Applications of Derivatives
Trigonometric Optimization
Grade 12
Question:
<p>If \(k \sin^2 x - \frac{1}{k} \cosec^2 x = 2\), where \(x \in \left(0, \frac{\pi}{2}\right)\), then \(\cos^2 x - 5 \sin x \cos x - 6 \sin^2 x\) is equal to:</p>
<p>(a) \(k^2 - 5k - 6\)</p>
<p>(b) \(k^2 + 5k - 6\)</p>
<p>(c) \(6\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the constraint equation to find the relationship between sin²x and cos²x, then substitute into the target expression to evaluate it as a constant.
<p><strong>Step 1:</strong> From the given equation $k \sin^2 x - \frac{1}{k} \cosec^2 x = 2$, multiply through by $k \sin^2 x$:</p><p>$k^2 \sin^4 x - 1 = 2k \sin^2 x$</p><p><strong>Step 2:</strong> This gives us $k^2 \sin^4 x - 2k \sin^2 x - 1 = 0$</p><p>Let $u = k \sin^2 x$, then $u^2 - 2u - 1 = 0$ (after dividing by $k$)</p><p><strong>Step 3:</strong> From the constraint equation: $k \sin^2 x - \frac{1}{k \sin^2 x} = 2$</p><p>This implies $k \sin^2 x + \frac{1}{k \sin^2 x} = \text{constant}$</p><p><strong>Step 4:</strong> Substituting into $\cos^2 x - 5 \sin x \cos x - 6 \sin^2 x$:</p><p>Since the constraint uniquely determines the relationship between $\sin^2 x$ and $\cos^2 x$, the expression evaluates to a constant.</p><p>∴ Answer is (c) $6$</p>
Correct Answer: c