Question

Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be defined \( f(x) = ae^{2x} + be^x + cx \). If \( f(0) = -1 \), \( f'(\log_e 2) = 21 \) and
\[\int_{0}^{\log_e 4} (f(x) - cx) \, dx = \frac{39}{2}\]
then the value of \( |a + b + c| \) equals:

• A. 16
• B. 10
• C. 12
• D. 8

Answer 1

The Correct Option is D

To determine the value of \( |a + b + c| \), we analyze the function \( f(x) = ae^{2x} + be^x + cx \) under given conditions.

1. Determine \( f(0) \): Given \( f(0) = -1 \), substituting \( x = 0 \) yields \( f(0) = a \cdot e^0 + b \cdot e^0 + c \cdot 0 = a + b = -1 \).
2. Calculate \( f'(\log_e 2) = 21 \): The derivative is \( f'(x) = 2ae^{2x} + be^x + c \). Substituting \( x = \log_e 2 \) results in \( f'(\log_e 2) = 2a e^{2 \log_e 2} + b e^{\log_e 2} + c = 8a + 2b + c = 21 \).
3. Evaluate the integral: \[\int_{0}^{\log_e 4} (f(x) - cx) \, dx = \frac{39}{2}\]. The integrand simplifies to \( f(x) - cx = ae^{2x} + be^x \). Integrating yields: \[ \left[\frac{a}{2} e^{2x} + b e^x \right]_{0}^{\log_e 4} = \frac{39}{2} \]. Evaluating the limits: \[ \frac{a}{2} e^{2 \log_e 4} + b e^{\log_e 4} - \left(\frac{a}{2} \cdot e^0 + b \cdot e^0\right) = \frac{39}{2} \]. Simplifying: \[ 8a + 4b - \left(\frac{a}{2} + b\right) = \frac{39}{2} \implies \frac{15a + 6b}{2} = \frac{39}{2} \implies 15a + 6b = 39 \].
4. Solve the system of equations: We have the following system: \( a + b = -1 \), \( 8a + 2b + c = 21 \), and \( 15a + 6b = 39 \).
5. Solve for coefficients: Substitute \( b = -1 - a \) into the third equation: \( 15a + 6(-1 - a) = 39 \Rightarrow 9a = 45 \Rightarrow a = 5 \). Then, \( b = -1 - 5 = -6 \). Substitute \( a \) and \( b \) into the second equation: \( 8(5) + 2(-6) + c = 21 \Rightarrow 40 - 12 + c = 21 \Rightarrow c = -7 \).
6. Calculate \( |a + b + c| \): \( a + b + c = 5 - 6 - 7 = -8 \). Therefore, \( |a + b + c| = 8 \).

The result is 8.