CBSE Sample Paper 2023-24 · Section A · 1 mark

Q4: $k$ for which $\frac{kx}{|x|}$ ( $x \lt 0$ ⁠), $3$ ( $x \ge 0$ ⁠) is continuous at 0

The question

f(x) = \begin{cases} \dfrac{kx}{|x|}, & \text{if } x < 0 \\ 3, & \text{if } x \ge 0 \end{cases}

is continuous at $x = 0$ ⁠, then the value of $k$ is

(a) $-3$

(b) $0$

(d) any real number

Exam answer

For $x < 0$ ⁠, $|x| = -x$ ⁠, so $f(x) = \frac{kx}{-x} = -k$ ⁠.

Why?

By definition, $|x| = x$ when $x \ge 0$ and $|x| = -x$ when $x < 0$ ⁠. For example, $\lvert -2\rvert = 2 = -(-2)$ ⁠.

Since $x \ne 0$ here, the $x$ cancels, leaving the constant $-k$ ⁠.

\begin{aligned} \frac{kx}{|x|} &= \frac{kx}{-x} \\ &= -\frac{kx}{x} \\ &= -k \quad (\text{divide by } x) \end{aligned}

\begin{aligned} \lim_{x \to 0^-}f(x) &= -k \\ \lim_{x \to 0^+}f(x) &= 3 = f(0) \end{aligned}

Why?

For $x \ge 0$ the function is the constant 3, so the right-hand limit and $f(0)$ are both 3. Note that $x = 0$ itself uses the rule for $x \ge 0$ ⁠.

For continuity at $x = 0$ ⁠: $-k = 3$ ⁠, so $k = -3$ ⁠.

Why?

$f$ is continuous at $x = a$ when

\lim_{x \to a^-}f = \lim_{x \to a^+}f = f(a)

Here that means $-k = 3$ ⁠. Multiply both sides by $-1$ ⁠:

\begin{aligned} -k &= 3 \\ k &= -3 \end{aligned}

Watch out: $k = 3$ gives a left-hand limit of $-3$ ⁠, not 3. The sign change comes from $|x| = -x$ ⁠.

Answer: (a) $-3$

Where this comes from

Chapter 5, Continuity and Differentiability: Continuity of a piecewise function at a point.

Question type: Find $k$ so that $f$ is continuous

How to spot it: click to see

Type 2. Find $k$ so that $f$ is continuous

Spot it: a function in pieces with an unknown $k$ (or $a$ and $b$ ⁠), and the words "find the value of $k$ ".

Here: Left of 0, $|x| = -x$ ⁠, so $\frac{kx}{|x|} = -k$ and LHL is $-k$ ⁠. The piece 3 gives RHL and $f(0)$ ⁠. Set $-k = 3$ ⁠: $k = -3$ ⁠.

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Continuity and Differentiability in past papers: question 27 of 35 · see them all

Q4: kkk for which kx∣x∣\frac{kx}{|x|}∣x∣kx​ (x<0x \lt 0x<0⁠), 333 (x≥0x \ge 0x≥0⁠) is continuous at 0