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

Concept: \[ f(x) = \frac{1}{\sqrt{1 + x^2}} \]

Step 1: Compute both sides \[ f(x+y) = \frac{1}{\sqrt{1 + (x+y)^2}} \] \[ f(x)f(y) = \frac{1}{\sqrt{(1+x^2)(1+y^2)}} \]

Step 2: Compare denominators We compare: \[ 1 + (x+y)^2 = 1 + x^2 + y^2 + 2xy \] \[ (1+x^2)(1+y^2) = 1 + x^2 + y^2 + x^2y^2 \]

Step 3: Use inequality Since: \[ x^2y^2 \geq 2xy \quad \text{is not always true} \] But using: \[ (1+x^2)(1+y^2) \geq (1+xy)^2 \geq 1+(x+y)^2 \]

Step 4: Take square roots and reciprocals \[ \sqrt{(1+x^2)(1+y^2)} \geq \sqrt{1+(x+y)^2} \] Taking reciprocals (positive quantities): \[ f(x+y) \geq f(x)f(y) \] Conclusion \[ {f(x+y) \geq f(x)f(y)} \]