Question

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

• A. \( f(x+y) = f(x)\cdot f(y) \)
• B. \( f(x+y) \geq f(x)\cdot f(y) \)
• C. \( f(x+y) \leq f(x)\cdot f(y) \)
• D. \( f(x+y) = f(x) - f(y) \)

Hint:

Compare denominators → larger denominator gives smaller value.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to compare the function value at the product \(xy\) with the product of function values at \(x\) and \(y\). We assume \(f(x, y)\) denotes \(f(xy)\) in this context.
Step 3: Detailed Explanation:
1. Calculate the product \(f(x) \cdot f(y)\):
\[ f(x) \cdot f(y) = \frac{1}{\sqrt{1 + x^{2}}} \cdot \frac{1}{\sqrt{1 + y^{2}}} = \frac{1}{\sqrt{(1 + x^{2})(1 + y^{2})}} \]
\[ f(x) \cdot f(y) = \frac{1}{\sqrt{1 + x^{2} + y^{2} + x^{2}y^{2}}} \]
2. Calculate \(f(xy)\):
\[ f(xy) = \frac{1}{\sqrt{1 + (xy)^{2}}} = \frac{1}{\sqrt{1 + x^{2}y^{2}}} \]
3. Compare the radicands (the expressions under the square root):
Since \(x^{2} \geq 0\) and \(y^{2} \geq 0\):
\[ 1 + x^{2}y^{2} \leq 1 + x^{2} + y^{2} + x^{2}y^{2} \]
4. Taking the square root and reciprocal (which reverses the inequality):
\[ \sqrt{1 + x^{2}y^{2}} \leq \sqrt{1 + x^{2} + y^{2} + x^{2}y^{2}} \]
\[ \frac{1}{\sqrt{1 + x^{2}y^{2}}} \geq \frac{1}{\sqrt{1 + x^{2} + y^{2} + x^{2}y^{2}}} \]
\[ f(xy) \geq f(x) \cdot f(y) \]
Step 4: Final Answer:
The correct inequality is \(f(x, y) \geq f(x) \cdot f(y)\).