The value of m, for which both roots of equation $x^2-mx+1=0$ are less then unity is 

Let $f\left(x\right)=x^2-mx+1$ as both roots of $f\left(x\right)=0$ less than 1

  $\Rightarrow D\ge 0,af\left(1\right)>0and\frac{-b}{2a}<1$

  i) $D\ge 0\Rightarrow {(-m)}^2-4\left(1\right)\left(1\right)\ge 0$

  $\Rightarrow (m+2)(m-2)\ge 0$

 $\Rightarrow m\in (-\infty ,-2]\cup [2,\infty )------\left(1\right)$

  ii) $af\left(1\right)>0\Rightarrow 1.(1-m+1)>0$

  $\Rightarrow m-2<0\Rightarrow m<2$

   $\Rightarrow m\in (-\infty ,2)--------\left(2\right)$

  iii) $\frac{-b}{2a}<1\Rightarrow \frac{m}{2}<1=m<2$

 $\Rightarrow m\in (-\infty ,2)--------\left(3\right)$

From (1) , (2) and (3) $\Rightarrow m\in (-\infty ,2)$