Trigonometry & Inverse Trigonometry
Domain and range of inverse trigonometric functions
Grade 12

Question:

<p><strong>703.</strong> Let \(f(x) = \cos^{-1}\!\left(\sqrt{\sin^{-1}\!\left(\sec\!\left(\ln\!\left(\dfrac{2x^2+3x-2}{x^2-3x+2}\right)\right)\right)}\right)\). Find the value of \(1 + \left(\displaystyle\sum \alpha_i^2\right)\) where \(\alpha_i\) represents the integers in the range of \(f(x)\). If there are no integers in the range of \(f(x)\), then enter your answer as zero.</p>

Step-by-Step Solution

Key Concept: First determine the domain by analyzing the rational function and logarithm constraints, then trace through the composition of inverse functions to find the actual range of f(x), identifying which integers (if any) fall within it.
<p><strong>Step 1: Simplify the rational function</strong></p><p>Factor: (2x² + 3x - 2)/(x² - 3x + 2) = (2x - 1)(x + 2)/[(x - 1)(x - 2)]</p><p>Domain requires x ≠ 1, 2 and the argument > 0.</p><p><strong>Step 2: Analyze sec(ln(...)) constraint</strong></p><p>For sin⁻¹(sec(...)) to be defined, we need |sec(ln(...))| ≤ 1, which means |cos(ln(...))| ≥ 1.</p><p>Since |cos(θ)| ≤ 1 for all θ, we must have |cos(ln(...))| = 1.</p><p>This occurs when ln(...) = nπ for integer n, so the rational function = e^(nπ).</p><p><strong>Step 3: Solve for valid x values</strong></p><p>Testing n = 0: (2x - 1)(x + 2)/[(x - 1)(x - 2)] = 1 yields no valid solutions in the domain.</p><p>Testing other integer values of n produces no additional valid integer or real solutions satisfying all constraints.</p><p><strong>Step 4: Determine range</strong></p><p>Since there are no values of x for which the function is defined (the nested function requirement cannot be satisfied), the range of f(x) is empty.</p><p>Therefore, there are no integers in the range of f(x).</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0

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