Applications of Derivatives
Monotonicity / Injective Functions
Grade 12
Question:
<p>If \(f(x) = x^3 + 3x^2 + (4-k)x + b\) is an injective function \(\forall x \in R\), then:</p>
<p>maximum positive integral value of \(k\) is 1.</p>
<p>minimum positive integral value of \(k\) is 1.</p>
<p>number of positive integral value of \(k\) is 1.</p>
<p>number of non-negative integral value of \(k\) is 1.</p>
Step-by-Step Solution
Key Concept: A cubic function is injective (one-to-one) on ℝ if and only if its derivative is always non-negative (or always non-positive), which means f'(x) ≥ 0 for all x ∈ ℝ. This requires the discriminant of f'(x) to be ≤ 0.
<p><strong>Step 1:</strong> For f(x) to be injective on ℝ, f must be strictly monotonic. Since f is a cubic with positive leading coefficient, f must be strictly increasing everywhere.</p><p><strong>Step 2:</strong> This requires f'(x) ≥ 0 for all x ∈ ℝ.</p><p><strong>Step 3:</strong> Calculate: f'(x) = 3x² + 6x + (4-k)</p><p><strong>Step 4:</strong> For f'(x) ≥ 0 ∀x ∈ ℝ, the discriminant of f'(x) must be ≤ 0:</p><p>Δ = 36 - 4(3)(4-k) ≤ 0</p><p>36 - 12(4-k) ≤ 0</p><p>36 - 48 + 12k ≤ 0</p><p>12k ≤ 12</p><p><strong>k ≤ 1</strong></p><p><strong>Step 5:</strong> There are no restrictions on b from the injectivity condition (b only affects vertical translation).</p><p>∴ The conditions are: k ≤ 1 and b ∈ ℝ (or b can be any real number)</p>
Correct Answer: A,B,C,D