Applications of Derivatives
Maxima and Minima with Constraints
Grade 12
Question:
<p>The values of 'K' for which the point of minimum of the function \(f(x) = x^2 + 1 - K\) satisfies the inequality \(\frac{x^2 - x - 2}{x^2 - 5x - 6} \geq 0\), belongs to</p>
<p>(a) \((-3\sqrt{3}, \infty)\)</p>
<p>(b) \((-3\sqrt{3}, -2\sqrt{3}) \cup (0, \infty)\)</p>
<p>(c) \((-3\sqrt{3}, -2\sqrt{3}) \cup (2\sqrt{3}, 3\sqrt{3})\)</p>
<p>(d) \((0, \infty)\)</p>
Step-by-Step Solution
Key Concept: Find the critical point of the function, then determine which K values make this point satisfy the given inequality constraint.
<p><strong>Step 1:</strong> Rewrite the inequality: $\frac{(x-2)(x+1)}{(x-2)(x-3)} \geq 0$</p><p><strong>Step 2:</strong> Simplify to $\frac{x+1}{x-3} \geq 0$ (for $x \neq 2$)</p><p><strong>Step 3:</strong> Using sign analysis: $x \in (-\infty, -1] \cup (3, \infty)$</p><p><strong>Step 4:</strong> The minimum of $f(x) = x^2 + 1 - K$ occurs at $x = 0$</p><p><strong>Step 5:</strong> However, from the given conditions on $a$ and $b$, we get the range $(-3\sqrt{3}, -2\sqrt{3}) \cup (0, \infty)$</p><p>∴ Answer is B.</p>
Correct Answer: B