Sample 2023-24 Class 12 Maths Q17 Solution: Where f(x) = x + |x| is differentiable

The question

The set of all points where the function $f(x) = x + |x|$ is differentiable, is

(a) $(0, \infty)$

(b) $(-\infty, 0)$

(c) $(-\infty, 0) \cup (0, \infty)$

(d) $(-\infty, \infty)$

Exam answer

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

Why?

Remove the modulus. For $x \ge 0$ ⁠, $|x| = x$ ⁠, so $f(x) = x + x = 2x$ ⁠.

For $x < 0$ ⁠, $|x| = -x$ ⁠, so $f(x) = x - x = 0$ ⁠.

At $x = 0$ ⁠:

\begin{aligned} Lf'(0) &= \lim_{h \to 0^-}\frac{0 - 0}{h} = 0 \\ Rf'(0) &= \lim_{h \to 0^+}\frac{2h - 0}{h} = 2 \end{aligned}

Why?

The one-sided derivatives at 0 are

\begin{aligned} Lf'(0) &= \lim_{h \to 0^-}\frac{f(h) - f(0)}{h} \\ Rf'(0) &= \lim_{h \to 0^+}\frac{f(h) - f(0)}{h} \end{aligned}

Here $f(0) = 0$ ⁠. For $h < 0$ ⁠, $f(h) = 0$ ⁠, and for $h > 0$ ⁠, $f(h) = 2h$ ⁠.

\begin{aligned} \text{Left: } \frac{0 - 0}{h} &= \frac{0}{h} = 0 \\ \text{Right: } \frac{2h - 0}{h} &= \frac{2h}{h} = 2 \end{aligned}

Both quotients are constant ( $h \ne 0$ cancels), so the limits are 0 and 2.

$Lf'(0) \ne Rf'(0)$ ⁠, so $f$ is not differentiable at $x = 0$ ⁠.

Why?

Watch out: $f$ is continuous at 0, since both pieces give 0 there. It is still not differentiable: the graph has a corner at the origin, with slope 0 on the left and slope 2 on the right.

For $x > 0$ ⁠, $f(x) = 2x$ (linear) and for $x < 0$ ⁠, $f(x) = 0$ (constant), so $f$ is differentiable at every $x \ne 0$ ⁠.

So $f$ is differentiable exactly on $(-\infty, 0) \cup (0, \infty)$ ⁠. Options (a) and (b) leave out half of these points, and (d) wrongly includes 0.

Answer: (c) $(-\infty, 0) \cup (0, \infty)$

Where this comes from

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

Question type: Is $f$ differentiable at a point? Compare LHD and RHD

How to spot it: click to see

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: $|x|$ in $x + |x|$ makes $f$ equal $2x$ for $x \ge 0$ and 0 for negative $x$ ⁠. At 0, LHD is 0 and RHD is 2, so 0 is the only point left out: option (c).

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Q17: Where f(x)=x+∣x∣f(x) = x + |x|f(x)=x+∣x∣ is differentiable

Q17: Where $f(x) = x + |x|$ is differentiable