If `Delta(x)=|(1,1,1),((e^(x)+e^(-x))^(2),(pi^(x)+pi^(-x))^(2),2),((e^(x)-e^(-x))^(2),(pi^(x)-pi^(-x))^(2),-2)|` then` Delta(x)` equals

Options

A. `x^(2)`
B. `x^(2)-1`
C. `e^(x^(2))-pi^(x^(2))`
D. `0`

Answer

To solve the determinant \( \Delta(x) = \begin{vmatrix} 1 & 1 & 1 \\ (e^x + e^{-x})^2 & (\pi^x + \pi^{-x})^2 & 2 \\ (e^x - e^{-x})^2 & (\pi^x - \pi^{-x})^2 & -2 \end{vmatrix} \), we will follow these steps:

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

\[ \Delta(x) = \begin{vmatrix} 1 & 1 & 1 \\ (e^x + e^{-x})^2 & (\pi^x + \pi^{-x})^2 & 2 \\ (e^x - e^{-x})^2 & (\pi^x - \pi^{-x})^2 & -2 \end{vmatrix} \]

### Step 2: Simplify the second row
Notice that \( (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x} \) and \( (e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x} \). We can denote \( a = e^x \) and \( b = \pi^x \) for simplicity.

Thus:
- \( (e^x + e^{-x})^2 = a^2 + 2 + \frac{1}{a^2} \)
- \( (e^x - e^{-x})^2 = a^2 - 2 + \frac{1}{a^2} \)
- \( (\pi^x + \pi^{-x})^2 = b^2 + 2 + \frac{1}{b^2} \)
- \( (\pi^x - \pi^{-x})^2 = b^2 - 2 + \frac{1}{b^2} \)

### Step 3: Perform row operations
We perform the row operation \( R_2 \leftarrow R_2 - R_3 \):

\[ R_2 = \begin{pmatrix} (e^x + e^{-x})^2 - (e^x - e^{-x})^2 & (\pi^x + \pi^{-x})^2 - (\pi^x - \pi^{-x})^2 & 2 - (-2) \end{pmatrix} \]

Calculating the first two elements:
- \( (e^x + e^{-x})^2 - (e^x - e^{-x})^2 = 4 \cdot e^x \cdot e^{-x} = 4 \)
- \( (\pi^x + \pi^{-x})^2 - (\pi^x - \pi^{-x})^2 = 4 \cdot \pi^x \cdot \pi^{-x} = 4 \)

Thus, the new second row becomes:
\[ R_2 = \begin{pmatrix} 4 & 4 & 4 \end{pmatrix} \]

### Step 4: Update the determinant
Now the determinant looks like:

\[ \Delta(x) = \begin{vmatrix} 1 & 1 & 1 \\ 4 & 4 & 4 \\ (e^x - e^{-x})^2 & (\pi^x - \pi^{-x})^2 & -2 \end{vmatrix} \]

### Step 5: Check for proportionality
Notice that the second row \( R_2 \) is proportional to the first row \( R_1 \) (both are multiples of 1). Thus, we can conclude that:

\[ \Delta(x) = 0 \]

### Final Answer
Therefore, the value of \( \Delta(x) \) is:

\[ \Delta(x) = 0 \]