Let f(x)= 1+x^2 then. - Vidyadip

MCQ

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

A
$\text{f(xy)} = \text{f(x)}.\text{f(y)}$
B
$\text{f(xy)} \geq \text{f(x)}.\text{f(y)}$
✓
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
D
None of these

✓

Answer

Correct option: C.

$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

Given that: $\text{f(x)}=\sqrt{1+\text{x}^2}$
$\Rightarrow\text{f(xy)}=\sqrt{1+\text{x}^2\text{y}^2}$
and $\text{f(x)}.\text{f(y)}=\sqrt{1+\text{x}^2}.\sqrt{1+\text{x}^2}=\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$
$\therefore\sqrt{1+\text{x}^2\text{y}^2}\leq\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

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