Quadratic Equations
Nature of Roots
Grade 11
Question:
<p><strong>Ex. 80:</strong> Given \(h(x) = (1 - \sin\theta)x^2 + 2(1 - \sin\theta)x - 3\sin\theta\) where \(\theta \in \mathbb{R} - \left\{(4n+1)\frac{\pi}{2}, n \in \mathbb{Z}\right\}\). If the quadratic equation \(h(x) = 0\) has both roots complex, then \(\theta\) belongs to:</p>
<p>(a) \(\left(-\frac{\pi}{2}, 0\right)\)</p>
<p>(b) \((0, \pi)\)</p>
<p>(c) \(\left(0, \frac{\pi}{2}\right)\)</p>
<p>(d) \(\left(-\pi, -\frac{\pi}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Complex roots occur when discriminant is negative; analyze the sign of D for the given quadratic.
<p>For the quadratic \(h(x) = 0\) to have complex roots, the discriminant must be negative.</p><p>\(D = 4(1-\sin\theta)^2 + 12\sin\theta(1-\sin\theta) < 0\)</p><p>\(= (1-\sin\theta)[4(1-\sin\theta) + 12\sin\theta] < 0\)</p><p>\(= (1-\sin\theta)(4 - 4\sin\theta + 12\sin\theta) < 0\)</p><p>\(= (1-\sin\theta)(4 + 8\sin\theta) < 0\)</p><p>This gives: \(\sin\theta > 1\) or \(\sin\theta < -\frac{1}{2}\)</p><p>Since \(\sin\theta \leq 1\), we need \(\sin\theta < -\frac{1}{2}\), which gives \(\theta \in \left(-\frac{\pi}{2}, 0\right)\) in the principal domain.</p>
Correct Answer: a