Question

Let f(x) = `sqrt(1 + x^2)`, then ______.

Options

• f(xy) = f(x) . f(y)

• f(xy) ≥ f(x) . f(y)

• f(xy) ≤ f(x) . f(y)

• None of these

Answer 1

Let f(x) = `sqrt(1 + x^2)`, then f(xy) ≤ f(x) . f(y).

Explanation:

Given that: f(x) = `sqrt(1 + x^2)`

⇒ f(xy) = `sqrt(1 + x^2y^2)`

And f(x) . f(y) = `sqrt(1 + x^2) * sqrt(1 + y^2)`

= `sqrt(1 + x^2 + y^2 + x^2y^2)`

∵ `sqrt(1 + x^2y^2) ≤ sqrt(1 + x^2 + y^2 + x^2y^2)`

⇒ f(xy) ≤ f(x) . f(y)