CBSE Board Examination 2023 · Section A · 1 mark

Q13: k for which $3x + 5$ ( $x \ge 2$ ⁠), $kx^2$ ( $x < 2$ ⁠) is continuous

Board 2023Ch 5 · Continuity and Differentiability

The question

The value of k for which $f(x) = \begin{cases} 3x + 5, & x \ge 2 \\ kx^2, & x < 2 \end{cases}$ is a continuous function, is :

(a) $-\frac{11}{4}$

(b) $\frac{4}{11}$

(d) $\frac{11}{4}$

Exam answer

Both pieces are polynomials, so f is continuous at every $x \ne 2$ ⁠. Check $x = 2$ ⁠:

Why?

Polynomials are continuous everywhere. Near any point other than 2, f is given by one polynomial only, so the only place where continuity can fail is the joining point $x = 2$ ⁠.

\begin{aligned} \lim_{x \to 2^-} kx^2 &= 4k \\ f(2) = 3(2) + 5 &= 11 \\ \lim_{x \to 2^+} (3x + 5) &= 11 \end{aligned}

Why?

Left of 2 means $x < 2$ ⁠, so use $kx^2$ ⁠: put $x = 2$ to get $k \cdot 2^2 = 4k$ ⁠.

The piece $3x + 5$ has $x \ge 2$ ⁠, which includes $x = 2$ ⁠. So it gives both $f(2)$ and the right-hand limit.

\begin{aligned} 3(2) + 5 &= 6 + 5 \\ &= 11 \end{aligned}

For continuity at $x = 2$ ⁠, the left-hand limit must equal $f(2)$ ⁠:

$\therefore\ 4k = 11 \Rightarrow k = \frac{11}{4}$

Why?

f is continuous at 2 when the left-hand limit, the right-hand limit and $f(2)$ are all equal. The right side already gives 11, so we need $4k = 11$ ⁠.

Divide both sides by 4:

\begin{aligned} 4k &= 11 \\ \frac{4k}{4} &= \frac{11}{4} \\ k &= \frac{11}{4} \end{aligned}

Watch out: dividing the wrong way gives $\frac{4}{11}$ ⁠, option (b).

Answer: (d) $\dfrac{11}{4}$

Where this comes from

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

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 2 the piece is $kx^2$ ⁠, so LHL is $4k$ ⁠; the piece $3x + 5$ gives $f(2) = 11$ and RHL 11. Set $4k = 11$ ⁠, so $k = \frac{11}{4}$ ⁠.

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Continuity and Differentiability in past papers: question 11 of 35 · see them all

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Q13: k for which $3x + 5$ ( $x \ge 2$ ⁠), $kx^2$ ( $x < 2$ ⁠) is continuous