Given (f(x) `=log_(10)` x and g(x) = `e^(piix)` .
`phi (x) =|{:(f(x).g(x),(f(x))^(g(x)),1),(f(x^(2)).g(x^(2)),(f(x^(2)))^(g(x^(2))),0),(f(x^(3)).g(x^(3)),(f(x^(3)))^(g(x^(3))),1):}|` the value of `phi` (10), is

Options

A. 1
B. 2
C. 0
D. None of these

Answer

To find the value of \( \phi(10) \), we will first evaluate the functions \( f(x) \) and \( g(x) \) at \( x = 10 \) and then substitute these values into the determinant structure defined by \( \phi(x) \).

### Step 1: Evaluate \( f(10) \)
Given:
\[ f(x) = \log_{10} x \]
Substituting \( x = 10 \):
\[ f(10) = \log_{10} 10 = 1 \]

### Step 2: Evaluate \( g(10) \)
Given:
\[ g(x) = e^{\pi i x} \]
Substituting \( x = 10 \):
\[ g(10) = e^{\pi i \cdot 10} = e^{10 \pi i} \]
Using Euler's formula \( e^{ix} = \cos x + i \sin x \):
\[ g(10) = \cos(10\pi) + i \sin(10\pi) \]
Since \( \cos(10\pi) = 1 \) and \( \sin(10\pi) = 0 \):
\[ g(10) = 1 \]

### Step 3: Evaluate \( f(10^2) \)
Substituting \( x = 10^2 = 100 \):
\[ f(10^2) = f(100) = \log_{10} 100 = \log_{10} (10^2) = 2 \]

### Step 4: Evaluate \( g(10^2) \)
Substituting \( x = 10^2 = 100 \):
\[ g(10^2) = g(100) = e^{\pi i \cdot 100} = e^{100 \pi i} \]
Using Euler's formula:
\[ g(100) = \cos(100\pi) + i \sin(100\pi) \]
Since \( \cos(100\pi) = 1 \) and \( \sin(100\pi) = 0 \):
\[ g(100) = 1 \]

### Step 5: Evaluate \( f(10^3) \)
Substituting \( x = 10^3 = 1000 \):
\[ f(10^3) = f(1000) = \log_{10} 1000 = \log_{10} (10^3) = 3 \]

### Step 6: Evaluate \( g(10^3) \)
Substituting \( x = 10^3 = 1000 \):
\[ g(10^3) = g(1000) = e^{\pi i \cdot 1000} = e^{1000 \pi i} \]
Using Euler's formula:
\[ g(1000) = \cos(1000\pi) + i \sin(1000\pi) \]
Since \( \cos(1000\pi) = 1 \) and \( \sin(1000\pi) = 0 \):
\[ g(1000) = 1 \]

### Step 7: Construct the determinant for \( \phi(10) \)
Now we can substitute these values into the determinant:
\[ \phi(10) = \begin{vmatrix} f(10) g(10) & (f(10))^{g(10)} & 1 \\ f(10^2) g(10^2) & (f(10^2))^{g(10^2)} & 0 \\ f(10^3) g(10^3) & (f(10^3))^{g(10^3)} & 1 \end{vmatrix} \]
Substituting the values:
\[ \phi(10) = \begin{vmatrix} 1 \cdot 1 & 1^1 & 1 \\ 2 \cdot 1 & 2^1 & 0 \\ 3 \cdot 1 & 3^1 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 2 & 0 \\ 3 & 3 & 1 \end{vmatrix} \]

### Step 8: Calculate the determinant
Using the determinant formula:
\[ \text{Det} = 1 \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} - 1 \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 2 \\ 3 & 3 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ = 1(2 \cdot 1 - 0 \cdot 3) - 1(2 \cdot 1 - 0 \cdot 3) + 1(2 \cdot 3 - 2 \cdot 3) \]
\[ = 1(2) - 1(2) + 1(0) = 2 - 2 + 0 = 0 \]

### Final Result
Thus, the value of \( \phi(10) \) is:
\[ \phi(10) = 0 \]