<p>For a real number <i>x</i>, let [<i>x</i>] denote the greatest integer less than or equal to <i>x</i>. Let <i>f</i>: ℝ → ℝ be defined by <i>f</i>(<i>x</i>) = 2<i>x</i> + [<i>x</i>] + sin <i>x</i> cos <i>x</i>. Then <i>f</i> is:</p>
<p>(a) One-one but not onto</p>
<p>(b) Onto but not one-one</p>
<p>(c) Both one-one and onto</p>
<p>(d) Neither one-one nor onto</p>
Step-by-Step Solution
Key Concept: To determine if f is one-one and onto, analyze the monotonicity of f and its range. The key is recognizing that f(x) = 2x + [x] + sin x cos x is strictly increasing (despite the oscillating trigonometric term being bounded), making it injective, and its range is all of ℝ, making it surjective.
<p><strong>Step 1: Analyze the function components</strong></p><p>f(x) = 2x + [x] + sin x cos x</p><p>Note that sin x cos x = ½ sin 2x, which is bounded: -½ ≤ sin x cos x ≤ ½</p><p><strong>Step 2: Examine monotonicity</strong></p><p>Consider the derivative where it exists (i.e., for non-integer x):</p><p>f'(x) = 2 + cos 2x ≥ 2 - 1 = 1 > 0</p><p>Since f'(x) > 0 wherever f is differentiable, f is strictly increasing on each interval [n, n+1) for integer n.</p><p><strong>Step 3: Verify strict monotonicity across all ℝ</strong></p><p>At integer points, [x] has a jump discontinuity (increases by 1), but the net change in f is continuous because:</p><p>- As x → n⁻ (from left): f(x) → 2n + (n-1) + sin n cos n = 3n - 1 + sin n cos n</p><p>- As x → n⁺ (from right): f(x) = 2n + n + sin n cos n = 3n + sin n cos n</p><p>The function remains strictly increasing overall since 2x + [x] is strictly increasing.</p><p><strong>Step 4: Check if f is one-one (injective)</strong></p><p>Since f is strictly increasing on ℝ, if x₁ < x₂, then f(x₁) < f(x₂).</p><p>Therefore, f(x₁) = f(x₂) ⟹ x₁ = x₂, so f is one-one.</p><p><strong>Step 5: Check if f is onto (surjective)</strong></p><p>Consider limits:</p><p>- As x → +∞: f(x) = 2x + [x] + sin x cos x → +∞ (since 2x + [x] → +∞ and sin x cos x is bounded)</p><p>- As x → -∞: f(x) = 2x + [x] + sin x cos x → -∞ (since 2x + [x] → -∞ and sin x cos x is bounded)</p><p>By the Intermediate Value Theorem and continuity of f on ℝ, f takes all real values.</p><p>Therefore, f is onto.</p><p><strong>Step 6: Conclusion</strong></p><p>Since f is both one-one and onto, f is bijective.</p><p><strong>∴ Answer: c</strong></p>
Correct Answer: c