Question

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.

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

Answer 1

Answer: (A)

-1 < x < 1

Answer 2

${\frac{x^2 + k^2}{k(6 + x)} \geq 1}$ $\Rightarrow \:\: {\frac{x^2 - kx + k^2 - 6k}{k(6 + x)} \geq 0}$ ......(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 ? (-1, 1) and hence the inequality is valid for all k > 8. For k = 0, the inequality is indeterminate.