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

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

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

Step 2: Compare.
\[ (1+x^2)(1+y^2) = 1 + x^2 + y^2 + x^2y^2 \ge 1 + x^2y^2 \]

Step 3: Conclusion.
\[ f(xy) \le f(x)f(y) \]