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 determine the value of \( a^2 + b^2 + e^2 \) given the polynomial \( f(x) = ax^3 + bx^2 + ex + 41 \) and the conditions \( f(1) = 40 \), \( f'(1) = 2 \), and \( f''(1) = 4 \), the following steps are taken:

1. Evaluate \( f(1) \):

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

This simplifies to: \[ a + b + e = -1 \] (Equation 1)

1. Calculate the first derivative \( f'(x) \):

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

Evaluate \( f'(1) \):

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

(Equation 2)

1. Calculate the second derivative \( f''(x) \):

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

Evaluate \( f''(1) \):

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

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

(Equation 3)

1. A system of three linear equations is established:
   • Equation 1: \( a + b + e = -1 \)
   • Equation 2: \( 3a + 2b + e = 2 \)
   • Equation 3: \( 3a + b = 2 \)
2. Eliminate \( e \) by subtracting Equation 1 from Equation 2:

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

(Equation 4)

1. Solve for \( a \) by subtracting Equation 4 from Equation 3:

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

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

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

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

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

1. Calculate the required sum of squares:

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

The value of \( a^2 + b^2 + e^2 \) is 51.