Let `f:RrarrR` be defined by `f(x)=sinx` and `g:RrarrR` be defined by `g(x)=x^(2)`, then fog is :

Options

A. `x^(2)sinx`
B. `(sinx)^(2)`
C. `sinx^(2)`
D. `(sinx)/(x^(2))`

Answer

To solve the problem, we need to find the composition of the functions \( f \) and \( g \), denoted as \( f \circ g \).

1. **Identify the functions**:
- The function \( f \) is defined as \( f(x) = \sin x \).
- The function \( g \) is defined as \( g(x) = x^2 \).

2. **Composition of functions**:
- The composition \( f \circ g \) means we will substitute \( g(x) \) into \( f(x) \).
- This can be written as \( f(g(x)) \).

3. **Substituting \( g(x) \) into \( f(x) \)**:
- We know \( g(x) = x^2 \).
- Therefore, we need to find \( f(g(x)) = f(x^2) \).

4. **Applying the function \( f \)**:
- Now, we substitute \( x^2 \) into the function \( f \):
\[ f(x^2) = \sin(x^2) \]

5. **Final result**:
- Thus, the composition \( f \circ g \) is:
\[ f \circ g = \sin(x^2) \]

### Final Answer:
\[ f \circ g = \sin(x^2) \]