Question

Let $f(x) = ax^3 + bx^2 + ex + 41$ be such that $f(1) = 40$, $f'(1) = 2$ and $f''(1) = 4$. Then $a^2 + b^2 + c^2$ is equal to:

• A. 60
• B. 73
• C. 54
• D. 51

Answer 1

Answer: (D)

51

Answer 2

Step 1: Derivatives of $f(x)$ The given function is:

$f(x) = ax^3 + bx^2 + cx + 41.$

The first derivative:

$f'(x) = 3ax^2 + 2bx + c.$

The second derivative:

$f''(x) = 6ax + 2b.$

Step 2: Use the given conditions

1. From $f'(1) = 2$:
2. From $f''(1) = 4$:
3. From $f(1) = 40$:

Step 3: Solve for $a$, $b$, $c$

From equation (2):

$6a + 2b = 4 \implies 3a + b = 2.$ (4)

From equations (1) and (4):

$3a + 2b + c = 2,$

$3a + b = 2.$

Subtract equation (4) from (1):

$(3a + 2b + c) - (3a + b) = 2 - 2,$

$b + c = 0.$ (5)

From equations (3) and (5):

$a + b + c = -1, \quad b + c = 0.$

Subtract:

$a = -1.$ (6)

Substitute $a = -1$ into equation (4):

$3(-1) + b = 2 \implies -3 + b = 2 \implies b = 5.$ (7)

From equation (5):

$b + c = 0 \implies 5 + c = 0 \implies c = -5.$ (8)

Step 4: Compute $a^2 + b^2 + c^2$

$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2 = 1 + 25 + 25 = 51.$

Final Answer: Option (4).