Find the value of m for which both roots of equation $x^{2} - m x + 1 = 0$ are less than unity.

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Solution

Let $f (x) = x^{2} - m x + 1 = 0$ as both roots of $f (x) = 0$ are less than $1$ ,
we can take $D \geq 0$ , $a f (1) > 0$ and $- \frac{b}{2 a} < 1$ .
Consider $D \geq 0 : {(- m)}^{2} - 4.1.1 \geq 0 \Rightarrow (m + 2) (m - 2) \geq 0 \Rightarrow m ϵ (- \infty, - 2) \cup (2, \infty)$ ...(1)
Consider $a f (1) > 0 : 1. (1 - m + 1) > 0 \Rightarrow m - 2 < 0 \Rightarrow m < 2 \Rightarrow m ϵ (- \infty, 2)$ ...(2)
Consider $- \frac{b}{2 a} < 1 : \frac{m}{2} < 1 \Rightarrow m - 2 \Rightarrow m ϵ (- \infty, 2)$ ...(3)
Hence, the value of $m$ satisfying (1),(2)and (3) at the same time are $m \in (- \infty, 2)$

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