If the slope of the curve $y=\frac{ax}{b-x}$ at the point $(1,1)$ is 2 then

Solution:

We have,

$\begin{array}{l}y=\frac{ax}{b-x}\Rightarrow \frac{dy}{dx}=\frac{(b-x)a+ax}{(b-x{)}^2}\Rightarrow \frac{dy}{dx}=\frac{ab}{(b-x{)}^2}\\ \therefore {\left(\frac{dy}{dx}\right)}_{(1,1)}=\frac{ab}{(b-1{)}^2}\Rightarrow \frac{ab}{(b-1{)}^2}=2\end{array}$

Since, $(1, 1)$ lies on (i). Therefore,

$1=\frac{a}{b-1}$

From (ii) and (iii), we have

$\frac{b}{b-1}=2\Rightarrow b=2$

Putting $b = 2$ in (iii), we get $a = 1.$

Hence, $a = 1.$ and $b = 2$