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

(A) 1     

(B) 2

(C) 3      

(D) none of these

This question is similar to Ex 5.1, 19 - 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/(𝑥−[𝑥] ) is not continuous is (A) 1 (B) 2 (C) 3 (D) none of these Given f(x) = 1/(𝑥 − [𝑥] ) 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/(𝑥 − [𝑥] ) Let d be any non integer point Now, f(x) is continuous at 𝑥=𝑑 if (𝐥𝐢𝐦)┬(𝐱→𝒅) 𝒇(𝒙)=𝒇(𝒅) (𝐥𝐢𝐦)┬(𝐱→𝒅) 𝒇(𝒙) = lim┬(x→𝑑) 1/(𝑥 − [𝑥] ) Putting x = d =1/(𝑑 − [𝑑] ) 𝒇(𝒅) =1/(𝑑 − [𝑑] ) Since lim┬(x→𝑑) 𝑓(𝑥)= 𝑓(𝑑) ∴ 𝑓(𝑥) 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 𝑥 =𝑐 if L.H.L = R.H.L = 𝑓(𝑐) if (𝐥𝐢𝐦)┬(𝐱→𝒄^− ) 𝒇(𝒙)=(𝐥𝐢𝐦)┬(𝐱→𝒄^+ ) " " 𝒇(𝒙)= 𝒇(𝒄) LHL at x → c (𝒍𝒊𝒎)┬(𝐱→𝒄^− ) f (x) = (𝒍𝒊𝒎)┬(𝐡→𝟎) f (c − h) = lim┬(h→0) 𝟏/((𝑐 − ℎ) − [𝒄 − 𝒉]) = lim┬(h→0) 𝟏/((𝑐 − ℎ) − (𝒄 − 𝟏)) = lim┬(h→0) 𝟏/(𝑐 − ℎ − 𝑐 + 1) = lim┬(h→0) 𝟏/(−ℎ + 1) = 𝟏/(0 + 1) = 𝟏/𝟏 = 1 RHL at x → c (𝒍𝒊𝒎)┬(𝐱→𝒄^+ ) f (x) = (𝒍𝒊𝒎)┬(𝐡→𝟎) f (c + h) = lim┬(h→0) (𝑐+ℎ)−[𝒄+𝒉] = lim┬(h→0) (𝑐−ℎ)−(𝒄) = lim┬(h→0) −ℎ = 𝟎 Since LHL ≠ 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/(𝑥−[𝑥] ) 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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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.