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

Answer: (D)

6

Answer 2

For $f(x)$ to be continuous at $x = 0$, we require:
$\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$.
Given: $f(0) = \frac{0}{3} = 0$.
Right-hand limit:

Consider:
$\lim_{x \to 0^+} \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{ax}}$.
To simplify, multiply the numerator and the denominator by the conjugate of the numerator:
$\lim_{x \to 0^+} \frac{(\sqrt{ax + b^2x^2} - \sqrt{ax}) \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}$.
This simplifies to:
$\lim_{x \to 0^+} \frac{ax + b^2x^2 - ax}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}$.
Simplifying further:
$\lim_{x \to 0^+} \frac{b^2x^2}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}$.
Canceling terms and simplifying:
$\lim_{x \to 0^+} \frac{bx}{\sqrt{a} \times 2 \sqrt{ax}} = \frac{b}{2a}$.
For continuity, we equate this limit to the value of $f(0)$:
$\frac{b}{2a} = 3 \implies \frac{b}{a} = 6$.
Therefore:
$\boxed{6}$.