The number of points at which the function f (x) = 1/(x-[x] ) is not continuous is

(A) 1Β  Β Β Β 

(B) 2

(C) 3Β  Β Β Β Β 

(D) none of these

This question is similar to Ex 5.1, 19 - Chapter 5 Class 12 - Continuity and Differentiability

[MCQ] The number of points at which the function f (x) = 1/ x - [x] is - NCERT Exemplar - MCQs

part 2 - Question 3 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability
part 3 - Question 3 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability part 4 - Question 3 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability part 5 - Question 3 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability

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Transcript

Question 3 The number of points at which the function f (x) = 1/(π‘₯βˆ’[π‘₯] ) is not continuous is (A) 1 (B) 2 (C) 3 (D) none of these Given f(x) = 1/(π‘₯ βˆ’ [π‘₯] ) Since Greatest Integer Function changes value on integer numbers Thus, we check continuity When x is not an integer When x is an integer Case 1 : When 𝒙 is not an integer f(x) = 1/(π‘₯ βˆ’ [π‘₯] ) Let d be any non integer point Now, f(x) is continuous at π‘₯=𝑑 if (π₯𝐒𝐦)┬(𝐱→𝒅) 𝒇(𝒙)=𝒇(𝒅) (π₯𝐒𝐦)┬(𝐱→𝒅) 𝒇(𝒙) = lim┬(x→𝑑) 1/(π‘₯ βˆ’ [π‘₯] ) Putting x = d =1/(𝑑 βˆ’ [𝑑] ) 𝒇(𝒅) =1/(𝑑 βˆ’ [𝑑] ) Since lim┬(x→𝑑) 𝑓(π‘₯)= 𝑓(𝑑) ∴ 𝑓(π‘₯) is continuous for all non-integer points Case 2 : When x is an integer f(x) = [x] Let c be any integer point Now, f(x) is continuous at π‘₯ =𝑐 if L.H.L = R.H.L = 𝑓(𝑐) if (π₯𝐒𝐦)┬(𝐱→𝒄^βˆ’ ) 𝒇(𝒙)=(π₯𝐒𝐦)┬(𝐱→𝒄^+ ) " " 𝒇(𝒙)= 𝒇(𝒄) LHL at x β†’ c (π’π’Šπ’Ž)┬(𝐱→𝒄^βˆ’ ) f (x) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) f (c βˆ’ h) = lim┬(hβ†’0) 𝟏/((𝑐 βˆ’ β„Ž) βˆ’ [𝒄 βˆ’ 𝒉]) = lim┬(hβ†’0) 𝟏/((𝑐 βˆ’ β„Ž) βˆ’ (𝒄 βˆ’ 𝟏)) = lim┬(hβ†’0) 𝟏/(𝑐 βˆ’ β„Ž βˆ’ 𝑐 + 1) = lim┬(hβ†’0) 𝟏/(βˆ’β„Ž + 1) = 𝟏/(0 + 1) = 𝟏/𝟏 = 1 RHL at x β†’ c (π’π’Šπ’Ž)┬(𝐱→𝒄^+ ) f (x) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) f (c + h) = lim┬(hβ†’0) (𝑐+β„Ž)βˆ’[𝒄+𝒉] = lim┬(hβ†’0) (π‘βˆ’β„Ž)βˆ’(𝒄) = lim┬(hβ†’0) βˆ’β„Ž = 𝟎 Since LHL β‰  RHL ∴ f(x) is not continuous at x = c Hence, f(x) is not continuous at all integral points. ∴ There are infinite number of points where f(x) = 1/(π‘₯βˆ’[π‘₯] ) is not continuous Since we need to find points where f(x) is not continuous And, our options are (A) 1 (B) 2 (C) 3 (D) none of these So, the correct answer is (D)

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