$∣ ∣ ∣ \frac{x^{2} + m x + 1}{x^{2} + x + 1} ∣ ∣ ∣ < 3$ , Find $m$

- 1 < m < 5

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0 < m < 4

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1 < m < 4

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0 < m < 5

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Solution

The correct option is A $- 1 < m < 5$
Since $x^{2} + x + 1 > 0$ for all real $x$ , the given inequality can be written as
$- 3 (x^{2} + x + 1) < x^{2} + m x + 1 < 3 (x^{2} + x + 1)$
$∴ 4 x^{2} + (m + 3) x + 4 > 0$ ...(1)
and $2 x^{2} - (m - 3) x + 2 > 0$ ...(2)
The inequality (1) will hold for all x if its
$Δ = {(m + 3)}^{2} - 64 < 0$
and the sign will be same as the coefficient of Ist term i.e., +ive.
or $(m + 11) (m - 5) < 0$ which imples that
$- 11 < m < 5.$ ...(3)
Similarly, (2) will hold for all x if its $Δ < 0$
$- 1 < m < 7$ ...(4)
Finally from (3) and (4), $- 1 < m < 5.$

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∣∣∣x2+mx+1x2+x+1∣∣∣<3, Find m

$∣ ∣ ∣ \frac{x^{2} + m x + 1}{x^{2} + x + 1} ∣ ∣ ∣ < 3$ , Find $m$