Let f(x) = (\sqrt{1 + x^{2}})then | MATH - FUNCTIONS | SolveeIt | Solveeit

Today, August 24

Question

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

A

f(xy) = f(x). f(y)

B

f(xy) ≥ f(x). f(y)

C

f(xy) ≤ f(x). f(y)

D

None of these

Answer

f(xy) ≤ f(x). f(y)

Explanation

Solution

f(xy) = $\sqrt { 1 + x ^ { 2 } y ^ { 2 } }$

⇒ f(x).f(y) =

=

^> f(xy) < f(x). f(y)