Let f: [0, 1] `to` [1, 2] defined as f(x) = 1 + x and g : [1, 2] `to` [0, 1] defined as g(x) = 2 – x then the composite function gof is

Options

A. injective as well as surjective
B. Surjective but not injective
C. Injective but non surjective
D. Neither injective nor surjective

Answer

To find the composite function \( g \circ f \), we need to follow these steps:

### Step 1: Define the functions
The functions are defined as follows:
- \( f: [0, 1] \to [1, 2] \) given by \( f(x) = 1 + x \)
- \( g: [1, 2] \to [0, 1] \) given by \( g(x) = 2 - x \)

### Step 2: Find \( g(f(x)) \)
To find the composite function \( g(f(x)) \), we substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(1 + x) \]
Now, substitute \( 1 + x \) into \( g(x) \):
\[ g(1 + x) = 2 - (1 + x) = 2 - 1 - x = 1 - x \]
Thus, we have:
\[ g \circ f(x) = 1 - x \]

### Step 3: Determine the domain of \( g \circ f \)
The function \( f(x) \) maps \( x \) from the interval \([0, 1]\) to the interval \([1, 2]\). Therefore, the output of \( f(x) \) is valid as input for \( g(x) \), which takes values from \([1, 2]\). Thus, the domain of \( g \circ f \) is:
\[ \text{Domain of } g \circ f = [0, 1] \]

### Step 4: Determine the codomain of \( g \circ f \)
The output of \( g(f(x)) = 1 - x \) will vary as \( x \) varies from \( 0 \) to \( 1 \):
- When \( x = 0 \), \( g(f(0)) = 1 - 0 = 1 \)
- When \( x = 1 \), \( g(f(1)) = 1 - 1 = 0 \)

Thus, the range of \( g \circ f \) is:
\[ \text{Range of } g \circ f = [0, 1] \]
Since the output also lies within \([0, 1]\), the codomain is also \([0, 1]\).

### Step 5: Check if \( g \circ f \) is injective (one-to-one)
To check if \( g \circ f \) is injective, we can analyze the function \( 1 - x \):
- The function \( 1 - x \) is a linear function with a negative slope, which means it is strictly decreasing.
- Since it is strictly decreasing, it is one-to-one.

### Step 6: Check if \( g \circ f \) is surjective (onto)
To check if \( g \circ f \) is surjective, we need to see if every element in the codomain \([0, 1]\) is hit by some \( x \) in the domain \([0, 1]\):
- The minimum value of \( g \circ f \) is \( 0 \) (when \( x = 1 \)).
- The maximum value of \( g \circ f \) is \( 1 \) (when \( x = 0 \)).
- Since every value in \([0, 1]\) can be achieved, the function is onto.

### Conclusion
The composite function \( g \circ f \) is:
\[ g \circ f(x) = 1 - x \]
with domain \([0, 1]\) and codomain \([0, 1]\). The function is both injective and surjective.

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