Relations & Functions
Range of a function
Grade 12
Question:
<p>If \(f(x) = \dfrac{3}{1+\tan^2 x} + \dfrac{9}{1+\cot^2 x}\), then:</p>
<p>(a) number of integers in the range of \(f(x)\) is 7.</p>
<p>(b) number of integers in the range of \(f(x)\) is 5.</p>
<p>(c) sum of the integers in the range of \(f(x)\) is 30.</p>
<p>(d) sum of the integers in the range of \(f(x)\) is 42.</p>
Step-by-Step Solution
Key Concept: Use the Pythagorean identities 1 + tan²x = sec²x and 1 + cot²x = csc²x to simplify, then express the result in terms of sin²x and cos²x to find the range and properties of f(x).
<p><strong>Step 1:</strong> Apply Pythagorean identities to simplify denominators.</p><p>Since 1 + tan²x = sec²x and 1 + cot²x = csc²x:</p><p>f(x) = 3/sec²x + 9/csc²x</p><p><strong>Step 2:</strong> Convert to sin and cos.</p><p>f(x) = 3cos²x + 9sin²x</p><p><strong>Step 3:</strong> Rewrite using sin²x + cos²x = 1.</p><p>f(x) = 3cos²x + 9sin²x = 3(cos²x + sin²x) + 6sin²x = 3 + 6sin²x</p><p><strong>Step 4:</strong> Find the range.</p><p>Since 0 ≤ sin²x ≤ 1, we have 3 ≤ f(x) ≤ 9.</p><p>f(x) = 3 when sin²x = 0 (x = nπ)</p><p>f(x) = 9 when sin²x = 1 (x = π/2 + nπ)</p><p><strong>Step 5:</strong> Determine properties.</p><p>• Range is [3, 9] ✓ (Option A/D likely)</p><p>• Function is periodic with period π ✓</p><p>• Neither purely even nor odd ✓</p><p>∴ Answer: A, D</p>
Correct Answer: A,D