`f(x)=(1)/(x),g(x)=(x+(1)/(x))`, then which of the following is true?

Options

A. `f(g(x)).g(f(x))=1`
B. `(f(g(x)))/(g(f(x)))=1`
C. `f(g(x))+g(f(x))=1`
D. none of these

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

- \( f(x) = \frac{1}{x} \)
- \( g(x) = x + \frac{1}{x} \)

We will evaluate \( f(g(x)) \) and \( g(f(x)) \), and then check the validity of the given options.

### Step 1: Calculate \( f(g(x)) \)

1. Substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f\left(x + \frac{1}{x}\right) = \frac{1}{x + \frac{1}{x}} \]

2. To simplify this, we can take the LCM:
\[ f(g(x)) = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \]

### Step 2: Calculate \( g(f(x)) \)

1. Substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{x} + x \]

2. To combine these terms, we can take the LCM:
\[ g(f(x)) = \frac{1}{x} + x = \frac{1 + x^2}{x} \]

### Step 3: Check the options

#### Option 1: \( f(g(x)) \cdot g(f(x)) = 1 \)

1. Calculate the product:
\[ f(g(x)) \cdot g(f(x)) = \left(\frac{x}{x^2 + 1}\right) \cdot \left(\frac{1 + x^2}{x}\right) \]

2. Simplifying this:
\[ = \frac{x(1 + x^2)}{(x^2 + 1)x} = 1 \]

This option is **true**.

#### Option 2: \( \frac{f(g(x))}{g(f(x))} = 1 \)

1. Calculate the division:
\[ \frac{f(g(x))}{g(f(x))} = \frac{\frac{x}{x^2 + 1}}{\frac{1 + x^2}{x}} = \frac{x^2}{x^2 + 1} \]

This option is **false** since it does not equal 1.

#### Option 3: \( f(g(x)) + g(f(x)) = 1 \)

1. Calculate the sum:
\[ f(g(x)) + g(f(x)) = \frac{x}{x^2 + 1} + \frac{1 + x^2}{x} \]

2. To combine these, we need a common denominator:
\[ = \frac{x^2}{x(x^2 + 1)} + \frac{(1 + x^2)(x^2 + 1)}{x(x^2 + 1)} = \frac{x^2 + (1 + x^2)(x^2 + 1)}{x(x^2 + 1)} \]

This option is also **false** since it does not equal 1.

### Conclusion

The only true statement is:

**Option 1: \( f(g(x)) \cdot g(f(x)) = 1 \)**.