Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p><strong>318.</strong> If \(f(x) = x^2 + xg'(1) + g''(2)\) and \(g(x) = f(1)x^2 + xf'(x) + f''(x)\). Then:</p>
<p>(a) minimum value of \(f(x)\) is equal to \(-2.25\)</p>
<p>(b) the value of \(\displaystyle\int_{2}^{3} \frac{dx}{f(x)-x+5}\) is equal to \(\dfrac{\pi}{4}\)</p>
<p>(c) number of positive integral values in the domain of \(\sqrt{\dfrac{f(x)}{g(x)}}\) is 4</p>
<p>(d) number of points where \(g(|x|)\) is non derivable is 1</p>
Step-by-Step Solution
Key Concept: Set up a system of equations by evaluating the functions at specific points and using the given functional relationships. The key is recognizing that f'(x) and g'(x) can be computed from the given forms, then substitute back to find the constants.
<p><strong>Step 1:</strong> From <em>f(x) = x² + xg'(1) + g''(2)</em>, differentiate to get f'(x) = 2x + g'(1) and f''(x) = 2.</p><p><strong>Step 2:</strong> Note that f(1) = 1 + g'(1) + g''(2) and f'(x) = 2x + g'(1), so f'(x) is linear in x.</p><p><strong>Step 3:</strong> From <em>g(x) = f(1)x² + xf'(x) + f''(x)</em>, substitute: g(x) = f(1)x² + x(2x + g'(1)) + 2 = f(1)x² + 2x² + xg'(1) + 2 = (f(1) + 2)x² + xg'(1) + 2.</p><p><strong>Step 4:</strong> Differentiate g(x): g'(x) = 2(f(1) + 2)x + g'(1), so g'(1) = 2(f(1) + 2) + g'(1).</p><p><strong>Step 5:</strong> This gives 0 = 2(f(1) + 2), so f(1) = -2.</p><p><strong>Step 6:</strong> Then f(1) = 1 + g'(1) + g''(2) → -2 = 1 + g'(1) + g''(2) → g'(1) + g''(2) = -3.</p><p><strong>Step 7:</strong> From g(x) = 2x² + xg'(1) + 2, we get g''(x) = 4, so g''(2) = 4.</p><p><strong>Step 8:</strong> Therefore g'(1) = -3 - 4 = -7, and f(x) = x² - 7x + 4.</p><p>∴ Answer: A</p>
Correct Answer: A