Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions and Their Properties
Grade 12
Question:
<p>Let <span>\(f(x) = \sin x + \cos x + \tan x + \arcsin x + \arccos x + \arctan x\)</span>. If <span>\(M\)</span> and <span>\(m\)</span> are maximum and minimum values of <span>\(f(x)\)</span>, then their arithmetic mean is equal to</p>
<p>(a) <span>\(\frac{\pi}{2} + \cos 1\)</span></p>
<p>(b) <span>\(\frac{\pi}{2} + \sin 1\)</span></p>
<p>(c) <span>\(\frac{\pi}{4} + \tan 1 + \cos 1\)</span></p>
<p>(d) <span>\(\frac{\pi}{4} + \tan 1 + \sin 1\)</span></p>
Step-by-Step Solution
Key Concept: Domain of inverse trigonometric functions combined with monotonicity analysis determines the arithmetic mean of extrema.
<p><strong>Step 1:</strong> Domain of <span>$f$</span> is <span>$[-1,1]$</span>.</p><p><strong>Step 2:</strong> <span>$f'(x) = \cos x - \sin x + \sec^2 x + \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} + \frac{1}{1+x^2}$</span></p><p><strong>Step 3:</strong> Hence, <span>$f'(x) > 0 \Rightarrow f$</span> is increasing.</p><p><strong>Step 4:</strong> <span>$\text{Range is } [f(-1), f(1)]$</span></p><p><strong>Step 5:</strong> <span>$f(x)|_{\min} = f(-1) = -\sin 1 + \cos 1 - \tan 1 + \left(-\frac{\pi}{2}\right) + \frac{\pi}{2} - \arctan(-1) = -\sin 1 + \cos 1 - \tan 1 + \frac{\pi}{4}$</span></p><p><strong>Step 6:</strong> <span>$f(x)|_{\max} = f(1) = \sin 1 + \cos 1 + \tan 1 + \frac{\pi}{2} + 0 + \frac{\pi}{4} = \sin 1 + \cos 1 + \tan 1 + \frac{3\pi}{4}$</span></p><p><strong>Step 7:</strong> <span>$\frac{M + m}{2} = \frac{\pi}{2} + \cos 1$</span></p><p>∴ Answer is (A).</p>
Correct Answer: A