NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

## (D) None of these

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

### Transcript

Question 4 For the function f (x) = x + 1/𝑥 , x ∈ [1, 3], the value of c for mean value theorem is 1 (B) √3 (C) 2 (D) None of these 𝑓(𝑥)="x + " 1/𝑥 in interval [1, 3] Checking conditions for Mean value Theorem Conditions of Mean value theorem 𝑓(𝑥) is continuous at {𝑎, 𝑏} 𝑓(𝑥) is differentiable at (𝑎 , 𝑏) If both conditions satisfied, then there exist some c in (𝑎 , 𝑏) such that 𝑓′(𝑐) = (𝑓(𝑏) − 𝑓(𝑎))/(𝑏 − 𝑎) Condition 1 We need to check if 𝑓(𝑥)="x + " 1/𝑥 is continuous in the interval [1, 3] Let 𝒈(𝒙)=𝒙 and 𝐡(𝒙)=𝟏/𝒙 We know that, 𝒈(𝒙)=𝑥 is continuous as it is a polynomial function And, 𝐡(𝒙)=1/𝑥 is continuous for all 𝑥 except for 𝒙=𝟎 ∴ h(𝑥)=1/𝑥 is continuous in the interval [1, 3] Hence, 𝒇(𝒙)=𝒈(𝒙)+𝒉(𝒙) is also continuous in the interval [1, 3] Condition 2 We need to check if 𝑓(𝑥)="x + " 1/𝑥 is differentiable at ("1, 3") A function is said to be differentiable if the derivative of the function exists. Differentiating 𝑓(𝑥) wrt 𝑥 𝒇^′ (𝒙)=𝟏−𝟏/𝒙^𝟐 Since, derivative of the given function exists Hence, 𝒇(𝒙) is differentiable at ("1, 3") Since both conditions are satisfied From Mean Value Theorem, There exists a c ∈ (1, 3) such that, 𝑓^′ (𝑐) = (𝑓(3) − 𝑓(1))/(3 −1) 𝟏−𝟏/𝒄^𝟐 = ((𝟑 + 𝟏/𝟑) − (𝟏 + 𝟏/𝟏))/𝟐 1−1/𝑐^2 = ((9 + 1)/3 − 2)/2 1−1/𝑐^2 = (10/3 − 2)/2 1−1/𝑐^2 = ((10 − 6)/3)/2 1−1/𝑐^2 = 4/(3 × 2 ) 𝟏−𝟏/𝒄^𝟐 = 2/3 𝟏−𝟐/𝟑= 𝟏/𝒄^𝟐 (𝟑 − 𝟐)/𝟑= 𝟏/𝒄^𝟐 𝟏/𝟑= 𝟏/𝒄^𝟐 𝑐^2=3 𝑐=±√3 𝒄=±𝟏.𝟕𝟑 Either, c = −1.73 But, −1.73 ∉(𝟏, 𝟑) Or, c = 1.73 And, 1.73 ∈ (𝟏, 𝟑) Therefore, 𝒄=√𝟑 So, the correct answer is (B)

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.