let `f(x)=sqrt(1+x^2)` then

Options

A. `f(xy)=f(x). f(y)`
B. `f(xy) ge f(x) . f(y)`
C. `f(xy) le f(x) . f(y)`
D. None of these

Answer

To solve the problem, we need to find the relationship between \( f(xy) \) and \( f(x) \cdot f(y) \), where \( f(x) = \sqrt{1 + x^2} \).

### Step 1: Calculate \( f(xy) \)

We start by substituting \( xy \) into the function \( f \):

\[ f(xy) = \sqrt{1 + (xy)^2} = \sqrt{1 + x^2y^2} \]

### Step 2: Calculate \( f(x) \cdot f(y) \)

Next, we calculate the product \( f(x) \cdot f(y) \):

\[ f(x) = \sqrt{1 + x^2} \]
\[ f(y) = \sqrt{1 + y^2} \]

Now, we multiply these two expressions:

\[ f(x) \cdot f(y) = \sqrt{1 + x^2} \cdot \sqrt{1 + y^2} = \sqrt{(1 + x^2)(1 + y^2)} \]

Expanding this product gives us:

\[ f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2} \]

### Step 3: Compare \( f(xy) \) and \( f(x) \cdot f(y) \)

Now we compare \( f(xy) \) and \( f(x) \cdot f(y) \):

- We have \( f(xy) = \sqrt{1 + x^2y^2} \)
- And \( f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2} \)

### Step 4: Establish the inequality

To compare these two expressions, we note that:

\[ 1 + x^2 + y^2 + x^2y^2 \geq 1 + x^2y^2 \]

This is true because \( x^2 \) and \( y^2 \) are both non-negative (since they are squares). Therefore, we can conclude that:

\[ f(x) \cdot f(y) \geq f(xy) \]

### Conclusion

Thus, we can write the final relation as:

\[ f(xy) \leq f(x) \cdot f(y) \]

### Final Answer

The correct relationship is:

\[ f(xy) \leq f(x) \cdot f(y) \]