If `f:R->R` and `g:R->R` is given by f(x) =|x| and g(x)=[x] for each `x in R` then `{x in R:g(f(x))le f(g(x))}`

Options

A. `Z cup (-oo,0) `
B. `(-oo,0)`
C. Z
D. R

Answer

To solve the problem, we need to analyze the functions \( f \) and \( g \) defined as follows:

- \( f(x) = |x| \) (the absolute value of \( x \))
- \( g(x) = [x] \) (the greatest integer less than or equal to \( x \))

We need to find the set of \( x \in \mathbb{R} \) such that:

\[ g(f(x)) \leq f(g(x)) \]

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

First, we compute \( g(f(x)) \):

\[ g(f(x)) = g(|x|) = [|x|] \]

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

Next, we compute \( f(g(x)) \):

\[ f(g(x)) = f([x]) = |[x]| \]

### Step 3: Set up the inequality

Now we need to set up the inequality:

\[ [g(f(x))] \leq [f(g(x))] \]

This translates to:

\[ [|x|] \leq ||[x]| \]

### Step 4: Analyze the cases for \( x \)

We will analyze two cases based on the sign of \( x \).

#### Case 1: \( x \geq 0 \)

If \( x \) is non-negative, then:

- \( |x| = x \)
- \( [|x|] = [x] \)

Thus, the inequality becomes:

\[ [x] \leq |[x]| \]

Since \( [x] \) is the greatest integer less than or equal to \( x \) and is non-negative for \( x \geq 0 \), we have:

\[ [x] \leq [x] \]

This is always true.

#### Case 2: \( x < 0 \)

If \( x \) is negative, then:

- \( |x| = -x \)
- \( [|x|] = [-x] \)

The inequality now becomes:

\[ [-x] \leq |[x]| \]

For \( x < 0 \), \( [x] \) is negative, so \( |[x]| = -[x] \). Therefore, the inequality can be rewritten as:

\[ [-x] \leq -[x] \]

This simplifies to:

\[ -x \leq -[x] \]

Multiplying through by -1 (which reverses the inequality):

\[ x \geq [x] \]

Since \( [x] \) is the greatest integer less than or equal to \( x \), this inequality is always true for \( x < 0 \).

### Conclusion

Since both cases show that the inequality holds for all \( x \in \mathbb{R} \), we conclude that:

\[ \{ x \in \mathbb{R} : g(f(x)) \leq f(g(x)) \} = \mathbb{R} \]

### Final Answer

The solution set is \( \mathbb{R} \).