## For the function f (x) = x + 1/x , x β [1, 3], the value of c for mean value theorem is

## (A)1Β

## (B) β3

## (C) 2Β

## (D) None of these

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Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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Last updated at Nov. 18, 2021 by Teachoo

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This question is similar to

Example 43 - Chapter 5 Class 12 -Continuity and Differentiability

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Question 26 For the function f (x) = x + 1/π₯ , x β [1, 3], the value of c for mean value theorem is 1 (B) β3 (C) 2 (D) None of these π(π₯)="x + " 1/π₯ in interval [1, 3] 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 πβ²(π) = (π(π) β π(π))/(π β π) Condition 1 We need to check if π(π₯)="x + " 1/π₯ is continuous in the interval [1, 3] Let π(π)=π and π‘(π)=π/π We know that, π(π)=π₯ is continuous as it is a polynomial function And, π‘(π)=1/π₯ is continuous for all π₯ except for π=π β΄ h(π₯)=1/π₯ is continuous in the interval [1, 3] Hence, π(π)=π(π)+π(π) is also continuous in the interval [1, 3] Condition 2 We need to check if π(π₯)="x + " 1/π₯ is differentiable at ("1, 3") A function is said to be differentiable if the derivative of the function exists. Differentiating π(π₯) wrt π₯ π^β² (π)=πβπ/π^π Since, derivative of the given function exists Hence, π(π) is differentiable at ("1, 3") Since both conditions are satisfied From Mean Value Theorem, There exists a c β (1, 3) such that, π^β² (π) = (π(3) β π(1))/(3 β1) πβπ/π^π = ((π + π/π) β (π + π/π))/π 1β1/π^2 = ((9 + 1)/3 β 2)/2 1β1/π^2 = (10/3 β 2)/2 1β1/π^2 = ((10 β 6)/3)/2 1β1/π^2 = 4/(3 Γ 2 ) πβπ/π^π = 2/3 πβπ/π= π/π^π (π β π)/π= π/π^π π/π= π/π^π π^2=3 π=Β±β3 π=Β±π.ππ Either, c = β1.73 But, β1.73 β(π, π) Or, c = 1.73 And, 1.73 β (π, π) Therefore, π=βπ So, the correct answer is (B)