Relations & Functions
Invertibility Conditions
Grade 12
Question:
<p>If <span>\(f : \mathbb{R} \to \mathbb{R}\)</span>, <span>\(f ( x ) = ax + \cos x\)</span> is an invertible function, then complete set of values of <span>\(a\)</span> is:</p>
<p>(a) <span>\(( -2 , - 1] \cup [1, 2 )\)</span></p>
<p>(b) <span>\([ -1, 1]\)</span></p>
<p>(c) <span>\(( -\infty , - 1] \cup [1, \infty)\)</span></p>
<p>(d) <span>\(( -\infty , - 2 ] \cup [2 , \infty)\)</span></p>
Step-by-Step Solution
Key Concept: A function f: ℝ → ℝ is invertible if and only if it is both injective (one-to-one) and surjective (onto). For a continuous function from ℝ to ℝ, this means f must be strictly monotonic. We need f'(x) to have constant sign throughout ℝ.
<p><strong>Step 1: Condition for Invertibility</strong></p><p>For f(x) = ax + cos x to be invertible from ℝ → ℝ, it must be strictly monotonic (either strictly increasing or strictly decreasing) on its entire domain.</p><p><strong>Step 2: Find the Derivative</strong></p><p>f'(x) = a - sin x</p><p>For strict monotonicity, f'(x) must never change sign, meaning f'(x) > 0 for all x ∈ ℝ, or f'(x) < 0 for all x ∈ ℝ.</p><p><strong>Step 3: Analyze f'(x) > 0</strong></p><p>We need: a - sin x > 0 for all x ∈ ℝ</p><p>This means: a > sin x for all x ∈ ℝ</p><p>Since sin x ∈ [-1, 1], we need: a ≥ 1</p><p><strong>Step 4: Analyze f'(x) < 0</strong></p><p>We need: a - sin x < 0 for all x ∈ ℝ</p><p>This means: a < sin x for all x ∈ ℝ</p><p>Since sin x ∈ [-1, 1], we need: a ≤ -1</p><p><strong>Step 5: Combine Conditions</strong></p><p>From Step 3: a ≥ 1, or</p><p>From Step 4: a ≤ -1</p><p>Therefore: a ∈ (-∞, -1] ∪ [1, ∞)</p><p><strong>Step 6: Verify Surjectivity</strong></p><p>When |a| ≥ 1, f is strictly monotonic and continuous, so it maps ℝ onto ℝ (by Intermediate Value Theorem). Both conditions for invertibility are satisfied.</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c