Question

For a, b > 0, let

$$f(x) = \begin{cases} \frac{\tan ((a+1)x) + b \tan x}{x} &, x<0 \\ 3 &, x = 0\\ \frac{\sqrt{ax+b^2x^2} - \sqrt{ax}}{b\sqrt{a} x\sqrt{x}} &, x > 0 \end{cases}$$

be a continous 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 continuity at $x=0$, $LHL = RHL = f(0)$.

LHL:

$$\lim_{x \to 0^-} \frac{\tan((a+1)x)}{x} + \frac{b\tan x}{x} = (a+1) + b$$

RHL:

$$\lim_{x \to 0^+} \frac{\sqrt{ax+b^2x^2} - \sqrt{ax}}{b\sqrt{a} x\sqrt{x}}$$

Multiply by conjugate:

$$\lim_{x \to 0^+} \frac{b^2x^2}{b\sqrt{a} x\sqrt{x}(\sqrt{ax+b^2x^2} + \sqrt{ax})} = \lim_{x \to 0^+} \frac{b\sqrt{x}}{\sqrt{a}(\sqrt{a+b^2x} + \sqrt{a})} = \frac{b}{2a}$$

Equating: $a+1+b = 3$ and $\frac{b}{2a} = 3$.

From $\frac{b}{2a} = 3$, $b=6a$.

Substitute into first equation: $a+1+6a=3 \Rightarrow 7a=2 \Rightarrow a=\frac{2}{7}$.

Then $b=6(\frac{2}{7}) = \frac{12}{7}$.

Finally, $\frac{b}{a} = \frac{12/7}{2/7} = 6$.