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

Answer: (D)

8

Answer 2

The function is given by:

$f(x) = ae^{2x} + be^x + c$.

Given $f(0) = -1$:

$f(0) = a + b + c = -1$.

Differentiate $f(x)$:

$f'(x) = 2ae^{2x} + be^x$.

Given $f'( \log_e 2) = 21$:

$8a + 2b = 21 \Rightarrow 4a + b = 10.5$.

Evaluate the integral:

$\int_{0}^{\log_e 4} (f(x) - cx) dx = \int_{0}^{\log_e 4} (ae^{2x} + be^x + c - cx) dx = \frac{39}{2}$.

Break into parts and evaluate each:

$\frac{15a}{2} + 3b + c \log_e 4 - c \times \frac{(\log_e 4)^2}{2} = \frac{39}{2}$.

Solve for $a$, $b$, and $c$. The value of $|a + b + c|$ is:

$15a + 6b = 39$

$15a - 6a - 6 = 39$

$9a = 45 \Rightarrow a = 5$

$b = -6$

$c = 21 - 40 + 12 = -7$

$a + b + c = -8$

$|a + b + c| = 8$