Question

If f(x) = \(\left\{ \begin{aligned}     & \frac{1-\cos Kx}{x\sin x} ,\ \ \text{if x}\neq0 \\     &\ \ \ \ \ \  \ \frac{1}{2}\ \ \ \ \ \ \ \ \  ,\ \ \  \text{if x=0} \end{aligned} \right.\) is continuous at x = 0, then the value of K is

• A. \(±\frac{1}{2}\)
• B. 0
• C. ±2
• D. ±1

Answer 1

The Correct Option is D

If \(f(x) = \begin{cases} \frac{1 - \cos(Kx)}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases}\) is continuous at x = 0, then we need to find the value of K.

For f(x) to be continuous at x = 0, we must have \(\lim_{x \to 0} f(x) = f(0)\).

So, \(\lim_{x \to 0} \frac{1 - \cos(Kx)}{x \sin x} = \frac{1}{2}\).

We can use the trigonometric identity \(1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})\):

\(\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x \sin x} = \frac{1}{2}\)

\(\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x^2} \cdot \frac{x}{\sin x} = \frac{1}{2}\)

We know that \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), so \(\lim_{x \to 0} \frac{x}{\sin x} = 1\).

\(\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x^2} \cdot 1 = \frac{1}{2}\)

\(2 \lim_{x \to 0} \frac{\sin^2(\frac{Kx}{2})}{x^2} = \frac{1}{2}\)

\(\lim_{x \to 0} \frac{\sin^2(\frac{Kx}{2})}{x^2} = \frac{1}{4}\)

\(\lim_{x \to 0} (\frac{\sin(\frac{Kx}{2})}{x})^2 = \frac{1}{4}\)

We can rewrite this as:

\(\lim_{x \to 0} (\frac{\sin(\frac{Kx}{2})}{\frac{Kx}{2}} \cdot \frac{K}{2})^2 = \frac{1}{4}\)

Since \(\lim_{x \to 0} \frac{\sin(\frac{Kx}{2})}{\frac{Kx}{2}} = 1\), we have:

\((1 \cdot \frac{K}{2})^2 = \frac{1}{4}\)

\(\frac{K^2}{4} = \frac{1}{4}\)

\(K^2 = 1\)

\(K = \pm 1\)

Therefore, the correct option is (D) \(\pm 1\).