Question

Let $f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3), x \in \mathbb{R}$. Then the value of $f'(5)$ is :

• A. $\frac{62}{5}$
• B. $\frac{657}{5}$
• C. $\frac{215}{5}$
• D. $\frac{117}{5}$

Hint:

In such functional equations, identify the derivatives at specific points as constants and solve a system of linear equations.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The values $f'(1), f''(2), f'''(3)$ are constants. Let $f'(1) = a, f''(2) = b, f'''(3) = c$.

Step 2: Detailed Explanation:
$f(x) = x^3 + ax^2 + 2bx + c$.
Differentiating repeatedly:
$f'(x) = 3x^2 + 2ax + 2b$
$f''(x) = 6x + 2a$
$f'''(x) = 6$.
So, $c = f'''(3) = 6$.
Now using $a = f'(1)$:
\[ a = 3(1)^2 + 2a(1) + 2b \implies a + 2b = -3 \quad \dots (i) \]
Using $b = f''(2)$:
\[ b = 6(2) + 2a \implies 2a - b = -12 \quad \dots (ii) \]
From (ii), $b = 2a + 12$. Substitute in (i):
\[ a + 2(2a + 12) = -3 \implies 5a + 24 = -3 \implies a = -27/5. \]
Then $b = 2(-27/5) + 60/5 = 6/5$.
Calculate $f'(5)$:
\[ f'(5) = 3(5)^2 + 2a(5) + 2b = 75 + 10a + 2b \]
\[ f'(5) = 75 + 10(-27/5) + 2(6/5) = 75 - 54 + 12/5 = 21 + 12/5 = 117/5. \]

Step 3: Final Answer:
$f'(5) = 117/5$.