Quadratic Equations
Rational Functions with Quadratics
Grade 11
Question:
<p>Let the expression <span class="math">\(\frac{ax^2 + bx + c}{dx^2 + ex + f}\)</span> take all real values when <span class="math">\(x\)</span> is real, <span class="math">a, b, c, d\)</span> are all distinct real parameters. Then which of the following is/are possible</p>
<p>(A) <span class="math">\(a^2 > b^2\)</span> and <span class="math">\(c^2 < d^2\)</span></p>
<p>(B) <span class="math">\(a^2 > b^2\)</span> and <span class="math">\(c^2 > d^2\)</span></p>
<p>(C) <span class="math">\(a^2 < b^2\)</span> and <span class="math">\(c^2 > d^2\)</span></p>
<p>(D) <span class="math">\(a^2 < b^2\)</span> and <span class="math">\(c^2 < d^2\)</span></p>
Step-by-Step Solution
Key Concept: If a rational function takes all real values, then for any real value y, the equation (ax² + bx + c)/(dx² + ex + f) = y must have real solutions. This translates to requiring the discriminant of the resulting quadratic to be non-negative for all y.
<p><strong>Step 1: Set up the condition for taking all real values</strong></p><p>Let y = (ax² + bx + c)/(dx² + ex + f). Rearranging: ax² + bx + c = y(dx² + ex + f)</p><p>(a - yd)x² + (b - ye)x + (c - yf) = 0</p><p><strong>Step 2: Apply the discriminant condition</strong></p><p>For this equation to have real solutions for ALL real values of y, the discriminant with respect to x must be non-negative for all y:</p><p>Δ = (b - ye)² - 4(a - yd)(c - yf) ≥ 0 for all real y</p><p><strong>Step 3: Expand and rearrange as a quadratic in y</strong></p><p>Expanding: b² - 2bye + y²e² - 4ac + 4ayf + 4cyd - 4y²df ≥ 0</p><p>Collecting terms: (e² - 4df)y² + (-2be + 4af + 4cd)y + (b² - 4ac) ≥ 0</p><p>For this to hold for all y, we need the coefficient of y² to be zero (otherwise the quadratic would eventually become negative):</p><p><strong>e² - 4df = 0, or e² = 4df</strong></p><p><strong>Step 4: Apply remaining conditions</strong></p><p>With e² = 4df, the linear term must satisfy: -2be + 4af + 4cd = 0</p><p>And: b² - 4ac ≥ 0</p><p><strong>Step 5: Interpret the constraint e² = 4df</strong></p><p>From e² = 4df, we have: e² > 0, so 4df > 0, meaning d and f have the same sign.</p><p>This gives us: |e| = 2√(df), which implies |e|² = 4df</p><p>Squaring: e² = 4df means |e|² = 4|d||f| when d, f > 0 or when d, f < 0</p><p><strong>Step 6: Check the options against e² = 4df</strong></p><p>Option A: a² > b² and c² < d² - No necessary connection to e² = 4df</p><p>Option B: a² > b² and c² > d² - Possible; these don't contradict e² = 4df</p><p>Option C: a² < b² and c² > d² - Possible; these don't contradict e² = 4df</p><p>Option D: a² < b² and c² < d² - This constrains the coefficients of the numerator and denominator</p><p><strong>Step 7: Verify Option B is consistent</strong></p><p>The key constraint is e² = 4df. The conditions on a, b, c, d in the options are independent of this constraint. Option B (a² > b² and c² > d²) is entirely possible while still satisfying e² = 4df with appropriate choices of e and f.</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B