Quadratic Equations
Roots and Expressions
Grade 11
Question:
<p>Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2 - 5x + 5 = 0\).<br>If \(b = \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}\) and \(t = x^2 - 4x + 3b - \dfrac{1}{5} + \dfrac{1}{x^2 - 4x + 9},\ x \in R\) then:</p>
<p>(a) minimum value of \((b+t)\) is 8</p>
<p>(b) maximum value of \(\log_{1/5}(t)\) is \(-1\)</p>
<p>(c) range of \(y = \cot^{-1}(\log_5 t)\) is \(\left(0, \dfrac{\pi}{4}\right]\)</p>
<p>(d) range of \(y = \cot^{-1}(\log_{1/5}(t))\) is \(\left[\dfrac{\pi}{4}, \pi\right)\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find α + β and αβ, then compute b = α/β + β/α = (α² + β²)/(αβ). Subsequently, recognize that the expression for t involves a substitution u = x² - 4x + 9, transforming it into a rational function whose range can be determined by analyzing critical points.
<p><strong>Step 1: Find α + β and αβ using Vieta's formulas</strong></p><p>For x² - 5x + 5 = 0:</p><p>α + β = 5 and αβ = 5</p><p><strong>Step 2: Calculate b</strong></p><p>b = α/β + β/α = (α² + β²)/(αβ)</p><p>α² + β² = (α + β)² - 2αβ = 25 - 10 = 15</p><p>∴ b = 15/5 = 3</p><p><strong>Step 3: Simplify the expression for t</strong></p><p>t = x² - 4x + 3(3) - 1/5 + 1/(x² - 4x + 9)</p><p>t = x² - 4x + 9 - 1/5 + 1/(x² - 4x + 9)</p><p>Let u = x² - 4x + 9. Since x² - 4x + 9 = (x - 2)² + 5 ≥ 5, we have u ≥ 5</p><p><strong>Step 4: Find range of t(u)</strong></p><p>t(u) = u - 1/5 + 1/u where u ≥ 5</p><p>dt/du = 1 - 1/u² = (u² - 1)/u²</p><p>For u ≥ 5: dt/du > 0, so t is strictly increasing</p><p>At u = 5: t(5) = 5 - 0.2 + 0.2 = 5</p><p>As u → ∞: t(u) → ∞</p><p>∴ t ∈ [5, ∞)</p><p><strong>Answer: A, B, D correspond to the correct range characterization statements</strong></p>
Correct Answer: ABD