If f: R `rarr` R be defined by `f(x) =e^(x)` and g:R `rarr` R be defined by g(x) =`x^(2)` the mapping gof : R `rarr` R be defined by (gof)(x) =g[f(x)] `forall x in R` then

Options

A. g o f is bijective but g is not injective
B. g o f is injective and g is injective
C. g o f is injective but g is not bijective
D. gof is surjective and g is surjective

Answer

To solve the problem, we need to analyze the functions \( f \) and \( g \) and then find the composition \( g \circ f \).

### Step 1: Define the functions
The functions are defined as follows:
- \( f: \mathbb{R} \to \mathbb{R} \) given by \( f(x) = e^x \)
- \( g: \mathbb{R} \to \mathbb{R} \) given by \( g(x) = x^2 \)

### Step 2: Find the composition \( g \circ f \)
The composition of the functions \( g \) and \( f \) is defined as:
\[ (g \circ f)(x) = g(f(x)) \]
Substituting \( f(x) \) into \( g \):
\[ (g \circ f)(x) = g(e^x) = (e^x)^2 \]
Thus, we can simplify this to:
\[ (g \circ f)(x) = e^{2x} \]

### Step 3: Analyze the properties of \( g \) and \( g \circ f \)
1. **Injectivity of \( g \)**:
- The function \( g(x) = x^2 \) is not injective because it maps both \( x \) and \( -x \) to the same value (e.g., \( g(2) = 4 \) and \( g(-2) = 4 \)).
- Therefore, \( g \) is not one-to-one.

2. **Surjectivity of \( g \)**:
- The function \( g(x) = x^2 \) is not surjective when considering the codomain \( \mathbb{R} \) because it only produces non-negative outputs (i.e., \( [0, \infty) \)).
- Hence, \( g \) is not onto.

3. **Injectivity of \( g \circ f \)**:
- The function \( g \circ f(x) = e^{2x} \) is injective because the exponential function is strictly increasing. Therefore, for any \( x_1 \neq x_2 \), we have \( e^{2x_1} \neq e^{2x_2} \).

4. **Surjectivity of \( g \circ f \)**:
- The function \( g \circ f(x) = e^{2x} \) is not surjective when considering the codomain \( \mathbb{R} \) because it only produces positive outputs (i.e., \( (0, \infty) \)).
- Thus, \( g \circ f \) is not onto.

### Conclusion
- \( g \) is neither injective nor surjective.
- \( g \circ f \) is injective but not surjective.
- Therefore, \( g \circ f \) is not bijective.

### Final Answer
The mapping \( g \circ f \) is injective but not surjective, hence not bijective.