If the function \( f(x)=\begin{cases}\displaystyle\frac{1-\cos kx}{\sin^2 x} & ,x\ne 0\\ \displaystyle\frac{1}{2} & ,x=0\end{cases} \) is continuous at \( x=0 \), then the value of \( k \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \pm 1 \)
For continuity at x = 0, the limit must equal f(0) = \( \frac{1}{2} \). Using \( 1-\cos kx=2\sin^2\frac{kx}{2} \):
\( \displaystyle \lim_{x\to 0}\frac{1-\cos kx}{\sin^2 x}=\lim_{x\to 0}\frac{2\sin^2\frac{kx}{2}}{\sin^2 x} \)
Writing each sine over its angle:
\( \displaystyle =\lim_{x\to 0}\frac{2\left(\frac{\sin\frac{kx}{2}}{\frac{kx}{2}}\right)^2\cdot\frac{k^2x^2}{4}}{\left(\frac{\sin x}{x}\right)^2\cdot x^2} \)
\( \displaystyle =\frac{2\cdot\frac{k^2}{4}}{1} \)
\( \displaystyle =\frac{k^2}{2} \)
\( \displaystyle \frac{k^2}{2}=\frac{1}{2} \)
\( \displaystyle k^2=1\ \Rightarrow\ k=\pm 1 \)
So the values of k are k = 1 and k = −1, i.e. ±1.
Tip: Because \( \cos(-kx)=\cos kx \), k and −k always give the same function here, so a single sign can never be the whole answer; the complete set of values is ±1.