Question

For $a, b > 0$, let $ f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0, \\ \frac{x}{3}, & x = 0, \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{a x \sqrt{x}}}, & x > 0 \end{cases} $ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to: 

• A. 5
• B. 4
• C. 8
• D. 6

Answer 1

The Correct Option is D

For \( f(x) \) to be continuous at \( x = 0 \), the condition \( \lim_{x \to 0} f(x) = f(0) = 3 \) must be satisfied.
Step 1: Evaluate the right-hand limit
Consider the limit:
\( \lim_{x \to 0^+} \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{ax} \sqrt{x}} = 3. \)
Step 2: Rationalize the numerator
Multiply the numerator and denominator by the conjugate of the numerator:
\( \lim_{x \to 0^+} \frac{(\sqrt{ax + b^2x^2} - \sqrt{ax})(\sqrt{ax + b^2x^2} + \sqrt{ax})}{b \sqrt{ax} \sqrt{x} (\sqrt{ax + b^2x^2} + \sqrt{ax})}. \)
Step 3: Simplify the expression
This simplifies to:
\( \lim_{x \to 0^+} \frac{ax + b^2x^2 - ax}{b x^{3/2} (\sqrt{ax + b^2x^2} + \sqrt{ax})}. \)
Step 4: Cancel terms and simplify further
\( \lim_{x \to 0^+} \frac{b^2}{b \sqrt{a} (\sqrt{a + b^2x} + \sqrt{a})}. \)
Step 5: Substitute \( x = 0 \)
\( \frac{b}{\sqrt{a} \cdot 2\sqrt{a}} = \frac{b}{2a}. \)
Step 6: Apply the continuity condition
Since the limit must equal \( f(0) = 3 \), we have:
\( \frac{b}{2a} = 3 \implies \frac{b}{a} = 6. \)