Let `D(x)=|{:(x^2+4x-3, 2x+4,13),(2x^2+5x-9,4x+5,26),(8x^2-16x+1, 16x-6, 104):}|=alphax^3+betax^2 + gammax+delta` then :

Options

A. `alpha+beta=0`
B. `beta+gamma=0`
C. `alpha+beta+gamma+delta=0`
D. `alpha + beta+ gamma =0`

Answer

To solve the given problem, we need to evaluate the determinant \( D(x) \) and express it in the form \( \alpha x^3 + \beta x^2 + \gamma x + \delta \). Let's go through the steps systematically.

### Step 1: Write down the determinant
We start with the determinant given in the problem:

\[ D(x) = \begin{vmatrix} x^2 + 4x - 3 & 2x + 4 & 13 \\ 2x^2 + 5x - 9 & 4x + 5 & 26 \\ 8x^2 - 16x + 1 & 16x - 6 & 104 \end{vmatrix} \]

### Step 2: Differentiate the determinant
To find \( D'(x) \), we will differentiate the determinant with respect to \( x \). We can use the property of determinants that allows us to differentiate each column:

1. Differentiate the first column:
- The derivative of \( x^2 + 4x - 3 \) is \( 2x + 4 \).
- The derivative of \( 2x^2 + 5x - 9 \) is \( 4x + 5 \).
- The derivative of \( 8x^2 - 16x + 1 \) is \( 16x - 16 \).

2. The second and third columns remain unchanged for this differentiation.

Thus, we have:

\[ D'(x) = \begin{vmatrix} 2x + 4 & 2x + 4 & 13 \\ 4x + 5 & 4x + 5 & 26 \\ 16x - 16 & 16x - 6 & 104 \end{vmatrix} \]

### Step 3: Simplify the determinant
Notice that the first two columns of the determinant are identical:

\[ D'(x) = \begin{vmatrix} 2x + 4 & 2x + 4 & 13 \\ 4x + 5 & 4x + 5 & 26 \\ 16x - 16 & 16x - 6 & 104 \end{vmatrix} \]

Since two columns of a determinant are the same, the value of the determinant is zero:

\[ D'(x) = 0 \]

### Step 4: Conclude about \( D(x) \)
Since the derivative \( D'(x) = 0 \), it implies that \( D(x) \) is a constant function. Therefore, we can express \( D(x) \) as:

\[ D(x) = C \]

where \( C \) is a constant.

### Step 5: Identify coefficients
From the expression \( D(x) = \alpha x^3 + \beta x^2 + \gamma x + \delta \), since \( D(x) \) is constant, we have:

- \( \alpha = 0 \)
- \( \beta = 0 \)
- \( \gamma = 0 \)
- \( \delta = C \)

### Step 6: Calculate the required sums
Now we can calculate the required sums:

1. \( \lambda + 1 = C + 1 \)
2. \( \lambda + \beta = C + 0 = C \)
3. \( \beta + \gamma = 0 + 0 = 0 \)
4. \( \alpha + \beta + \gamma = 0 + 0 + 0 = 0 \)

### Final Answer
Thus, the values of the sums are:

- \( \lambda + 1 = C + 1 \)
- \( \lambda + \beta = C \)
- \( \beta + \gamma = 0 \)
- \( \alpha + \beta + \gamma = 0 \)