Question

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

Answer 1

Answer: 3.00

We have, $y=\frac{ax}{b-x}$
$\Rightarrow \frac{dy}{dx}=\frac{(b-x)a-ax\cdot (-1)}{(b-x{)}^2}=\frac{ab}{(b-x{)}^2}$
$\therefore {\left[\frac{dy}{dx}\right]}_{(1,1)}=\frac{ab}{(b-1{)}^2}=2$ (given) .....(i)
Since the curve passes through the point $(1,1)$, therefore,
$1=\frac{a}{b-1}$
$\Rightarrow a=b-1$
On putting $a=b-1$ in equation (i), we get
$\frac{(b-1)b}{(b-1{)}^2}=2$
$\Rightarrow b=2$
$\therefore a=2-1=1$
Hence, $a=1,b=2$