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

<p>$$f(x)=\left\{\begin{array}{cc} \frac{\tan ((a+1) x)+b \tan x}{x}, & x<0 \\ 3 & x=0 \\ \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, & x>0 \end{array}\right.$$</p> <p>$f(x)$ is continuous at $x=0$</p> <p>$$\begin{aligned} & \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\ & \lim _{x \rightarrow 0^{-}} f(x)=3 \\ & \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \\ & \Rightarrow a+1+b=3 \\ & \Rightarrow a+b=2 \quad \text{..... (1)} \end{aligned}$$</p> <p>also, $$\lim _\limits{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3$$</p> <p>$$\begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \\ & =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \\ & \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \\ & \Rightarrow \frac{b}{2 a}=3 \\ & \Rightarrow \frac{b}{a}=6 \quad \text{..... (2)} \end{aligned}$$</p>