let `f(x)=sqrt(1+x^2)` then (i) f(xy)=f(x)*f(y)`(ii) f(xy)gef(x)*f(y)`(iii) `f(xy)lef(x)*f(y)`(iv) None of these

Options

A. ` f(xy)=f(x)*f(y)`
B. `f(xy)gef(x)*f(y)`
C. `f(xy)lef(x)*f(y)`
D. None of these

Explanation

To solve the problem, we need to analyze the function \( f(x) = \sqrt{1 + x^2} \) and check the relationships between \( f(xy) \) and \( f(x) \cdot f(y) \).

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function given in the question:
\[ f(x) = \sqrt{1 + x^2} \]

2. **Calculate \( f(xy) \)**:
We need to find \( f(xy) \):
\[ f(xy) = \sqrt{1 + (xy)^2} = \sqrt{1 + x^2y^2} \]

3. **Calculate \( f(x) \cdot f(y) \)**:
Next, we calculate \( f(x) \cdot f(y) \):
\[ f(x) \cdot f(y) = \sqrt{1 + x^2} \cdot \sqrt{1 + y^2} = \sqrt{(1 + x^2)(1 + y^2)} \]
Expanding this, we get:
\[ f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2} \]

4. **Compare \( f(xy) \) and \( f(x) \cdot f(y) \)**:
Now we need to compare \( f(xy) \) and \( f(x) \cdot f(y) \):
\[ f(xy) = \sqrt{1 + x^2y^2} \]
\[ f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2} \]

5. **Establish the Inequality**:
We can see that:
\[ 1 + x^2 + y^2 + x^2y^2 \geq 1 + x^2y^2 \]
This is because \( x^2 + y^2 \geq 0 \) for all real numbers \( x \) and \( y \). Therefore, we can conclude:
\[ \sqrt{1 + x^2y^2} \leq \sqrt{1 + x^2 + y^2 + x^2y^2} \]
which implies:
\[ f(xy) \leq f(x) \cdot f(y) \]

6. **Final Conclusion**:
Thus, we have established that:
\[ f(xy) \leq f(x) \cdot f(y) \]
Therefore, the correct option is:
\[ (iii) \quad f(xy) \leq f(x) \cdot f(y) \]