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

Last updated at August 12, 2026 by

If f (x) = x 2 sin 1/x, 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 Β 

(B) – 1

(C) 1Β 

(D) none of these

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

[MCQ] If f(x) = x2 sin 1/x, where x β‰  0, then value of function at x=0 - NCERT Exemplar - MCQs

part 2 - Question 16 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability
part 3 - Question 16 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability

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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)

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