Question

If the values of $a$ for which one root of equation $(a-5)x^2-2ax+a-4=0$ is smaller than $1$ and the other greater than $2$ belong to set $(c,b)$, then $\frac{b}{c+1}$ is

Answer 1

Let $f\left(x\right)=(a-5)x^2-2ax+a-4(a\ne 5)$
As 1 and 2 lie between the roots of $f\left(x\right)=0$, we have, $D>0$, and $(a-5)f\left(1\right)<0$ and $(a-5)f\left(2\right)<0.$
I. Consider $D>0$ ${(-2a)}^2-4.(a-5).(a-4)>0$.
$\Rightarrow 9a-20>0\Rightarrow a\in \left(20/9,\infty \right)$
II. Consider $(a-5)f\left(1\right)<0$ $(a-5)(a-5-2a+a-4)<0$ $\Rightarrow$ $(a-5)(-9)<0$.
$\Rightarrow$ $a-5>0$
$\Rightarrow a\in (5,\infty )$.............(2)
III. Consider $(a-5)f\left(2\right)<0$ $(a-5)\left(4\right(a-5)-4a+a-4)<0$.
$\Rightarrow$ $(a-5)(a-24)<0$
$\Rightarrow a\in (5,24)$ ..................(3)
Hence, the values of a satisfying (1),(2) and (3) at the same time are $a\in (5,24)$.
$\Rightarrow \frac{b}{c+1}=\frac{24}{5+1}=4$.