The question is as follows:
For what values of '$a$'($\in\mathbb R$) is the line $x+y=3$ a tangent to the curve $y=\dfrac{ax}{1+x}$?
The answer: No such '$a$' exists.
How do I solve this question?
What I've tried:
If the line $x+y=3$ is a tangent to the curve $y=\dfrac{ax}{1+x}$ at some point $(x_1,y_1)$ on the curve, then we know that the slope of the tangent $(-1)$ is equal to $\dfrac{dy}{dx}$ at the point $(x_1,y_1)$.
$$\implies\dfrac{d}{dx}\left(\dfrac{ax}{1+x}\right)\Bigg|_{x=x_1}=-1$$ $$\implies\dfrac{a}{(1+x_1)^2}=-1$$ $$\implies (1+x_1)^2=-a$$
Which is not possible for $a\in\mathbb R^+$.
My working is only halfway to the solution, what am I doing wrong?
Suggestions for other ways of solving this question will be appreciated as well.
Let us find the abscissa of intersection
$$\dfrac{ax}{1+x}=3-x$$
$$\implies x^2+x(a-2)-3=0$$
For tangency both roots must be same, $\implies$ the discriminant must be $0$
But the discriminant $=(a-2)^2+12\ge12>0$ for real $a$