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Q.
Find the values of k for which $\left|\frac{x^{2} + kx + 1}{x^{2} + x + 1}\right|<2\, \forall\, x \, \in \, R$
Linear Inequalities
Solution:
$| x | < a \Rightarrow - a < x < + a $
$\therefore $ the given inequality implies
$- 2 < \frac{x^{2}+kx + 1}{x^{2} + x + 1}< 2$ ...(i)
Now $x^{2} + x + 1 = \left( x + \frac{1}{2} \right)^2 + \frac{3}{4}$ is positive for all values of x.
Multiplying (i) $by \, x^2 + x + 1 - 2(x^2 + x + 1) < x^2 + kx + 1 < 2 (x^2 + x + 1) $
This yields two inequations $3x^2 + (2 + k)x + 3 > 0$ and $x^2 + (2 - k) x + 1 > 0$
For these quadratic expressions to be positive for all values of x, their discriminants must be negative
i.e., $(2 + k)^2 - 36 < 0$ and $(2 - k)^2 - 4 < 0 $ ... (ii)
$(k + 8) (k - 4) < 0$ and $k (k - 4) < 0$ ... (iii)
$\therefore $ - 8 < k < 4 and 0 < k < 4
For both these conditions to be satisfied, 0 < k < 4.