<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 expressing f and g in terms of each other's derivatives, then use the self-referential nature to find concrete values for all constants (g'(1), g''(2), f(1), f'(x), f''(x)).
<p><strong>Step 1:</strong> Let g'(1) = a and g''(2) = b. Then f(x) = x² + ax + b</p><p><strong>Step 2:</strong> Find f'(x) = 2x + a and f''(x) = 2. Also f(1) = 1 + a + b</p><p><strong>Step 3:</strong> Substitute into g(x) = f(1)x² + xf'(x) + f''(x):<br/>g(x) = (1 + a + b)x² + x(2x + a) + 2<br/>g(x) = (1 + a + b)x² + 2x² + ax + 2<br/>g(x) = (3 + a + b)x² + ax + 2</p><p><strong>Step 4:</strong> Find g'(x) = 2(3 + a + b)x + a, so g'(1) = 2(3 + a + b) + a = 6 + 3a + 2b</p><p><strong>Step 5:</strong> Since g'(1) = a: a = 6 + 3a + 2b → -2a + 2b = 6 → b - a = 3 ... (i)</p><p><strong>Step 6:</strong> Find g''(x) = 2(3 + a + b), so g''(2) = 2(3 + a + b)</p><p><strong>Step 7:</strong> Since g''(2) = b: b = 2(3 + a + b) = 6 + 2a + 2b → -b = 6 + 2a → b + 2a = -6 ... (ii)</p><p><strong>Step 8:</strong> Solve (i) and (ii): From (i): b = a + 3. Substitute into (ii): (a + 3) + 2a = -6 → 3a = -9 → a = -3, b = 0</p><p><strong>Step 9:</strong> Therefore: f(x) = x² - 3x, g(x) = -2x² - 3x + 2, f(1) = -2, g'(1) = -3, g''(2) = 0</p><p>∴ Answer: All derived values satisfy the system: f(x) = x² - 3x, g(x) = -2x² - 3x + 2</p>
Correct Answer: A,B,C,D