If A is a square matrix and `e^a` is defined as `e^A=1+A^2/(2!)+A^3/(3!)...........oo=1/2[f(x) ,g(x) and g(x) ,f(x)],` where `A=[(x,x),(x,x)].` and I being the identity matrix then `int (g(x))/(f(x))dx=`

Options

A. `(e^(x))/(2)(sinx-cosx)`
B. `(e^(2x))/(5)(2sinx-cosx)`
C. `(e^(x))/(5)(sin2x-cos2x)`
D. none of these

Explanation

To solve the given problem, we need to find the integral of \( \frac{g(x)}{f(x)} \, dx \) where \( A = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \) and \( e^A \) is defined as:

\[ e^A = I + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots \]

### Step 1: Calculate \( A^2 \)

First, we calculate \( A^2 \):

\[ A^2 = A \cdot A = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \cdot \begin{pmatrix} x & x \\ x & x \end{pmatrix} \]

Calculating the product:

\[ A^2 = \begin{pmatrix} x^2 + x^2 & x^2 + x^2 \\ x^2 + x^2 & x^2 + x^2 \end{pmatrix} = \begin{pmatrix} 2x^2 & 2x^2 \\ 2x^2 & 2x^2 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)

Next, we calculate \( A^3 \):

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 2x^2 & 2x^2 \\ 2x^2 & 2x^2 \end{pmatrix} \cdot \begin{pmatrix} x & x \\ x & x \end{pmatrix} \]

Calculating the product:

\[ A^3 = \begin{pmatrix} 2x^2 \cdot x + 2x^2 \cdot x & 2x^2 \cdot x + 2x^2 \cdot x \\ 2x^2 \cdot x + 2x^2 \cdot x & 2x^2 \cdot x + 2x^2 \cdot x \end{pmatrix} = \begin{pmatrix} 4x^3 & 4x^3 \\ 4x^3 & 4x^3 \end{pmatrix} \]

### Step 3: General Form of \( A^n \)

From the pattern, we can see that:

\[ A^n = \begin{pmatrix} 2^{n-1} x^n & 2^{n-1} x^n \\ 2^{n-1} x^n & 2^{n-1} x^n \end{pmatrix} \]

### Step 4: Compute \( e^A \)

Now we can compute \( e^A \):

\[ e^A = I + \sum_{n=1}^{\infty} \frac{A^n}{n!} = I + \sum_{n=1}^{\infty} \frac{1}{n!} \begin{pmatrix} 2^{n-1} x^n & 2^{n-1} x^n \\ 2^{n-1} x^n & 2^{n-1} x^n \end{pmatrix} \]

This leads to:

\[ e^A = \begin{pmatrix} 1 + \sum_{n=1}^{\infty} \frac{2^{n-1} x^n}{n!} & \sum_{n=1}^{\infty} \frac{2^{n-1} x^n}{n!} \\ \sum_{n=1}^{\infty} \frac{2^{n-1} x^n}{n!} & 1 + \sum_{n=1}^{\infty} \frac{2^{n-1} x^n}{n!} \end{pmatrix} \]

### Step 5: Identify \( f(x) \) and \( g(x) \)

From the matrix, we can identify:

\[ f(x) = 1 + e^{2x} - 1 = e^{2x} \]
\[ g(x) = e^{2x} - 1 \]

### Step 6: Calculate \( \int \frac{g(x)}{f(x)} \, dx \)

Now we need to compute:

\[ \int \frac{g(x)}{f(x)} \, dx = \int \frac{e^{2x} - 1}{e^{2x}} \, dx = \int \left( 1 - \frac{1}{e^{2x}} \right) \, dx \]

This can be separated into two integrals:

\[ \int 1 \, dx - \int e^{-2x} \, dx \]

Calculating these integrals:

\[ = x - \left(-\frac{1}{2} e^{-2x}\right) + C = x + \frac{1}{2} e^{-2x} + C \]

### Final Answer

Thus, the final answer for the integral is:

\[ \int \frac{g(x)}{f(x)} \, dx = x + \frac{1}{2} e^{-2x} + C \]