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

The Correct Option is D

To solve for \( a^2 + b^2 + e^2 \) given the function \( f(x) = ax^3 + bx^2 + ex + 41 \) with the conditions \( f(1) = 40 \), \( f'(1) = 2 \), and \( f''(1) = 4 \), we proceed as follows:

1. First, compute \( f(1) \):

\[ f(1) = a(1)^3 + b(1)^2 + e(1) + 41 = a + b + e + 41 = 40 \]

Solving this gives us: \[ a + b + e = 40 - 41 = -1 \] (Equation 1)

1. Next, compute the first derivative \( f'(x) \):

\[ f'(x) = 3ax^2 + 2bx + e \]

Then, at \( x = 1 \),

\[ f'(1) = 3a(1)^2 + 2b(1) + e = 3a + 2b + e = 2 \]

(Equation 2)

1. Now, compute the second derivative \( f''(x) \):

\[ f''(x) = 6ax + 2b \]

Then, at \( x = 1 \),

\[ f''(1) = 6a(1) + 2b = 6a + 2b = 4 \]

This simplifies to: \( 3a + b = 2 \)

(Equation 3)

1. We now have the following system of equations:
   • Equation 1: \( a + b + e = -1 \)
   • Equation 2: \( 3a + 2b + e = 2 \)
   • Equation 3: \( 3a + b = 2 \)
2. Subtract Equation 1 from Equation 2 to eliminate \( e \):

\[ (3a + 2b + e) - (a + b + e) = 2 - (-1) \\ 2a + b = 3 \]

(Equation 4)

1. Subtract Equation 4 from Equation 3 to find \( a \):

\[ (3a + b) - (2a + b) = 2 - 3 \\ a = -1 \]

1. Substitute \( a = -1 \) into Equation 3 to find \( b \):

\[ 3(-1) + b = 2 \\ -3 + b = 2 \Rightarrow b = 5 \]

1. Finally, substitute \( a = -1 \) and \( b = 5 \) into Equation 1 to find \( e \):

\[ -1 + 5 + e = -1 \\ 4 + e = -1 \Rightarrow e = -5 \]

1. Now compute \( a^2 + b^2 + e^2 \):

\[ a^2 + b^2 + e^2 = (-1)^2 + 5^2 + (-5)^2 \\ = 1 + 25 + 25 = 51 \]

Thus, the value of \( a^2 + b^2 + e^2 \) is 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).