CBSE Sample Paper 2022-23 · Section A · 1 mark

Q4: $k$ so that $\frac{1 - \cos 4x}{8x^2}$ is continuous at $0$

Sample 2022-23Ch 5 · Continuity and Differentiability

The question

The value of 'k' for which the function

f(x) = \begin{cases} \dfrac{1 - \cos 4x}{8x^2}, & \text{if } x \ne 0 \\ k, & \text{if } x = 0 \end{cases}

is continuous at x = 0 is

(a) $0$

(b) $-1$

(d) $2$

Exam answer

f is continuous at $x = 0$ if $\lim_{x \to 0}f(x) = f(0)$ ⁠, and $f(0) = k$ ⁠.

Why?

Continuity at a point means the limit there exists and equals the value of the function there. The value at $x = 0$ comes from the second line of f, which is $k$ ⁠. The limit uses the first line, because $x \ne 0$ as $x \to 0$ ⁠.

\begin{aligned} &\lim_{x \to 0}f(x) \\ &= \lim_{x \to 0}\frac{1 - \cos 4x}{8x^2} \\ &= \lim_{x \to 0}\frac{2\sin^2 2x}{8x^2} \end{aligned}

Why?

Putting $x = 0$ straight in gives $\frac{0}{0}$ ⁠, so rewrite first. Use the double-angle identity with $\theta = 2x$ ⁠:

1 - \cos 2\theta = 2\sin^2\theta

So $1 - \cos 4x = 2\sin^2 2x$ ⁠.

\begin{aligned} &= \lim_{x \to 0}\frac{\sin^2 2x}{4x^2} \\ &= \lim_{x \to 0}\left(\frac{\sin 2x}{2x}\right)^2 = 1 \end{aligned}

Why?

Cancel 2 from $\frac{2}{8}$ to get $\frac{1}{4}$ ⁠. Then $4x^2 = (2x)^2$ ⁠, so the whole fraction is a square.

\begin{aligned} \frac{2\sin^2 2x}{8x^2} &= \frac{2 \div 2}{8 \div 2} \cdot \frac{\sin^2 2x}{x^2} \\ &= \frac{\sin^2 2x}{4x^2} \\ &= \frac{(\sin 2x)^2}{(2x)^2} \\ &= \left(\frac{\sin 2x}{2x}\right)^2 \end{aligned}

The standard limit, with $\theta = 2x \to 0$ ⁠:

\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1

Watch out: the angle and the denominator must match, $2x$ with $2x$ ⁠. That is why we write $4x^2$ as $(2x)^2$ ⁠.

So the limit is $(1)^2 = 1$ ⁠.

$\therefore\ k = 1$

Why?

Continuity needs the limit to equal $f(0)$ ⁠:

\begin{aligned} \lim_{x \to 0}f(x) &= f(0) \\ 1 &= k \end{aligned}

Answer: (c) $1$

Where this comes from

Chapter 5, Continuity and Differentiability: Continuity at a point: finding k.

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: $f(0) = k$ ⁠, and away from 0 the piece is $\frac{1 - \cos 4x}{8x^2}$ ⁠. With $1 - \cos 4x = 2\sin^2 2x$ the limit is $\left(\frac{\sin 2x}{2x}\right)^2 \to 1$ ⁠, so $k = 1$ ⁠.

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Q4: $k$ so that $\frac{1 - \cos 4x}{8x^2}$ is continuous at $0$