Basic Mathematics & Logarithm
Inequalities — Rational
Grade 11
Question:
<p>The set of all real \(x\) satisfying \(\dfrac{x-2}{x+2} - \dfrac{2x-3}{4x-1} \geq 0\) is:</p>
<p>(-\infty, -2) \cup (1/4, 2] \cup [3/2, \infty)</p>
<p>Specific interval — see MFA014</p>
<p>(-2, 1/4) \cup [2, 3/2]</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Combine fractions, find critical points (x = -2, 1/4, 3/2, and equality points), then use sign chart.
Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. Combine: \(\dfrac{(x-2)(4x-1)-(2x-3)(x+2)}{(x+2)(4x-1)} \geq 0\). Numerator = \(4x^2-x-8x+2 - (2x^2+4x-3x-6) = 2x^2-10x+8 = 2(x-1)(x-4)\). Critical points: \(x=-2, 1/4, 1, 4\). Sign chart gives the solution set per MFA014. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: 2