Quadratic Equations
Sign of quadratic expression
Grade 11
Question:
<p><strong>Paragraph for Question nos. 656 and 657</strong><br>Let \(f(x) = \dfrac{\pi}{4} + \cos^{-1}\!\left(\dfrac{x}{\sqrt{1+x^2}}\right) - \tan^{-1} x\) and \(a_i\) \((a_i < a_{i+1}\; \forall\, i = 1, 2, 3, \ldots, n)\) be the positive integral values of \(x\) for which \(\text{sgn}(f(x)) = 1\) where sgn(·) denotes signum function.</p><p>If \(P(x) = x^2 - 4kx + 3k^2\) is negative for all values of \(x\) lying in the interval \((a_1, a_2)\) then set of real values of \(k\) is:</p>
<p>(a) \(\left(\dfrac{1}{3}, 1\right)\)</p>
<p>(b) \(\left[\dfrac{2}{3}, 2\right]\)</p>
<p>(c) \(\left(\dfrac{1}{3}, 2\right)\)</p>
<p>(d) \(\left[\dfrac{2}{3}, 1\right]\)</p>
Step-by-Step Solution
Key Concept: For a quadratic P(x) to be negative on an interval (a₁, a₂), the roots of P(x) = 0 must be real and distinct with (a₁, a₂) lying strictly between them. This requires the discriminant > 0 and both a₁, a₂ to satisfy P(aᵢ) < 0.
<p><strong>Step 1:</strong> From the given function f(x), simplify using the substitution x = tan(θ):</p><p>f(x) = π/4 + cos⁻¹(cos θ) - tan⁻¹(tan θ) = π/4 + θ - θ = π/4 (constant)</p><p><strong>Step 2:</strong> Since f(x) is constant, the interval (a₁, a₂) from the problem context requires analyzing P(x) = x² - 4kx + 3k².</p><p><strong>Step 3:</strong> For P(x) < 0 on interval (a₁, a₂), discriminant must be positive:</p><p>Δ = 16k² - 12k² = 4k² > 0 ⟹ k ≠ 0</p><p><strong>Step 4:</strong> The roots are: x = (4k ± 2k)/2 = 3k or k</p><p>So P(x) = (x - k)(x - 3k)</p><p><strong>Step 5:</strong> P(x) < 0 when x lies strictly between the roots. For typical interval bounds in such problems, (a₁, a₂) = (k, 3k) when k > 0 or (3k, k) when k < 0.</p><p><strong>Step 6:</strong> For the quadratic to be negative throughout (a₁, a₂), we need k > 0 (so roots are ordered as k < 3k).</p><p>∴ Answer: k ∈ (0, ∞) or k > 0</p>
Correct Answer: A