Let `ax^(3) + bx^(2) + cx + d`
`|(x+1,2x,3x),(2x+3,x+1,x),(2-x,3x+4,5x-1)|`, then
What is the value of c ?

Options

A. -1
B. 34
C. 35
D. 50

Answer

To solve the problem, we need to find the value of \( c \) in the polynomial \( ax^3 + bx^2 + cx + d \) such that it is less than the determinant given by:

\[ \begin{vmatrix} x + 1 & 2x & 3x \\ 2x + 3 & x + 1 & x \\ 2 - x & 3x + 4 & 5x - 1 \end{vmatrix} \]

### Step 1: Set Up the Determinant
We start by writing the determinant explicitly:

\[ D = \begin{vmatrix} x + 1 & 2x & 3x \\ 2x + 3 & x + 1 & x \\ 2 - x & 3x + 4 & 5x - 1 \end{vmatrix} \]

### Step 2: Differentiate the Determinant
To find \( c \), we need to differentiate the determinant \( D \) with respect to \( x \). We denote the derivative of the determinant as \( D' \).

### Step 3: Calculate the Derivative
We can differentiate the determinant using the properties of determinants. We will differentiate each row with respect to \( x \) and evaluate at \( x = 0 \).

1. Differentiate the first row:
\[ \frac{d}{dx}(x + 1) = 1, \quad \frac{d}{dx}(2x) = 2, \quad \frac{d}{dx}(3x) = 3 \]
So the first row becomes \( (1, 2, 3) \).

2. Differentiate the second row:
\[ \frac{d}{dx}(2x + 3) = 2, \quad \frac{d}{dx}(x + 1) = 1, \quad \frac{d}{dx}(x) = 1 \]
So the second row becomes \( (2, 1, 1) \).

3. Differentiate the third row:
\[ \frac{d}{dx}(2 - x) = -1, \quad \frac{d}{dx}(3x + 4) = 3, \quad \frac{d}{dx}(5x - 1) = 5 \]
So the third row becomes \( (-1, 3, 5) \).

### Step 4: Evaluate the Determinant at \( x = 0 \)
Now we evaluate the determinant \( D' \) at \( x = 0 \):

\[ D' = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ -1 & 3 & 5 \end{vmatrix} \]

### Step 5: Calculate the Determinant
Calculating the determinant:

\[ D' = 1 \begin{vmatrix} 1 & 1 \\ 3 & 5 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ -1 & 5 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ 3 & 5 \end{vmatrix} = (1)(5) - (1)(3) = 5 - 3 = 2 \)

2. \( \begin{vmatrix} 2 & 1 \\ -1 & 5 \end{vmatrix} = (2)(5) - (1)(-1) = 10 + 1 = 11 \)

3. \( \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} = (2)(3) - (1)(-1) = 6 + 1 = 7 \)

Putting it all together:

\[ D' = 1(2) - 2(11) + 3(7) = 2 - 22 + 21 = 1 \]

### Step 6: Relate \( D' \) to \( c \)
Since we have shown that \( D' = c \), we conclude:

\[ c = 1 \]

### Conclusion
The value of \( c \) is \( 1 \).