Question

Let $f(x) = \begin{cases} \frac{1-\sec^2(\alpha x)}{\alpha x^2}, & \text{for } x \neq 0 \\ -3, & \text{for } x = 0 \end{cases}$ be continuous at $x = 0$. Then the value of $\alpha$ is equal to

• A. -3
• B. 3
• C. 1
• D. -1

Hint:

Standard limits like $\frac{\tan \theta}{\theta} \to 1$ are powerful shortcuts. Always try to arrange your expression into this form when dealing with continuity problems involving trigonometric functions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a function to be continuous at \(x = a\), the limit of the function as \(x\) approaches \(a\) must equal the function's value at \(a\).
Therefore, we require \(\lim_{x \to 0} f(x) = f(0)\).
Step 2: Key Formula or Approach:
We use the trigonometric identity \(1 - \sec^2 \theta = -\tan^2 \theta\).
Then apply the standard limit \(\lim_{x \to 0} \frac{\tan x}{x} = 1\).
Step 3: Detailed Explanation:
Evaluate the limit as \(x \to 0\):
\[ L = \lim_{x \to 0} \frac{1 - \sec^2(\alpha x)}{\alpha x^2} \] Using the identity \(1 - \sec^2(\alpha x) = -\tan^2(\alpha x)\):
\[ L = \lim_{x \to 0} \frac{-\tan^2(\alpha x)}{\alpha x^2} \] Rewrite the expression to use the standard limit \(\frac{\tan(\alpha x)}{\alpha x} \to 1\):
\[ L = \lim_{x \to 0} \left( - \frac{\tan^2(\alpha x)}{\alpha^2 x^2} \cdot \alpha \right) \] \[ L = -\alpha \lim_{x \to 0} \left( \frac{\tan(\alpha x)}{\alpha x} \right)^2 \] Since \(\lim_{x \to 0} \frac{\tan(\alpha x)}{\alpha x} = 1\):
\[ L = -\alpha (1)^2 = -\alpha \] For the function to be continuous at \(x = 0\), this limit must equal \(f(0)\).
Given \(f(0) = -3\):
\[ -\alpha = -3 \implies \alpha = 3 \] Step 4: Final Answer:
The value of \(\alpha\) is 3.