<p>If <span class="math">\(p(x)\)</span> be a polynomial of degree three that has a local maximum value 8 at <span class="math">\(x = 1\)</span> and a local minimum value 4 at <span class="math">\(x = 2\)</span>; then <span class="math">\(p(0)\)</span> is equal to</p>
Step-by-Step Solution
Key Concept: For a cubic polynomial with extrema at given points, use p'(x) = 0 at those points to construct p'(x), then integrate and use the extrema values to find the polynomial.
<p><strong>Step 1:</strong> Since <span class="math">$p(x)$</span> is a cubic polynomial with local maximum at <span class="math">$x = 1$</span> and local minimum at <span class="math">$x = 2$</span>, we have <span class="math">$p'(1) = 0$</span> and <span class="math">$p'(2) = 0$</span>.</p><p><strong>Step 2:</strong> Let <span class="math">$p'(x) = a(x - 1)(x - 2)$</span> where <span class="math">$a$</span> is a constant. Since <span class="math">$p'(x)$</span> is a quadratic (derivative of cubic), we can write:</p><p><span class="math">$p'(x) = a(x - 1)(x - 2) = a(x^2 - 3x + 2)$</span></p><p><strong>Step 3:</strong> Integrating: <span class="math">$p(x) = a\left(\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right) + c$</span></p><p><strong>Step 4:</strong> Use the conditions: <span class="math">$p(1) = 8$</span> and <span class="math">$p(2) = 4$</span> to find constants <span class="math">$a$</span> and <span class="math">$c$</span>.</p><p><strong>Step 5:</strong> Solving the system yields <span class="math">$p(0) = -12$</span>.</p><p>∴ Answer is (d).</p>
Correct Answer: d