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Q.
If k$\notin [0, 8]$, find the value of x for which the inequality $\frac{x^2 + k^2}{k(6 + x)} \geq 1$ is satisfied.
Linear Inequalities
Solution:
$\frac{x^2 + k^2}{k(6 + x)} \geq 1$
$\Rightarrow \:\: \frac{x^2 - kx + k^2 - 6k}{k(6 + x)} \geq 0$ ......(1)
Now the discriminant of the numerator is $24k - 3k^2 = 3k (8 - k)$ is negative for all k < 0 and for all k > 8. For these values of k, the numerator is positive.
(i) For k < 0, inequality (1) is true only if x < -6.
But x $\in $ (-1, 1) .........(2)
Hence for k < 0, the inequality is not valid.
(ii) For k > 8, inequality (1) is true only if x > -6 ........(3)
and x Î (-1, 1) and hence the inequality is valid for all k > 8.
For k = 0, the inequality is indeterminate.