Q.Prove that the greatest integer function defined by f(x)=[x], 0<x<3, is not differentiable at x=1 and x=2.
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Continuity of the Greatest Integer Function
The greatest integer function f(x)=⌊x⌋ returns the largest integer not exceeding x: ⌊2.3⌋=2, ⌊−1.2⌋=−2, ⌊4⌋=4. Its graph is a staircase — flat segments that jump up by 1 at every integer.
The intuition
Walk along the graph from left to right. Near a non-integer such as x=1.5 the function is flat at 1; nudge x a little either way and the value does not change, so nothing is broken there. But as you approach an integer like x=2 from the left the value is stuck at 1, and the instant you reach x=2 it leaps to 2. That sudden leap is a break.
⌊x⌋ is continuous at every non-integer and discontinuous at every integer.
Why integers fail
At an integer n the one-sided limits disagree:
limx→n−⌊x⌋=n−1,limx→n+⌊x⌋=n,⌊n⌋=n.
Since the left- and right-hand limits differ, limx→n⌊x⌋ does not exist, so continuity fails. This is a jump discontinuity, and the jump is always exactly 1. At a non-integer c there is a whole small interval on which f is constant equal to ⌊c⌋, so the limit exists and matches f(c) — the function is continuous.
How to test it …
A function that is not continuous at a point cannot be differentiable there. The greatest integer function f(x)=[x] jumps at each integer.
At x=1: limx→1−[x]=0 but limx→1+[x]=1, so the limit does not exist and f is discontinuous — hence not differentiable. Checking derivatives with f(1)=1: the right-hand derivative limh→0+h[1+h]−1=limh→0+h0=0, while the left-hand derivative limh→0−h[1+h]−1=limh→0−h−1→+∞; the …
[x] has a jump at each integer, so it is discontinuous — and therefore not differentiable — at x=1 and x=2; the one-sided derivatives there also disagree.
On 0<x<3 the greatest integer function is a staircase: [x]=0 on (0,1), [x]=1 on [1,2), [x]=2 on [2,3). Differentiability requires continuity first, and where the graph jumps it cannot be continuous.
Discontinuity forces non-differentiability at x=1
Left: for x just below 1, [x]=0, so limx→1−[x]=0.
Right: for x just above 1, [x]=1, so limx→1+[x]=1.
The one-sided limits differ, so limx→1[x] does not exist — f is discontinuous, hence not differentiable at x=1.
Confirm with the derivative definition at x=1
Using f(1)=[1]=1:
f+′(1)=limh→0+h[1+h]−1=limh→0+h1−1=0,
f−′(1)=limh→0−h[1+h]−1=limh→0−h0−1=limh→0−h−1→+∞.
The right-hand derivative is 0 and the left-hand derivative diverges, so they are unequal — f′(1) does not exist. …
Method: Showing Non-Differentiability at a Jump Discontinuity
Use this method for functions like the greatest integer (floor) function that jump abruptly at certain points — this route is shorter than the corner-point method above because differentiability can be ruled out immediately once a jump is shown.
Steps
Step 1: Recall that differentiability requires continuity first
If a function is not even continuous at a point, it cannot be differentiable there — there is no need to compute a derivative limit at all. This shortcut saves work whenever a jump can be shown directly.
Step 2: Compute the left-hand and right-hand limits of the function itself (not yet the derivative) at the point in question
For the greatest integer function [x] at an integer n, evaluate [x] for x slightly less than n and slightly greater than n separately.
Step 3: Compare the two one-sided limits to the function's actual value
If limx→n−f(x)=limx→n+f(x), the two-sided limit does not exist, so f is discontinuous at n — and, by Step 1's logic, therefore automatically not differentiable there. …
Common Mistakes
Mistake 1: Trying to prove non-differentiability directly from the derivative limit without first checking continuity
Why it's wrong: it's more work, and easy to make sign errors, to jump straight into computing limh→0h[n+h]−n from both sides without first noticing the simpler fact that [x] isn't even continuous at n. Correct approach: always check continuity first at a suspected trouble point — if it fails, non-differentiability follows immediately and no derivative computation is required.
Mistake 2: Evaluating [1+h] or [2+h] incorrectly for negative h
Why it's wrong: for h a small negative number, 1+h is just below 1 (e.g. 0.99), so [1+h]=0, not 1 — students sometimes assume the floor value doesn't change until h crosses a whole unit. Correct approach: pick a concrete small value (like h=−0.01) and evaluate [1+h] numerically before generalizing. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Let [x] denote the greatest integer less than or equal to x. Then limx→0−[x]sin[x] is equal to (A) 1 (B) −sin1 (C) −2 (D) 0 (E) sin1
›Reveal solutionSolution
Evaluate [x] just below 0: it equals −1, a constant, so the limit is direct.
As x→0−, [x]=−1. Therefore …
- KEAM 2026Set eng-2026-04224 marksMCQQ.The range of f(x)=[cosx], where [x] is the greatest integer less than or equal to x, is (A) (−1,1) (B) {−1,0} (C) {1,0} (D) {−1,0,1} (E) [−1,1]
›Reveal solutionSolution
Greatest-integer of cosx takes values {−1,0,1}.
Since −1≤cosx≤1:
- cosx=1 (at x=2nπ) gives [cosx]=1,
- 0≤cosx<1 gives [cosx]=0, …
- KEAM 2026Set eng-2026-04224 marksMCQQ.The value of limx→2+x−2[x]−2 is (A) −2 (B) 4 (C) 2 (D) 1 (E) 0
›Reveal solutionSolution
Just right of 2 the greatest-integer part is exactly 2, killing the numerator.
For x slightly greater than 2 (but <3), [x]=2.
So the numerator [x]−2=0 while the denominator x−2→0+ is nonzero for x>2. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.If [x] denotes the greatest integer less than or equal to x for x∈R, then the value of limx→0−(2[x]−∣x∣x) is equal to (A) 1 (B) −1 (C) −3 (D) 3 (E) 0
›Reveal solutionSolution
Evaluate the two pieces from the left: [x]→−1 and ∣x∣x→−1, giving −1.
For x approaching 0 from the left, x is a small negative number, so the greatest integer ≤x is [x]=−1.
Also for x<0, ∣x∣=−x, hence ∣x∣x=−xx=−1. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.Let f(x)=[x],x∈(0,6), where [x] is the greatest integer function. Then the number of discontinuities of f(x) (A) 1 (B) 2 (C) 3 (D) 4 (E) 5
›Reveal solutionSolution
The greatest integer function is discontinuous at each integer; within the open interval (0,6) these are 1,2,3,4,5, giving 5. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.If [x] is the greatest integer less than or equal to x, then limx→0−[x]sin[x] is equal to (A) 1 (B) sin1 (C) −1 (D) 0 (E) −sin1
›Reveal solutionSolution
As x→0− the greatest integer [x]=−1; substituting gives −1sin(−1)=sin1.
For x slightly less than 0 (e.g. −0.001), [x]=−1 (constant). Therefore …
- KEAM 2025Set eng-2025-04284 marksMCQQ.If f(x)=[2x], where [x] denotes the greatest integer function in x, then the image of {−2.3,2.9} is (A) {−5,3} (B) {−5,5} (C) {−4,5} (D) {−3,2} (E) {−4,6}
›Reveal solutionSolution
Apply f(x)=[2x]: [2(−2.3)]=[−4.6]=−5 and [2(2.9)]=[5.8]=5.
The greatest integer function [x] gives the largest integer ≤x.
For x=−2.3: 2x=−4.6, so [−4.6]=−5 (the greatest integer ≤−4.6). …
- KEAM 2024Set eng-2024-06054 marksMCQQ.Let [x] be the greatest integer less than or equal to x. Then limx→0−∣x∣x([x]+∣x∣) is equal to (A) −1 (B) −2 (C) 0 (D) 1 (E) 2
›Reveal solutionSolution
Approaching 0 from the left, [x]=−1 and ∣x∣=−x; simplifying gives limit 1.
For x→0− (small negative x), the greatest integer [x]=−1 and ∣x∣=−x. So …
- KEAM 2024Set eng-2024-06084 marksMCQQ.If f(x)=[x], where [x] denotes the greatest integer function, and if the domain of f is {−3.01, 2.99}, then the range of f is (A) {−3, 3} (B) {−4, 3} (C) {−3, 2} (D) {−4, 2} (E) {−2, 3}
›Reveal solutionSolution
Applying the greatest-integer function to the domain {−3.01, 2.99} gives [−3.01]=−4 and [2.99]=2, so the range is {−4,2}.
The greatest integer function [x] returns the largest integer ≤x.
For x=−3.01: the integers ≤−3.01 have greatest value −4 (since −3>−3.01), so
[−3.01]=−4. …
- KEAM 2024Set eng-2024-06094 marksMCQQ.Let f(x)=x−[x], x∈(−1,2), where [⋅] denotes the greatest integer function. The number of points at which the function is not continuous is (A) 1 (B) 2 (C) 3 (D) 4 (E) 0
›Reveal solutionSolution
The fractional-part function x−[x] jumps at every integer; the integers inside (−1,2) are 0 and 1, giving 2 discontinuities.
The function f(x)=x−[x] is the fractional part; it is continuous everywhere except at integer values, where the greatest-integer term [x] jumps.
Within the open interval (−1,2) the integers present are x=0 and x=1. …
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.Let f(x)=[x],x∈R, where [x] denotes the greatest integer ≤x. Then the images of the elements −4.6 and 2.7 are respectively (A) −5,2 (B) −5,3 (C) −4,2 (D) −3,3 (E) −4,3
›Reveal solutionSolution
The images of −4.6 and 2.7 under the floor function are −5 and 2.
Concept and Intuition
The greatest integer (floor) function [x] returns the largest integer that is less than or equal to x. For negative non-integers this rounds downward (more negative), which is the usual trap.
Step-by-Step Solution
- [−4.6]: the integers ≤−4.6 are …,−6,−5; the greatest is −5.
- [2.7]: the integers ≤2.7 are …,1,2; the greatest is 2. …
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