Question

If $f(x)$ is a non-constant polynomial satisfying $f(x) = f'(x)f''(x)$ and $f(0) = 0$. Then the value of $\int_0^2 f(x) dx + f'(2) + f''(2)$ is :

• A. $\frac{14}{9}$
• B. $\frac{14}{11}$
• C. $\frac{11}{14}$
• D. $\frac{9}{14}$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We determine the degree of the polynomial by comparing the degrees of both sides. Then, solve for coefficients using the given functional equation.
Step 2: Key Formula or Approach:
If $\deg(f) = n$, then $\deg(f'f'') = (n-1) + (n-2) = 2n - 3$.
Step 3: Detailed Explanation:
1. Finding degree: $n = 2n - 3 \implies n = 3$.
2. Let $f(x) = ax^3 + bx^2 + cx + d$.
$f(0) = 0 \implies d = 0$.
3. Compare coefficients in $ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b)$.
Coeff of $x^3: a = 18a^2 \implies a = 1/18$ (as $f$ is non-constant).
Coeff of $x^2: b = 6ab + 12ab = 18ab = 18(1/18)b = b$ (Always true).
Coeff of constant: $2bc = 0$.
Coeff of $x: c = 4b^2 + 6ac = 4b^2 + 6(1/18)c \implies c = 4b^2 + c/3 \implies 2c/3 = 4b^2$.
Since $2bc=0$ and $2c/3=4b^2$, we must have $b=0$ and $c=0$.
Thus $f(x) = \frac{1}{18}x^3$.
4. Calculate values:
$\int_0^2 \frac{1}{18}x^3 dx = \frac{1}{18} \cdot \left[ \frac{x^4}{4} \right]_0^2 = \frac{16}{72} = \frac{2}{9}$.
$f'(x) = \frac{1}{6}x^2 \implies f'(2) = \frac{4}{6} = \frac{2}{3} = \frac{6}{9}$.
$f''(x) = \frac{1}{3}x \implies f''(2) = \frac{2}{3} = \frac{6}{9}$.
5. Sum $= \frac{2}{9} + \frac{6}{9} + \frac{6}{9} = \frac{14}{9}$.
Step 4: Final Answer:
The value is $\frac{14}{9}$.