1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
  3. NCERT Exemplar - MCQs
Check sibling questions

Question 16 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Dec. 16, 2024 by Teachoo

If f (x) = x 2 sin 1/x, where x ≠ 0, then the value of the function f at x = 0, so that the function is continuous at x = 0 , is

(A) 0  

(B) – 1

(C) 1 

(D) none of these

This question is similar to Ex 5.1, 24 - Chapter 5 Class 12 - Continuity and Differentiability

Transcript

Question 16 If f (x) = x2 sin 1/π‘₯, where x β‰  0, then the value of the function f at x = 0, so that the function is continuous at x = 0, is (A) 0 (B) – 1 (C) 1 (D) none of these Given f (x) = x2 sin 1/π‘₯, when x β‰  0 To find f (0) f(x) is continuous at π‘₯=0 if L.H.L = R.H.L = 𝑓(0) if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) (βˆ’β„Ž)^2 π’”π’Šπ’β‘γ€–πŸ/((βˆ’π’‰))γ€— = lim┬(hβ†’0) β„Ž^2 π’Œ = 02 .π‘˜ = 0 RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) β„Ž^2 π’”π’Šπ’β‘γ€–πŸ/𝒉〗 = lim┬(hβ†’0) β„Ž^2 π’Œ = 02. π‘˜ = 0 Since, L.H.L = R.H.L = 𝑓(0) 0=𝑓(0) ∴ f (0) =𝟎 So, the correct answer is (A)
  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

    3. NCERT Exemplar - MCQs

    Rolle's and Mean Value Theorem →
Rolle's and Mean Value Theorem →
Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

About the Author

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 and Computer Science at Teachoo