Sample 2020-21 Class 12 Maths Q21 Solution: Values of k that make (1 − cos kx)/(x sin x) continuous at x = 0

The question

Find the value(s) of k so that the following function is continuous at $x = 0$

f(x) = \begin{cases} \dfrac{1 - \cos kx}{x\sin x} & \text{if } x \ne 0 \\ \dfrac12 & \text{if } x = 0 \end{cases}

Exam answer

Using $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$ ⁠,

Why?

Putting $x = 0$ straight in gives $\frac{0}{0}$ ⁠, so the expression must be rewritten first. The standard limit we can use is

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

so we turn the cosine into a sine. The identity comes from $\cos\theta = 1 - 2\sin^2\frac{\theta}{2}$ ⁠. Here $\theta = kx$ ⁠, so the half angle is $\frac{kx}{2}$ ⁠.

\begin{aligned} &\lim_{x \to 0}\frac{1 - \cos kx}{x\sin x} \\ &= \lim_{x \to 0}\frac{2\sin^2\left(\frac{kx}{2}\right)}{x\sin x} \\ &= \lim_{x \to 0}\frac{\frac{2\sin^2\left(\frac{kx}{2}\right)}{x^2}}{\frac{x\sin x}{x^2}} \\ &= \frac{\displaystyle\lim_{x \to 0} 2\left(\frac{\sin\frac{kx}{2}}{\frac{kx}{2}}\right)^2 \times \left(\frac{k}{2}\right)^2}{\displaystyle\lim_{x \to 0}\frac{\sin x}{x}} \\ &= \frac{2 \times 1 \times \frac{k^2}{4}}{1} = \frac{k^2}{2} \end{aligned}

Why?

Dividing the numerator and the denominator by $x^2$ leaves the fraction unchanged, and turns each part into a standard limit:

\begin{aligned} \frac{x\sin x}{x^2} &= \frac{\sin x}{x} \\ \frac{\sin^2\frac{kx}{2}}{x^2} &= \left(\frac{\sin\frac{kx}{2}}{\frac{kx}{2}}\right)^2\frac{k^2}{4} \end{aligned}

The second line works because $x = \frac{kx}{2} \cdot \frac{2}{k}$ ⁠, so

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

Then put in the limits, top and bottom:

\begin{aligned} \frac{2 \times 1^2 \times \frac{k^2}{4}}{1} &= \frac{2k^2}{4} \\ &= \frac{2k^2 \div 2}{4 \div 2} = \frac{k^2}{2} \end{aligned}

As $x \to 0$ ⁠, also $\frac{kx}{2} \to 0$ ⁠, so each ratio of the form $\frac{\sin\theta}{\theta}$ tends to 1.

Watch out: the factor is $\left(\frac{k}{2}\right)^2 = \frac{k^2}{4}$ ⁠, not $\frac{k}{2}$ ⁠.

Since $f(x)$ is continuous at $x = 0$ ⁠,

\lim_{x \to 0} f(x) = f(0)

Why?

A function is continuous at a point when its limit there equals its value there. The limit is the $\frac{k^2}{2}$ found above. The value $f(0) = \frac12$ is given by the second line of the definition of $f$ ⁠.

\Rightarrow \frac{k^2}{2} = \frac12 \Rightarrow k^2 = 1

Why?

\begin{aligned} \frac{k^2}{2} &= \frac12 \\ k^2 &= 1 \quad (\times 2) \\ k &= \pm\sqrt{1} = \pm 1 \end{aligned}

$k^2 = 1$ has two roots, $k = 1$ and $k = -1$ ⁠. Both work, because $\cos(-kx) = \cos kx$ ⁠: the function is the same for $k$ and $-k$ ⁠.

Watch out: do not drop $k = -1$ ⁠. The question asks for the value(s).

$\therefore\ k = \pm 1$

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: Away from 0 the piece is $\frac{1 - \cos kx}{x\sin x}$ ⁠; with $1 - \cos kx = 2\sin^2\frac{kx}{2}$ its limit is $\frac{k^2}{2}$ ⁠. Set this equal to $f(0) = \frac12$ ⁠: $k = \pm1$ ⁠.

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Q21: Values of kkk that make 1−cos⁡kxxsin⁡x\frac{1 - \cos kx}{x\sin x}xsinx1−coskx​ continuous at x=0x = 0x=0

Q21: Values of $k$ that make $\frac{1 - \cos kx}{x\sin x}$ continuous at $x = 0$