Applications of Derivatives
Monotonic Functions
Grade 12
Question:
<p>\(f(x) = \tan^{-1}(\sin x + \cos x)\) is an increasing function in</p>
<p>(a) \((0, \pi/4)\)</p>
<p>(b) \((0, \pi/2)\)</p>
<p>(c) \((-\pi/4, \pi/4)\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Find where f'(x) > 0 by differentiating the composite function and analyzing the sign of the derivative. The function increases when the derivative of the inner function (sin x + cos x) remains positive throughout the domain where the outer function is defined.
<p><strong>Step 1:</strong> Find f'(x) using chain rule.</p><p>f(x) = tan⁻¹(sin x + cos x)</p><p>f'(x) = 1/(1 + (sin x + cos x)²) · (cos x - sin x)</p><p><strong>Step 2:</strong> Analyze the sign of f'(x).</p><p>Since 1 + (sin x + cos x)² > 0 always, the sign of f'(x) depends on (cos x - sin x).</p><p>f'(x) > 0 when cos x - sin x > 0, i.e., cos x > sin x</p><p><strong>Step 3:</strong> Solve cos x > sin x.</p><p>This occurs when tan x < 1, giving x ∈ (-π/2 + 2nπ, π/4 + 2nπ) for integer n.</p><p>Or more commonly written: x ∈ (-π/2, π/4) ∪ (3π/2, 9π/4) etc.</p><p>∴ Answer: A</p>
Correct Answer: A