Question

Let \(f\) be a real polynomial of degree \(n\) such that \(f(x) = f'(x)f''(x)\), for all \(x \in \mathbb{R}\). If \(f(0) = 0\), then \(36(f''(2) + f''(2) + \int_0^2 f(x)\,dx)\) is equal to:

• A. 42
• B. 46
• C. 56
• D. 66

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
By equating the degrees of the polynomials on both sides of the equation \(f(x) = f'(x)f''(x)\), we can determine the degree of the polynomial \(f(x)\).
Step 2: Key Formula or Approach:
If \(\text{deg}(f) = n\), then \(\text{deg}(f') = n-1\) and \(\text{deg}(f'') = n-2\).
Equating the degrees: \(n = (n-1) + (n-2) \implies n = 2n - 3 \implies n = 3\).
Thus, \(f(x)\) is a cubic polynomial. Let \(f(x) = ax^3 + bx^2 + cx + d\).
Step 3: Detailed Explanation:
Given \(f(0) = 0\), we have \(d = 0\), so \(f(x) = ax^3 + bx^2 + cx\).
Calculate the derivatives:
\[ f'(x) = 3ax^2 + 2bx + c \] \[ f''(x) = 6ax + 2b \] Substitute into the given equation \(f(x) = f'(x)f''(x)\):
\[ ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b) \] Expand the RHS:
\[ RHS = 18a^2 x^3 + (6ab + 12ab)x^2 + (4b^2 + 6ac)x + 2bc = 18a^2 x^3 + 18ab x^2 + (4b^2 + 6ac)x + 2bc \] Comparing coefficients of corresponding powers of \(x\):
1. \(x^3\): \(a = 18a^2 \implies a = \frac{1}{18}\) (since \(a \neq 0\) for a cubic polynomial).
2. \(x^2\): \(b = 18ab \implies b = 18(\frac{1}{18})b \implies b = b\) (consistent, no new info).
3. \(x^1\): \(c = 4b^2 + 6ac \implies c = 4b^2 + 6(\frac{1}{18})c \implies c = 4b^2 + \frac{1}{3}c \implies \frac{2}{3}c = 4b^2 \implies c = 6b^2\).
4. Constant: \(0 = 2bc\).
From \(2bc = 0\), since \(c = 6b^2\), we have \(2b(6b^2) = 0 \implies 12b^3 = 0 \implies b = 0\).
Since \(b = 0\), it follows that \(c = 6(0)^2 = 0\).
Therefore, the polynomial is \(f(x) = \frac{1}{18}x^3\).
Now calculate the required terms at \(x = 2\):
\[ f'(x) = \frac{3}{18}x^2 = \frac{1}{6}x^2 \implies f'(2) = \frac{1}{6}(4) = \frac{2}{3} \] \[ f''(x) = \frac{1}{3}x \implies f''(2) = \frac{2}{3} \] \[ \int_0^2 f(x) dx = \int_0^2 \frac{1}{18}x^3 dx = \frac{1}{18} \left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{18} \left( \frac{16}{4} \right) = \frac{4}{18} = \frac{2}{9} \] Step 4: Final Answer:
Calculate the expression:
\[ 36 \left( f'(2) + f''(2) + \int_0^2 f(x) dx \right) = 36 \left( \frac{2}{3} + \frac{2}{3} + \frac{2}{9} \right) \] \[ = 36 \left( \frac{4}{3} + \frac{2}{9} \right) = 36 \left( \frac{12 + 2}{9} \right) = 36 \left( \frac{14}{9} \right) = 4 \times 14 = 56 \]