Let f be a function on R given by f(x) = `x^(2)` and let E = `{x in R: - 1 le x le 0 } and F = { x in R , 0 le x le 1 }`
then which of the following is fase ?

Options

A. f(E) = f(F)
B. E `cap F sup f (E) cap f (F)`
C. `E cup F sup f(E) cup f (F)`.
D. ` f (E cap F) = {0}`

Answer

To solve the problem, we need to analyze the function \( f(x) = x^2 \) and the sets \( E \) and \( F \).

### Step 1: Define the Sets
- The set \( E \) is defined as \( E = \{ x \in \mathbb{R} : -1 \leq x \leq 0 \} \).
- The set \( F \) is defined as \( F = \{ x \in \mathbb{R} : 0 \leq x \leq 1 \} \).

### Step 2: Find \( f(E) \)
To find \( f(E) \), we need to apply the function \( f(x) = x^2 \) to every element in the set \( E \).

- For \( x \in E \), \( x \) ranges from -1 to 0.
- The square of any number in this range will be non-negative.
- The minimum value occurs at \( x = -1 \) (which gives \( f(-1) = 1 \)) and the maximum value occurs at \( x = 0 \) (which gives \( f(0) = 0 \)).
- Therefore, \( f(E) = \{ x^2 : -1 \leq x \leq 0 \} = [0, 1] \).

### Step 3: Find \( f(F) \)
Next, we find \( f(F) \).

- For \( x \in F \), \( x \) ranges from 0 to 1.
- The square of any number in this range will also be non-negative.
- The minimum value occurs at \( x = 0 \) (which gives \( f(0) = 0 \)) and the maximum value occurs at \( x = 1 \) (which gives \( f(1) = 1 \)).
- Therefore, \( f(F) = \{ x^2 : 0 \leq x \leq 1 \} = [0, 1] \).

### Step 4: Compare \( f(E) \) and \( f(F) \)
From the previous steps, we have:
- \( f(E) = [0, 1] \)
- \( f(F) = [0, 1] \)

Since \( f(E) = f(F) \), this statement is true.

### Step 5: Find \( E \cap F \)
Now, we find the intersection of sets \( E \) and \( F \).

- The set \( E \) contains values from -1 to 0.
- The set \( F \) contains values from 0 to 1.
- The only common element is \( 0 \).
- Therefore, \( E \cap F = \{ 0 \} \).

### Step 6: Find \( f(E \cap F) \)
Now we find \( f(E \cap F) \).

- Since \( E \cap F = \{ 0 \} \), we have \( f(E \cap F) = f(0) = 0^2 = 0 \).

### Step 7: Find \( f(E) \cap f(F) \)
Now we find the intersection of \( f(E) \) and \( f(F) \).

- Since both \( f(E) \) and \( f(F) \) are \( [0, 1] \), their intersection is also \( [0, 1] \).
- Therefore, \( f(E) \cap f(F) = [0, 1] \).

### Step 8: Determine if \( f(E \cap F) \subseteq f(E) \cap f(F) \)
We need to check if \( f(E \cap F) = \{ 0 \} \) is a subset of \( f(E) \cap f(F) = [0, 1] \).

- Since \( 0 \in [0, 1] \), this statement is true.

### Conclusion
Now we can summarize the findings:
1. \( f(E) = f(F) \) is true.
2. \( E \cap F = \{ 0 \} \) and \( f(E \cap F) = \{ 0 \} \) is true.
3. \( f(E) \cap f(F) = [0, 1] \) is true.

Since all statements are true, the only false statement is:
- **B**: \( f(E) \cap f(F) \) is not a subset of \( f(E \cap F) \).

### Final Answer
The false statement is **B**.