This is a question which I want to solve, taken from this sample question paper for an exam I'm appearing for tommorow:
If a line, parallel to, but not identical with, x- axis cuts the graph of the curve $$y={(x-1)}/({(x-2)(x-3)})$$ at $x=a$ and $x=b$ then evaluate $$(a-b)(b-1)$$
Here is what I did:
Let the line be $y=c$ (since it is parallel to x-axis) .
Then $c=(a-1)/((a-2)(a-3))$ and similarly for b .
But then what ? Please help.
Assuming $(x-2)(x-3)\ne0$
So, $$c=\frac{x-1}{(x-2)(x-3)}\implies cx^2-(5c+1)x+6c+1=0$$ which is a Quadratic Equation in $x$ whose roots are $a,b$
If $c=0,x=1$ for finite $x\implies a=b$
Else $$ a+b=\frac{5c+1}c=5+\frac1c\text{ and }ab=\frac{6c+1}c=6+\frac1c$$
I think proposition in the question should be the one available by eliminating the foreign(which we have introduced in the question) element $c$