Question

If \(f(x)=\begin{cases} \frac{1-\cos4x}{x^2} & x\ne0 \\ k & x=0 \end{cases}\) is continuous at x = 0, then the value of k is :

• A. 8
• B. 7
• C. 6
• D. 4

Answer 1

The Correct Option is A

To determine the value of \( k \) such that \( f(x) \) is continuous at \( x = 0 \), we must ensure that \(\lim_{x \to 0} f(x) = f(0)\). Given: \( f(x)=\begin{cases} \frac{1-\cos4x}{x^2} & x\ne0 \\ k & x=0 \end{cases} \).

1. Calculate the limit \( \lim_{x \to 0} \frac{1-\cos4x}{x^2} \):

Using the Taylor series, \(\cos4x \approx 1 - \frac{(4x)^2}{2} + \cdots\), so \(1-\cos4x \approx \frac{16x^2}{2} = 8x^2\).

2. Substitute this back into the limit:

\(\lim_{x \to 0} \frac{1-\cos4x}{x^2} = \lim_{x \to 0} \frac{8x^2}{x^2} = \lim_{x \to 0} 8 = 8\).

3. Set the limit equal to \( f(0) \) for continuity:

\( \lim_{x \to 0} f(x) = f(0) = k \Rightarrow 8 = k \).

Hence, the value of \( k \) is 8.