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Q.
If both the roots of the equation $x^{2}-mx+1=0$ are less than unity, then
NTA AbhyasNTA Abhyas 2022
Solution:
$f\left(x\right)=x^{2}-mx+1$ and $\alpha \leq \beta $ are the roots of $f\left(x\right)=0.$
Now, $\alpha < \beta < 1$ implies that $f\left(1\right)$ and the coefficients of $x^{2}$ have the same sign.
This gives $f\left(1\right)>0$
$\Rightarrow 1-m+1>0\Rightarrow m < 2$
Also, the discriminant is $m^{2}-4\geq 0.$
So $m\leq -2$ or $m\geq +2.$
Taking intersection, we get,
$m\leq -2.$
Also, note that if $m=-2,$ the roots are $-1,-1.$