For [a, b0], let [f(x)= cases ((a+1) x)+b xx, & x 0cases] be a continuous function at x=0…

Question

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

Accepted Answer

$$f(x)=\left\{\[\begin{array}{cc} \frac{\tan ((a+1) x)+b \tan x}{x}, & x0 \end{array}$$\right.\]

$f(x)$ is continuous at $x=0$

$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \
& \lim _{x \rightarrow 0^{-}} f(x)=3 \
& \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \
& \Rightarrow a+1+b=3 \
& \Rightarrow a+b=2 \quad \text{..... (1)}
\end{aligned}$$

also, $$\lim _\limits{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3$$

$$\begin{aligned}
& =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \
& =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \
& =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \
& \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \
& \Rightarrow \frac{b}{2 a}=3 \
& \Rightarrow \frac{b}{a}=6 \quad \text{..... (2)}
\end{aligned}$$

Solutions · 1