`a#b=(-1)^(ab)(a^(b)+b^(a))`
`f(x)=x^(2)-2x if x ge 0`
`=0if x lt0`
`g(x)=2x, if x ge0`
`=1, if x lt0`
Find the value of `f(2#3)#g(3#4)`:

Options

A. 2045
B. `(17)^(18)`
C. `(180)^(11)`
D. none of these

Answer

To solve the problem step by step, we will follow the defined operations and functions carefully.

### Step 1: Calculate \( 2 \# 3 \)

Using the formula given for \( a \# b \):

\[ a \# b = (-1)^{ab} (a^b + b^a) \]

Substituting \( a = 2 \) and \( b = 3 \):

\[ 2 \# 3 = (-1)^{2 \cdot 3} (2^3 + 3^2) \]

Calculating \( (-1)^{6} \):

\[ (-1)^{6} = 1 \]

Now calculate \( 2^3 \) and \( 3^2 \):

\[ 2^3 = 8 \quad \text{and} \quad 3^2 = 9 \]

So,

\[ 2 \# 3 = 1 \cdot (8 + 9) = 1 \cdot 17 = 17 \]

### Step 2: Calculate \( 3 \# 4 \)

Now, we calculate \( 3 \# 4 \) using the same formula:

\[ 3 \# 4 = (-1)^{3 \cdot 4} (3^4 + 4^3) \]

Calculating \( (-1)^{12} \):

\[ (-1)^{12} = 1 \]

Now calculate \( 3^4 \) and \( 4^3 \):

\[ 3^4 = 81 \quad \text{and} \quad 4^3 = 64 \]

So,

\[ 3 \# 4 = 1 \cdot (81 + 64) = 1 \cdot 145 = 145 \]

### Step 3: Calculate \( g(3 \# 4) \)

Now we need to find \( g(145) \). The function \( g(x) \) is defined as:

\[ g(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ 1 & \text{if } x < 0 \end{cases} \]

Since \( 145 \geq 0 \):

\[ g(145) = 2 \cdot 145 = 290 \]

### Step 4: Calculate \( f(2 \# 3) \)

Next, we need to find \( f(17) \). The function \( f(x) \) is defined as:

\[ f(x) = \begin{cases} x^2 - 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} \]

Since \( 17 \geq 0 \):

\[ f(17) = 17^2 - 2 \cdot 17 \]

Calculating \( 17^2 \):

\[ 17^2 = 289 \]

Calculating \( 2 \cdot 17 \):

\[ 2 \cdot 17 = 34 \]

So,

\[ f(17) = 289 - 34 = 255 \]

### Step 5: Calculate \( f(2 \# 3) \# g(3 \# 4) \)

Now we need to calculate \( 255 \# 290 \):

\[ 255 \# 290 = (-1)^{255 \cdot 290} (255^{290} + 290^{255}) \]

Calculating \( (-1)^{255 \cdot 290} \):

Since \( 255 \cdot 290 \) is even, we have:

\[ (-1)^{255 \cdot 290} = 1 \]

Now we need to compute \( 255^{290} + 290^{255} \).

Since these are large numbers, we can denote the result as:

\[ 255 \# 290 = 1 \cdot (255^{290} + 290^{255}) = 255^{290} + 290^{255} \]

### Final Result

Thus, the final answer is:

\[ f(2 \# 3) \# g(3 \# 4) = 255^{290} + 290^{255} \]