CBSE Board Examination 2025 · Section A · 1 mark

Q5: Is $f(x) = |x| + |x - 1|$ continuous and differentiable at $0$ and $1$ ⁠?

Board 2025Ch 5 · Continuity and Differentiability

The question

If $f(x) = |x| + |x - 1|$ ⁠, then which of the following is correct ?

(A) $f(x)$ is both continuous and differentiable, at $x = 0$ and $x = 1$ ⁠.

(B) $f(x)$ is differentiable but not continuous, at $x = 0$ and $x = 1$ ⁠.

(D) $f(x)$ is neither continuous nor differentiable, at $x = 0$ and $x = 1$ ⁠.

Exam answer

f(x) = \begin{cases} 1 - 2x, & x < 0 \\ 1, & 0 \le x < 1 \\ 2x - 1, & x \ge 1 \end{cases}

Why?

Each modulus changes its formula where its inside is 0: $|x|$ at $x = 0$ and $|x - 1|$ at $x = 1$ ⁠. Use $|x| = -x$ for $x < 0$ ⁠, and $|x - 1| = 1 - x$ for $x < 1$ ⁠. Then add on each interval:

\begin{aligned} x < 0&: -x + 1 - x = 1 - 2x \\ 0 \le x < 1&: x + 1 - x = 1 \\ x \ge 1&: x + x - 1 = 2x - 1 \end{aligned}

At $x = 0$ ⁠: LHL $= 1 - 0 = 1$ ⁠, RHL $= 1$ ⁠, $f(0) = 1$ ⁠.

At $x = 1$ ⁠: LHL $= 1$ ⁠, RHL $= 2 - 1 = 1$ ⁠, $f(1) = 1$ ⁠.

So f is continuous at $x = 0$ and $x = 1$ ⁠.

Why?

f is continuous at a point when LHL = RHL = the value there. The LHL uses the piece on the left of the point, and the RHL the piece on the right.

At $x = 0$ ⁠: left piece $1 - 2x$ ⁠, right piece 1.

\begin{aligned} \text{LHL} &= 1 - 2(0) = 1 - 0 = 1 \\ \text{RHL} &= 1 \\ f(0) &= 1 \quad (\text{middle piece}) \end{aligned}

At $x = 1$ ⁠: left piece 1, right piece $2x - 1$ ⁠.

\begin{aligned} \text{LHL} &= 1 \\ \text{RHL} &= 2(1) - 1 = 2 - 1 = 1 \\ f(1) &= 2(1) - 1 = 1 \end{aligned}

This also follows at once: $|x|$ and $|x - 1|$ are continuous everywhere, and a sum of continuous functions is continuous.

At $x = 0$ ⁠, with $h > 0$ ⁠:

\begin{aligned} \text{LHD} &= \lim_{h \to 0}\frac{f(-h) - f(0)}{-h} \\ &= \lim_{h \to 0}\frac{1 + 2h - 1}{-h} = -2 \\ \text{RHD} &= \lim_{h \to 0}\frac{f(h) - f(0)}{h} \\ &= \lim_{h \to 0}\frac{1 - 1}{h} = 0 \end{aligned}

Why?

The one-sided derivatives at $x = a$ are

\begin{aligned} \text{LHD} &= \lim_{h \to 0}\frac{f(a - h) - f(a)}{-h} \\ \text{RHD} &= \lim_{h \to 0}\frac{f(a + h) - f(a)}{h} \end{aligned}

with $h > 0$ ⁠. Since $-h < 0$ ⁠, use $1 - 2x$ ⁠: $f(-h) = 1 + 2h$ ⁠. For small h, h lies in $(0, 1)$ ⁠, so $f(h) = 1$ ⁠.

\begin{aligned} f(-h) &= 1 - 2(-h) = 1 + 2h \\[4pt] \frac{f(-h) - f(0)}{-h} &= \frac{1 + 2h - 1}{-h} \\ &= \frac{2h}{-h} = -2 \\[4pt] \frac{f(h) - f(0)}{h} &= \frac{1 - 1}{h} \\ &= \frac{0}{h} = 0 \end{aligned}

The h cancels before the limit is taken, so the limits are $-2$ and 0.

At $x = 1$ ⁠:

\begin{aligned} \text{LHD} &= \lim_{h \to 0}\frac{f(1 - h) - f(1)}{-h} \\ &= \lim_{h \to 0}\frac{1 - 1}{-h} = 0 \\ \text{RHD} &= \lim_{h \to 0}\frac{f(1 + h) - f(1)}{h} \\ &= \lim_{h \to 0}\frac{2(1 + h) - 1 - 1}{h} \\ &= \lim_{h \to 0}\frac{2h}{h} = 2 \end{aligned}

Why?

Here $a = 1$ and $f(1) = 1$ ⁠. For small h, $1 - h$ lies in $(0, 1)$ ⁠, so $f(1 - h) = 1$ ⁠. And $1 + h > 1$ ⁠, so use $2x - 1$ ⁠:

\begin{aligned} f(1 + h) &= 2(1 + h) - 1 \\ &= 2 + 2h - 1 = 1 + 2h \\[4pt] f(1 + h) - f(1) &= 1 + 2h - 1 = 2h \\[4pt] \frac{2h}{h} &= 2 \end{aligned}

LHD $\ne$ RHD at both points, so f is not differentiable at $x = 0$ and $x = 1$ ⁠.

Why?

Watch out: continuous does not mean differentiable. The graph has no break at 0 or 1, but it has sharp corners there: the slope jumps from $-2$ to 0 at $x = 0$ ⁠, and from 0 to 2 at $x = 1$ ⁠.

Answer: (C) $f(x)$ is continuous but not differentiable, at $x = 0$ and $x = 1$ ⁠.

Where this comes from

Chapter 5, Continuity and Differentiability: Continuity and differentiability of |x| + |x − 1|.

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

How to spot it: click to see

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

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

Here: First, rewrite $|x| + |x - 1|$ in three pieces, with joints at 0 and 1. At each joint LHL, RHL and $f$ are all 1, so $f$ is continuous there.

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

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

Here: Then the slopes: at 0, LHD is $-2$ and RHD is 0; at 1, LHD is 0 and RHD is 2. They differ, so neither point is differentiable: option (C).

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Q5: Is $f(x) = |x| + |x - 1|$ continuous and differentiable at $0$ and $1$ ⁠?