Applications of Derivatives
Monotonic Functions and Derivatives
Grade 12
Question:
<p>Let <strong>f</strong>(<strong>x</strong>) = <strong>x</strong> cos<sup>−1</sup>(−sin|<strong>x</strong>|), <strong>x</strong> ∈ [−π/2, π/2]. Which of the following is true?</p>
<p>(a) <strong>f</strong>′ is decreasing in (−π/2, 0) and increasing in (0, π/2)</p>
<p>(b) <strong>f</strong>′ is increasing in (−π/2, 0) and decreasing in (0, π/2)</p>
<p>(c) <strong>f</strong> is not differentiable at <strong>x</strong> = 0</p>
<p>(d) <strong>f</strong>′(0) = −π/2</p>
Step-by-Step Solution
Key Concept: Use the inverse trigonometric identity cos⁻¹(−x) = π − cos⁻¹(x) to simplify the function, then analyze the monotonicity of f′(x) by studying the sign of the auxiliary function g(x).
<p><strong>Given function:</strong></p><p><strong>f</strong>(<strong>x</strong>) = <strong>x</strong> cos<sup>−1</sup>(−sin|<strong>x</strong>|), <strong>x</strong> ∈ [−π/2, π/2]</p><p>Using the property cos<sup>−1</sup>(−<strong>x</strong>) = π − cos<sup>−1</sup>(<strong>x</strong>):</p><p><strong>f</strong>(<strong>x</strong>) = <strong>x</strong>(π − cos<sup>−1</sup>(sin|<strong>x</strong>|)) = <strong>x</strong>[π − (π/2 − sin<sup>−1</sup>|<strong>x</strong>|)] = <strong>x</strong>[π/2 + sin<sup>−1</sup>|<strong>x</strong>|]</p><p><strong>Step 1:</strong> Define auxiliary function:</p><p><strong>g</strong>(<strong>x</strong>) = <strong>x</strong> − (1 + <strong>x</strong>)log<sub>e</sub>(1 + <strong>x</strong>)</p><p><strong>g</strong>′(<strong>x</strong>) = 1 − 1 − log<sub>e</sub>(1 + <strong>x</strong>) = −log<sub>e</sub>(1 + <strong>x</strong>)</p><p><strong>Step 2:</strong> For <strong>x</strong> ∈ (−1, 0): <strong>g</strong>′(<strong>x</strong>) > 0, so <strong>g</strong>(<strong>x</strong>) is increasing. Since <strong>g</strong>(0) = 0, we have <strong>g</strong>(<strong>x</strong>) < 0 for all <strong>x</strong> ∈ (−1, 0).</p><p>Therefore, <strong>f</strong>′(<strong>x</strong>) = <strong>g</strong>(<strong>x</strong>)/<strong>x</strong><sup>2</sup> < 0 for <strong>x</strong> ∈ (−1, 0).</p><p>This means <strong>f</strong>(<strong>x</strong>) is decreasing on (−π/2, 0).</p><p><strong>Step 3:</strong> For <strong>x</strong> ∈ (0, ∞): <strong>g</strong>′(<strong>x</strong>) < 0, so <strong>g</strong>(<strong>x</strong>) is decreasing. Since <strong>g</strong>(0) = 0, we have <strong>g</strong>(<strong>x</strong>) < 0 for all <strong>x</strong> ∈ (0, ∞).</p><p>Therefore, <strong>f</strong>′(<strong>x</strong>) = <strong>g</strong>(<strong>x</strong>)/<strong>x</strong><sup>2</sup> < 0 for <strong>x</strong> ∈ (0, ∞).</p><p>This means <strong>f</strong>(<strong>x</strong>) is decreasing on (0, π/2).</p><p>∴ The answer is <strong>(a)</strong>.</p>
Correct Answer: a