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CBSE Sample Paper 2024-25 · Section A · 1 mark

Q17: The greatest integer function f(x)=[x]f(x) = [x] at x=2.5x = 2.5

Sample 2024-25Ch 5 · Continuity and Differentiability
The question

The function f:R→Zf : R \to Z defined by f(x)=[x]f(x) = [x]⁠; where [.][.] denotes the greatest integer function, is

(A) Continuous at x=2.5x = 2.5 but not differentiable at x=2.5x = 2.5

(B) Not Continuous at x=2.5x = 2.5 but differentiable at x=2.5x = 2.5

(C) Not Continuous at x=2.5x = 2.5 and not differentiable at x=2.5x = 2.5

(D) Continuous as well as differentiable at x=2.5x = 2.5

Exam answer

For 2<x<32 < x < 3⁠, f(x)=[x]=2f(x) = [x] = 2⁠, a constant.

Why?

[x][x] is the greatest integer less than or equal to xx⁠. For example [2.1]=2[2.1] = 2⁠, [2.5]=2[2.5] = 2 and [2.9]=2[2.9] = 2⁠. The value jumps only at the integers, and 2.5 is not an integer.

In particular f(2.5)=[2.5]=2f(2.5) = [2.5] = 2⁠, and

lim⁡x→2.5−f(x)=2lim⁡x→2.5+f(x)=2\begin{aligned} \lim_{x \to 2.5^-}f(x) &= 2 \\ \lim_{x \to 2.5^+}f(x) &= 2 \end{aligned} lim⁡x→2.5f(x)=2=f(2.5)\lim_{x \to 2.5}f(x) = 2 = f(2.5)

So ff is continuous at x=2.5x = 2.5⁠.

Why?

A function is continuous at a point when the left limit, the right limit and the value there are all equal. Points close to 2.5 on either side still lie in (2,3)(2, 3)⁠, where f=2f = 2⁠.

f′(2.5)=lim⁡h→0f(2.5+h)−f(2.5)h\begin{aligned} &f'(2.5) \\ &= \lim_{h \to 0}\frac{f(2.5 + h) - f(2.5)}{h} \end{aligned} f′(2.5)=lim⁡h→02−2h=0f'(2.5) = \lim_{h \to 0}\frac{2 - 2}{h} = 0
Why?

For ∣h∣<0.5|h| < 0.5⁠, 2.5+h2.5 + h stays in (2,3)(2, 3)⁠, so f(2.5+h)=2f(2.5 + h) = 2⁠. The top of the fraction is 0 for every such hh⁠, so the limit is 0 from both sides.

Watch out: [x][x] fails to be continuous only at integers. Away from them it is a constant, and a constant has derivative 0.

So ff is continuous and differentiable at x=2.5x = 2.5⁠.

Answer: (D) Continuous as well as differentiable at x=2.5x = 2.5

Where this comes from

Chapter 5, Continuity and Differentiability: Continuity and differentiability of [x].

Question type: Is ff continuous at the joint? Compare three numbers + Is ff differentiable at a point? Compare LHD and RHD (combines 2 types)

How to spot it: click to see

Type 1. Is ff continuous at the joint? Compare three numbers

Spot it: ff is given in pieces, and you are asked where it is continuous, or where it breaks.

Here: Between 2 and 3, [x][x] is the constant 2. So at x=2.5x = 2.5⁠, LHL, RHL and f(2.5)f(2.5) are all 2: continuous.

Type 3. Is ff differentiable at a point? Compare LHD and RHD

Spot it: a modulus such as ∣x−1∣|x - 1|⁠, or a function in pieces, and the word "differentiable".

Here: Then the slope: f(2.5+h)−f(2.5)=2−2=0f(2.5 + h) - f(2.5) = 2 - 2 = 0 for small hh on either side, so LHD and RHD are both 0: differentiable, option (D).

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