Question

If $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2} & , \text{if } x<0 \\ a & , \text{if } x = 0 \\ \frac{(16+\sqrt{x})^{\frac{1}{2}}-4}{\sqrt{x}} & , \text{if } x>0 \end{cases}$ is continuous at $x = 0$, then a =}

• A. 4
• B. 8
• C. -4
• D. -8

Hint:

$\lim_{x \to 0} \frac{1-\cos kx}{x^2} = \frac{k^2}{2}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a function $f(x)$ to be continuous at $x = 0$, the left-hand limit (LHL), the right-hand limit (RHL), and the value of the function at that point must all be equal.
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = a \] Step 2: Key Formula or Approach:
Use the standard trigonometric limit $\lim_{\theta \to 0} \frac{1 - \cos k\theta}{\theta^2} = \frac{k^2}{2}$.
For the right-hand limit, we will use rationalization or binomial approximation for small values.
Step 3: Detailed Explanation:
1. Calculate the Left-Hand Limit (LHL):
\[ \text{LHL} = \lim_{x \to 0^-} \frac{1 - \cos 4x}{x^2} = \frac{4^2}{2} = \frac{16}{2} = 8 \] 2. Since the function is continuous at $x = 0$, the value $f(0) = a$ must equal the LHL.
\[ a = 8 \] 3. (Verification) Calculate the Right-Hand Limit (RHL):
\[ \text{RHL} = \lim_{x \to 0^+} \frac{\sqrt{16+\sqrt{x}}-4}{\sqrt{x}} = \lim_{x \to 0^+} \frac{(16+\sqrt{x})-16}{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)} \] \[ \text{RHL} = \frac{1}{\sqrt{16}+4} = \frac{1}{4+4} = \frac{1}{8} \] Step 4: Final Answer:
The value of a is 8.