Question

If $(m-1)x^2+mx+1=0$ has roots less than $2$ then

Answer 1

Let $f\left(x\right)=(m-1)x^2+mx+1$.

$f\left(2\right)>0$ and $\frac{-m}{2(m-1)}<2$

$f\left(2\right)=4(m-1)+2m+1>0$

$6m-4+1>0$

$6m-4+1>0$

$6m-3>0$

$6m>3$

$m>1/2$.

$\frac{-m}{2(m-1)}<2\Rightarrow \frac{m}{m-1}<4$

$\frac{m}{m-1}-4<0\frac{m-4(m-1)}{m-1}<0$

$\frac{-3m+4}{m-1}<0$

( or )

$\frac{3m-4}{m-1}>0$

$\Rightarrow m<1$ ( or ) $m>\frac{4}{3}$

Also $\Delta \ge 0$.

$m^2-4(m-1)\ge 0$

$m^2-4m+4\ge 0\forall m$.

Hence $m\in (\frac{1}{2},1)\cup (\frac{4}{3},\infty )$.