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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Last updated at April 16, 2024 by Teachoo
This question is similar to Ex 5.1, 19 - Chapter 5 Class 12 - Continuity and Differentiability
Question 3 The number of points at which the function f (x) = 1/(π₯β[π₯] ) is not continuous is (A) 1 (B) 2 (C) 3 (D) none of these Given f(x) = 1/(π₯ β [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 π is not an integer f(x) = 1/(π₯ β [π₯] ) Let d be any non integer point Now, f(x) is continuous at π₯=π if (π₯π’π¦)β¬(π±βπ ) π(π)=π(π ) (π₯π’π¦)β¬(π±βπ ) π(π) = limβ¬(xβπ) 1/(π₯ β [π₯] ) Putting x = d =1/(π β [π] ) π(π ) =1/(π β [π] ) Since limβ¬(xβπ) π(π₯)= π(π) β΄ π(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₯ =π if L.H.L = R.H.L = π(π) if (π₯π’π¦)β¬(π±βπ^β ) π(π)=(π₯π’π¦)β¬(π±βπ^+ ) " " π(π)= π(π) LHL at x β c (πππ)β¬(π±βπ^β ) f (x) = (πππ)β¬(π‘βπ) f (c β h) = limβ¬(hβ0) π/((π β β) β [π β π]) = limβ¬(hβ0) π/((π β β) β (π β π)) = limβ¬(hβ0) π/(π β β β π + 1) = limβ¬(hβ0) π/(ββ + 1) = π/(0 + 1) = π/π = 1 RHL at x β c (πππ)β¬(π±βπ^+ ) f (x) = (πππ)β¬(π‘βπ) f (c + h) = limβ¬(hβ0) (π+β)β[π+π] = limβ¬(hβ0) (πββ)β(π) = limβ¬(hβ0) ββ = π Since LHL β RHL β΄ f(x) is not continuous at x = c Hence, f(x) is not continuous at all integral points. β΄ There are infinite number of points where f(x) = 1/(π₯β[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)