Applications of Derivatives
Maxima and minima
Grade 12
Question:
<p>Find the range of the function <span>\(f(x) = \frac{1}{|\sin x|} + \frac{1}{|\cos x|}\)</span></p>
<p>(a) <span>\([2\sqrt{2}, \infty)\)</span></p>
<p>(b) <span>\((\sqrt{2}, 2\sqrt{2})\)</span></p>
<p>(c) <span>\((0, 2\sqrt{2})\)</span></p>
<p>(d) <span>\((2\sqrt{2}, 4)\)</span></p>
Step-by-Step Solution
Key Concept: Use Cauchy-Schwarz inequality or Lagrange multipliers to find the minimum value of the sum of reciprocals subject to a constraint.
<p><strong>Step 1:</strong> Let <span>$|\sin x| = a$</span> and <span>$|\cos x| = b$</span> where <span>$a^2 + b^2 = 1$</span>, <span>$0 < a, b < 1$</span></p><p><strong>Step 2:</strong> We need to minimize <span>$f = \frac{1}{a} + \frac{1}{b}$</span> subject to <span>$a^2 + b^2 = 1$</span></p><p><strong>Step 3:</strong> By Cauchy-Schwarz inequality: <span>$\left(\frac{1}{a} + \frac{1}{b}\right)(a + b) \geq 4$</span></p><p><strong>Step 4:</strong> The minimum value is <span>$2\sqrt{2}$</span>, achieved when <span>$a = b = \frac{1}{\sqrt{2}}$</span></p><p><strong>Step 5:</strong> As either <span>$a$</span> or <span>$b$</span> approaches 0, the function approaches infinity.</p><p>∴ Answer is (a).</p>
Correct Answer: a