Question

If k$\notin [0,8]$, find the value of x for which the inequality $\frac{x^2+k^2}{k(6+x)}\ge 1$ is satisfied.

• A. -1 < x < 1
• B. -1 < x < 2
• C. -2 < x < 1
• D. -3 < x < 1

Answer 1

Answer: (A)

$\frac{x^2+k^2}{k(6+x)}\ge 1$
$\Rightarrow \frac{x^2-kx+k^2-6k}{k(6+x)}\ge 0$ ......(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.