If `f(x)=ae^(2x)+be^(x)+cx` satisfies the condition `f(0)=ae^(2x)+be^(x)+cx` satisfies the condition `f(0)=-1, f'(log2)=31,int_(0)^(log4)(f(x)-cx)dx=(39)/(2)`, then

Options

A. `a=-5, b=6, c=3`
B. `a=5, b=-6, c=3`
C. `a=-5, b=6, c=3`
D. none of these

Answer

To solve the problem step by step, we will analyze the given function and the conditions provided.

### Given:
The function is defined as:
\[ f(x) = ae^{2x} + be^{x} + cx \]

We have the following conditions:
1. \( f(0) = -1 \)
2. \( f'(\log 2) = 31 \)
3. \( \int_{0}^{\log 4} (f(x) - cx) \, dx = \frac{39}{2} \)

### Step 1: Find \( f(0) \)
Substituting \( x = 0 \) into the function:
\[ f(0) = ae^{0} + be^{0} + c(0) = a + b \]
Given that \( f(0) = -1 \):
\[ a + b = -1 \quad \text{(Equation 1)} \]

### Step 2: Find \( f'(x) \)
We differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(ae^{2x}) + \frac{d}{dx}(be^{x}) + \frac{d}{dx}(cx) \]
Using the chain rule:
\[ f'(x) = 2ae^{2x} + be^{x} + c \]

### Step 3: Evaluate \( f'(\log 2) \)
Substituting \( x = \log 2 \):
\[ f'(\log 2) = 2ae^{2\log 2} + be^{\log 2} + c \]
Using \( e^{\log 2} = 2 \):
\[ f'(\log 2) = 2a(2^2) + b(2) + c = 8a + 2b + c \]
Given that \( f'(\log 2) = 31 \):
\[ 8a + 2b + c = 31 \quad \text{(Equation 2)} \]

### Step 4: Evaluate the integral
We need to compute:
\[ \int_{0}^{\log 4} (f(x) - cx) \, dx \]
This can be rewritten as:
\[ \int_{0}^{\log 4} (ae^{2x} + be^{x}) \, dx \]
Calculating the integral:
\[ \int ae^{2x} \, dx = \frac{a}{2} e^{2x} \]
\[ \int be^{x} \, dx = be^{x} \]
Thus,
\[ \int_{0}^{\log 4} (ae^{2x} + be^{x}) \, dx = \left[ \frac{a}{2} e^{2x} + be^{x} \right]_{0}^{\log 4} \]
Calculating the limits:
\[ = \left( \frac{a}{2} e^{2\log 4} + b e^{\log 4} \right) - \left( \frac{a}{2} e^{0} + b e^{0} \right) \]
Using \( e^{2\log 4} = 16 \) and \( e^{\log 4} = 4 \):
\[ = \left( \frac{a}{2} \cdot 16 + 4b \right) - \left( \frac{a}{2} + b \right) \]
\[ = 8a + 4b - \frac{a}{2} - b = \frac{15a}{2} + 3b \]
Setting this equal to \( \frac{39}{2} \):
\[ \frac{15a}{2} + 3b = \frac{39}{2} \]
Multiplying through by 2:
\[ 15a + 6b = 39 \quad \text{(Equation 3)} \]

### Step 5: Solve the system of equations
We have three equations:
1. \( a + b = -1 \) (Equation 1)
2. \( 8a + 2b + c = 31 \) (Equation 2)
3. \( 15a + 6b = 39 \) (Equation 3)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = -1 - a \]

Substituting \( b \) into Equation 3:
\[ 15a + 6(-1 - a) = 39 \]
\[ 15a - 6 - 6a = 39 \]
\[ 9a - 6 = 39 \]
\[ 9a = 45 \]
\[ a = 5 \]

Now substituting \( a \) back into Equation 1:
\[ 5 + b = -1 \]
\[ b = -6 \]

### Step 6: Find \( c \)
Substituting \( a \) and \( b \) into Equation 2:
\[ 8(5) + 2(-6) + c = 31 \]
\[ 40 - 12 + c = 31 \]
\[ c = 31 - 28 \]
\[ c = 3 \]

### Final Values:
Thus, we have:
- \( a = 5 \)
- \( b = -6 \)
- \( c = 3 \)

### Summary:
The values are:
- \( a = 5 \)
- \( b = -6 \)
- \( c = 3 \)