2πβ2= ((2^2 β 2(2)) β (1^2 β 2(1)))/1 ππβπ = ((π β π) β (π β π))/π ππβπ = (π β (βπ))/π 2πβ2 = 1 2π = 1+2 2π = 3 π=π/π Since c = π/π β(π, π) So, the correct answer is (A)