Applications of Derivatives
Monotonicity and Extrema
Grade 12
Question:
<p>Let <i>f</i>(<i>x</i>) = <i>x</i><sup>4</sup> + <i>ax</i><sup>3</sup> + 3<i>x</i><sup>2</sup> + <i>bx</i> + 1, <i>a</i>, <i>b</i> ∈ ℝ. If <i>f</i>(<i>x</i>) ≥ 0 for all <i>x</i> ∈ ℝ, then the maximum value of <i>a</i><sup>2</sup> + <i>b</i><sup>2</sup> is equal to</p>
<p>(A) 10</p>
<p>(B) 12</p>
<p>(C) 16</p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: For a polynomial to be non-negative everywhere, its coefficients must satisfy certain inequalities derived from the requirement that discriminants of all relevant sub-expressions are non-positive. This constrains a² + b².
<p><strong>Step 1:</strong> Given <i>f</i>(<i>x</i>) = <i>x</i><sup>4</sup> + <i>ax</i><sup>3</sup> + 3<i>x</i><sup>2</sup> + <i>bx</i> + 1 and <i>f</i>(<i>x</i>) ≥ 0 for all <i>x</i> ∈ ℝ.</p><p><strong>Step 2:</strong> By the theory of non-negative polynomials and Cauchy-Schwarz inequality applied to the coefficients and discriminant conditions, we obtain constraints on <i>a</i> and <i>b</i>.</p><p><strong>Step 3:</strong> The maximum value of <i>a</i><sup>2</sup> + <i>b</i><sup>2</sup> subject to the non-negativity condition is 12.</p><p>∴ Answer is (B) 12.</p>
Correct Answer: B