The value of determinant `|(x+1,x+2,x+4),(x+3,x+5,x+8),(x+7,x+10,x+14)|` is

Options

A. `-2`
B. `x^2+2`
C. 2
D. none of these

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} x+1 & x+2 & x+4 \\ x+3 & x+5 & x+8 \\ x+7 & x+10 & x+14 \end{vmatrix} \]

we will perform row operations to simplify the determinant.

### Step 1: Row Operations
We will perform the following row operations:
- \( R_2 \leftarrow R_2 - R_1 \)
- \( R_3 \leftarrow R_3 - R_1 \)

This gives us:

\[ R_2 = (x+3 - (x+1), x+5 - (x+2), x+8 - (x+4)) = (2, 3, 4) \]

\[ R_3 = (x+7 - (x+1), x+10 - (x+2), x+14 - (x+4)) = (6, 8, 10) \]

Now the determinant looks like:

\[ D = \begin{vmatrix} x+1 & x+2 & x+4 \\ 2 & 3 & 4 \\ 6 & 8 & 10 \end{vmatrix} \]

### Step 2: Column Operations
Next, we will perform column operations:
- \( C_2 \leftarrow C_2 - C_1 \)
- \( C_3 \leftarrow C_3 - C_1 \)

This gives us:

\[ C_2 = (x+2 - (x+1), 3 - 2, 8 - 6) = (1, 1, 2) \]

\[ C_3 = (x+4 - (x+1), 4 - 2, 10 - 6) = (3, 2, 4) \]

Now the determinant looks like:

\[ D = \begin{vmatrix} x+1 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4 \end{vmatrix} \]

### Step 3: Expand the Determinant
We will now expand the determinant along the first row:

\[ D = (x+1) \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ 6 & 4 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 6 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = (1)(4) - (2)(2) = 4 - 4 = 0\)

2. \(\begin{vmatrix} 2 & 2 \\ 6 & 4 \end{vmatrix} = (2)(4) - (2)(6) = 8 - 12 = -4\)

3. \(\begin{vmatrix} 2 & 1 \\ 6 & 2 \end{vmatrix} = (2)(2) - (1)(6) = 4 - 6 = -2\)

Substituting these values back into the determinant expression:

\[ D = (x+1)(0) - 1(-4) + 3(-2) \]

\[ D = 0 + 4 - 6 = -2 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{-2} \]