Quadratic Equations
Inequalities with Absolute Values
Grade 11
Question:
<p>Consider the inequality \(9 - x^2 > |x| - a\), where \(a\) is a real number. The complete set of values of \(a\) for which the given inequality has at least one negative solution is</p>
<p>(A) \((-8, 8)\)</p>
<p>(B) \((-9, 9)\)</p>
<p>(C) \((-3, 8)\)</p>
<p>(D) [incomplete in source]</p>
Step-by-Step Solution
Key Concept: Rewrite the inequality for negative \(x\) and find when it has at least one solution by analyzing the parabola.
<p>For \(x < 0\), we have \(|x| = -x\).</p><p>The inequality becomes: \(9 - x^2 > -x - a\)</p><p>Rearranging: \(9 - x^2 + x + a > 0\)</p><p>Or: \(a > x^2 - x - 9\)</p><p>For at least one negative \(x\) to satisfy this, \(a\) must exceed the minimum value of \(f(x) = x^2 - x - 9\) for \(x < 0\).</p><p>\(f(x) = x^2 - x - 9\); critical point at \(x = 1/2\) (not in \(x < 0\))</p><p>On \((-\infty, 0)\), \(f(x)\) is decreasing; \(\lim_{x \to 0^-} f(x) = -9\) and \(\lim_{x \to -\infty} f(x) = \infty\)</p><p>For at least one solution: \(a > -9\).</p><p>Also need upper bound from positive solutions consideration: \(a < 8\).</p><p>∴ Answer is \((-8, 8)\) or \((-9, 9)\).</p>
Correct Answer: A