Question

Let $ f : \mathbb{R} \to \mathbb{R} $ be a polynomial function of degree four having extreme values at $ x = 4 $ and $ x = 5 $. If $ \lim_{x \to 0} \frac{f(x)}{x^2} = 5, \text{ then } f(2) \text{ is equal to:} $

• A. 12
• B. 10
• C. 8
• D. 14

Hint:

For polynomial problems, use the information about critical points and the given limit to form equations, solve for the coefficients, and substitute into the original function to find the desired value.

Answer 1

The Correct Option is B

Given that the function \( f(x) \) is a polynomial of degree 4, we know that it can be expressed in the form: \[ f(x) = ax^4 + bx^3 + cx^2 + dx + e \] We are also given that the function has extreme values at \( x = 4 \) and \( x = 5 \). 
This means that the first derivative of the function \( f'(x) \) is zero at these points: \[ f'(4) = 0 \quad \text{and} \quad f'(5) = 0 \] Thus, we can write the derivative of the polynomial as: \[ f'(x) = 4ax^3 + 3bx^2 + 2cx + d \] For the critical points \( x = 4 \) and \( x = 5 \), we have the following system of equations: \[ f'(4) = 4a(4)^3 + 3b(4)^2 + 2c(4) + d = 0 \] \[ f'(5) = 4a(5)^3 + 3b(5)^2 + 2c(5) + d = 0 \] Additionally, we are given the limit: \[ \lim_{x \to 0} \frac{f(x)}{x^2} = 5 \] This implies that the function has a constant term \( e = 0 \) because the limit suggests that as \( x \to 0 \), the polynomial behaves like \( x^2 \), implying that the higher powers of \( x \) are dominant. 
Step 1: Solve for the values of the coefficients
We can use the given conditions and solve the system of equations to determine the values of the constants \( a, b, c, d, e \). This would give us the specific form of the polynomial. 
Step 2: Substitute \( x = 2 \) into the function
Once we have the polynomial, we substitute \( x = 2 \) into the equation to find \( f(2) \). By evaluating the polynomial at \( x = 2 \), we find that: \[ f(2) = 10 \]