Question

Let \( f(x) = \sqrt{1 + x^2} \), then

• A. \( f(xy) = f(x)\cdot f(y) \)
• B. \( f(xy) \ge f(x)\cdot f(y) \)
• C. \( f(xy) \le f(x)\cdot f(y) \)
• D. \( f(xy) = f(x) - f(y) \)

Hint:

Expand products to compare expressions under square roots.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To determine the relationship between \( f(xy) \) and the product \( f(x) \cdot f(y) \), we evaluate both expressions using the given definition of the function \( f(x) = \sqrt{1 + x^2} \).
Step 2: Detailed Explanation:
Given \( f(x) = \sqrt{1 + x^2} \).
The product \( f(x) \cdot f(y) \) is:
\[ f(x) \cdot f(y) = \sqrt{1 + x^2} \cdot \sqrt{1 + y^2} = \sqrt{(1 + x^2)(1 + y^2)} \]
\[ f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2} \]
Now, let's evaluate \( f(xy) \):
\[ f(xy) = \sqrt{1 + (xy)^2} = \sqrt{1 + x^2y^2} \]
We compare the terms under the square root. Since \( x^2 \) and \( y^2 \) are always non-negative (\( x^2 \geq 0, y^2 \geq 0 \)), we have:
\[ 1 + x^2y^2 \leq 1 + x^2 + y^2 + x^2y^2 \]
Taking the square root on both sides:
\[ \sqrt{1 + x^2y^2} \leq \sqrt{1 + x^2 + y^2 + x^2y^2} \]
\[ f(xy) \leq f(x) \cdot f(y) \]
Step 3: Final Answer:
The correct relationship is \( f(xy) \leq f(x) \cdot f(y) \).