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 $f(x)$ to be continuous at a point $x = a$, the limit as $x \to a$ must exist and be equal to the value of the function at that point, i.e., $\lim_{x \to a} f(x) = f(a)$.

Step 2: Detailed Explanation:
1. Given $f(0) = -3$.
2. For continuity at $x = 0$, we must have $\lim_{x \to 0} \frac{1-\sec^2(\alpha x)}{\alpha x^2} = -3$.
3. Use the identity $1 - \sec^2 \theta = -\tan^2 \theta$:
\[ \lim_{x \to 0} \frac{-\tan^2(\alpha x)}{\alpha x^2} = -3 \]
4. Multiply and divide by $\alpha$:
\[ -\lim_{x \to 0} \frac{\tan^2(\alpha x)}{\alpha^2 x^2} \cdot \frac{\alpha^2}{\alpha} = -3 \]
\[ -\lim_{x \to 0} \left( \frac{\tan(\alpha x)}{\alpha x} \right)^2 \cdot \alpha = -3 \]
5. Since $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$:
\[ -(1)^2 \cdot \alpha = -3 \]
\[ -\alpha = -3 \implies \alpha = 3 \]

Step 3: Final Answer:
The value of $\alpha$ is 3.