Let `f(x) = ax^(3) + 5x^(2) - bx + 1`. If f(x) when divied by 2x + 1 leaves 5 as remainder, and f'(x) is divisible by 3x - 1, then

Options

A. `a = 26, b = 10`
B. `a = 24, b = 12`
C. `a = 26, b = 12`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given polynomial \( f(x) = ax^3 + 5x^2 - bx + 1 \) and use the information provided about the remainder and the derivative.

### Step 1: Use the Remainder Theorem
According to the Remainder Theorem, if \( f(x) \) is divided by \( 2x + 1 \), the remainder is \( f\left(-\frac{1}{2}\right) \). We know that this remainder is 5.

So, we calculate \( f\left(-\frac{1}{2}\right) \):
\[ f\left(-\frac{1}{2}\right) = a\left(-\frac{1}{2}\right)^3 + 5\left(-\frac{1}{2}\right)^2 - b\left(-\frac{1}{2}\right) + 1 \]
Calculating each term:
\[ = a\left(-\frac{1}{8}\right) + 5\left(\frac{1}{4}\right) + \frac{b}{2} + 1 \]
\[ = -\frac{a}{8} + \frac{5}{4} + \frac{b}{2} + 1 \]
Now, we set this equal to 5:
\[ -\frac{a}{8} + \frac{5}{4} + \frac{b}{2} + 1 = 5 \]
To simplify, we can convert everything to a common denominator (8):
\[ -\frac{a}{8} + \frac{10}{8} + \frac{4b}{8} + \frac{8}{8} = 5 \]
\[ -\frac{a + 4b - 18}{8} = 0 \]
Thus, we have:
\[ -a + 4b = 22 \quad \text{(Equation 1)} \]

### Step 2: Differentiate \( f(x) \)
Next, we differentiate \( f(x) \):
\[ f'(x) = 3ax^2 + 10x - b \]
We know that \( f'(x) \) is divisible by \( 3x - 1 \). This means that \( f'\left(\frac{1}{3}\right) = 0 \).

Calculating \( f'\left(\frac{1}{3}\right) \):
\[ f'\left(\frac{1}{3}\right) = 3a\left(\frac{1}{3}\right)^2 + 10\left(\frac{1}{3}\right) - b \]
\[ = 3a\left(\frac{1}{9}\right) + \frac{10}{3} - b \]
\[ = \frac{a}{3} + \frac{10}{3} - b \]
Setting this equal to 0 gives us:
\[ \frac{a + 10 - 3b}{3} = 0 \]
Thus, we have:
\[ a + 10 - 3b = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have two equations:
1. \( -a + 4b = 22 \) (Equation 1)
2. \( a - 3b = -10 \) (Equation 2)

We can solve these equations simultaneously. From Equation 2, we can express \( a \):
\[ a = 3b - 10 \]
Substituting this into Equation 1:
\[ -(3b - 10) + 4b = 22 \]
\[ -3b + 10 + 4b = 22 \]
\[ b + 10 = 22 \]
\[ b = 12 \]

Now substituting \( b = 12 \) back into Equation 2 to find \( a \):
\[ a - 3(12) = -10 \]
\[ a - 36 = -10 \]
\[ a = 26 \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = 26, \quad b = 12 \]