Question

The function defined by \[ f(x) = \begin{cases} \frac{\sin x}{x} + \cos x & \text{if } x>0 \\ -5k & \text{if } x = 0 \\ \frac{4(1 - \sqrt{1 - x})}{x} & \text{if } x<0 \end{cases} \] is continuous at

• A. \( \frac{-2}{5} \)
• B. -2
• C. 2
• D. \( \frac{-5}{2} \)

Hint:

For continuity, ensure the left-hand limit, right-hand limit, and the function value at the point match.

Answer 1

The Correct Option is A

To check the continuity of the function \( f(x) \) at \( x = 0 \), we need to ensure that: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] From the problem, we know that: - \( f(0) = -5k \) - \( \lim_{x \to 0^+} f(x) = \frac{\sin x}{x} + \cos x = 1 + 1 = 2 \) - \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{4(1 - \sqrt{1 - x})}{x} \) We will now compute \( \lim_{x \to 0^-} \frac{4(1 - \sqrt{1 - x})}{x} \) using the binomial expansion for \( \sqrt{1 - x} \): \[ \sqrt{1 - x} \approx 1 - \frac{x}{2} \quad \text{for small } x \] Thus: \[ 1 - \sqrt{1 - x} \approx \frac{x}{2} \] So: \[ \lim_{x \to 0^-} \frac{4(1 - \sqrt{1 - x})}{x} = \lim_{x \to 0^-} \frac{4 \times \frac{x}{2}}{x} = 2 \] For continuity at \( x = 0 \), we need: \[ 2 = -5k \] Solving for \( k \): \[ k = \frac{-2}{5} \] Thus, the correct answer is \( \frac{-2}{5} \).