If `f : R to R`, `f(x)=x^(2)` and `g(x)=2x+1`, then

Options

A. `fog=gof`
B. `fog ne gof`
C. `fof=gog`
D. none of these

Answer

To solve the problem, we need to analyze the functions \( f \) and \( g \) given by:

- \( f(x) = x^2 \)
- \( g(x) = 2x + 1 \)

We will compute \( f(g(x)) \) and \( g(f(x)) \) and check their equality, as well as compute \( f(f(x)) \) and \( g(g(x)) \).

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

To find \( f(g(x)) \), we substitute \( g(x) \) into \( f(x) \):

\[ f(g(x)) = f(2x + 1) \]

Now, substitute \( 2x + 1 \) into \( f(x) \):

\[ f(2x + 1) = (2x + 1)^2 \]

Expanding this, we get:

\[ (2x + 1)^2 = 4x^2 + 4x + 1 \]

So,

\[ f(g(x)) = 4x^2 + 4x + 1 \]

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

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

\[ g(f(x)) = g(x^2) \]

Now, substitute \( x^2 \) into \( g(x) \):

\[ g(x^2) = 2(x^2) + 1 = 2x^2 + 1 \]

### Step 3: Compare \( f(g(x)) \) and \( g(f(x)) \)

Now we compare the two results obtained:

- \( f(g(x)) = 4x^2 + 4x + 1 \)
- \( g(f(x)) = 2x^2 + 1 \)

Clearly, \( f(g(x)) \) is not equal to \( g(f(x)) \) since:

\[ 4x^2 + 4x + 1 \neq 2x^2 + 1 \]

### Step 4: Compute \( f(f(x)) \)

Now we compute \( f(f(x)) \):

\[ f(f(x)) = f(x^2) \]

Substituting \( x^2 \) into \( f(x) \):

\[ f(x^2) = (x^2)^2 = x^4 \]

### Step 5: Compute \( g(g(x)) \)

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

\[ g(g(x)) = g(2x + 1) \]

Substituting \( 2x + 1 \) into \( g(x) \):

\[ g(2x + 1) = 2(2x + 1) + 1 = 4x + 2 + 1 = 4x + 3 \]

### Step 6: Compare \( f(f(x)) \) and \( g(g(x)) \)

Now we compare \( f(f(x)) \) and \( g(g(x)) \):

- \( f(f(x)) = x^4 \)
- \( g(g(x)) = 4x + 3 \)

Clearly, \( f(f(x)) \) is not equal to \( g(g(x)) \) since:

\[ x^4 \neq 4x + 3 \]

### Conclusion

From our calculations, we conclude that:

- \( f(g(x)) \neq g(f(x)) \)
- \( f(f(x)) \neq g(g(x)) \)

Thus, the correct option is that \( f(g) \) is not equal to \( g(f) \).

### Final Answer

The final answer is: **f(g) is not equal to g(f)**.

---