Question

The number of integral values of $m$ such that the roots of $x^2-(m-3)x+m=0$ lie in the interval $(1,2),$ is

Answer 1

Let $f\left(x\right)=x^2-(m-3)x+m$, whose roots are $\alpha ,\beta$
Given: $1<\alpha ,\beta <2$


Conditions:
$\left(i\right)\Delta \ge 0\Rightarrow (m-3{)}^2-4m\ge 0\Rightarrow m^2-10m+9\ge 0$
$\Rightarrow (m-1)(m-9)\ge 0\Rightarrow m\in (-\infty ,1]\cup [9,\infty )\dots \left(1\right)$

$\left(ii\right)f\left(1\right)>0\Rightarrow 4>0$
Always true.

$\left(ii\right)f\left(2\right)>0\Rightarrow 4-2(m-3)+m>0\Rightarrow 10-m>0\Rightarrow m<10\dots \left(2\right)$

$\left(iv\right)1<-\frac{b}{2a}<2\Rightarrow 1<\frac{m-3}{2}<2\Rightarrow 2<m-3<4\Rightarrow m\in (5,7)\dots \left(3\right)$

From $\left(1\right),\left(2\right)\text{and}\left(3\right)$ we get,
$m\in \varphi$
Hence, no integral value of $m$ is posssible.