Question

Let \( a, b \in \mathbb{R}, b \neq 0 \). Define a function \[ f(x) = \begin{cases} a \sin \frac{\pi}{2}(x - 1), & \text{for } x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & \text{for } x > 0 \end{cases} \] If \( f \) is continuous at \( x = 0 \), then \( 10 - ab \) is equal to ________.

Hint:

Remember the useful expansion or limit: $\tan \theta - \sin \theta \approx \frac{1}{2} \theta^3$ as $\theta \to 0$. This can save time in limit evaluations.

Answer 1

Correct Answer: 14

Step 1: Understanding the Concept:
For a function to be continuous at a point $x = c$, the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the value of the function at that point must be equal.
Step 2: Detailed Explanation:
1. Left-Hand Limit (LHL) at $x = 0$:
\[ \text{LHL} = \lim_{x \to 0^-} a \sin \frac{\pi}{2}(x - 1) = a \sin \left( -\frac{\pi}{2} \right) = -a \]
2. Right-Hand Limit (RHL) at $x = 0$:
\[ \text{RHL} = \lim_{x \to 0^+} \frac{\tan 2x - \sin 2x}{bx^3} \]
Using the identity $\tan 2x - \sin 2x = \sin 2x \left( \frac{1}{\cos 2x} - 1 \right) = \sin 2x \frac{1 - \cos 2x}{\cos 2x}$:
\[ \text{RHL} = \lim_{x \to 0^+} \frac{\sin 2x (2 \sin^2 x)}{bx^3 \cos 2x} \]
Applying standard limits $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
\[ \text{RHL} = \frac{1}{b} \lim_{x \to 0^+} \left[ \frac{\sin 2x}{2x} \cdot 2 \cdot \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{\cos 2x} \right] = \frac{1}{b} (1 \cdot 2 \cdot 1^2 \cdot 1) \cdot \frac{2x \cdot x^2}{x^3} = \frac{4}{b} \]
3. Equating LHL and RHL for continuity:
\[ -a = \frac{4}{b} \implies ab = -4 \]
4. Calculating the required value:
\[ 10 - ab = 10 - (-4) = 14 \]
Step 3: Final Answer:
The value of $10 - ab$ is 14.