Question

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

• A. Noneof these
• B. $f\left(xy\right)\ge f\left(x\right)\cdot f\left(y\right)$
• C. $f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)$
• D. $f\left(xy\right)\le f\left(x\right)\cdot f\left(y\right)$

Answer 1

The correct option is D $f\left(xy\right)\le f\left(x\right)\cdot f\left(y\right)$

Given: $f\left(x\right)=\sqrt{1+x^2}$
Define $f\left(xy\right)andf\left(y\right)f\left(xy\right)=\sqrt{1+x^2{y}^2}\dots \left(i\right)$
and$f\left(y\right)=\sqrt{1+y^2}$\
Now $f\left(x\right)\cdot f\left(y\right)=\sqrt{1+x^2}\sqrt{1+y^2}f\left(x\right)\cdot f\left(y\right)=\sqrt{1+x^2+y^2+x^2{y}^2}\dots \left(ii\right)$
$(\because \sqrt{a}\cdot \sqrt{b}=\sqrt{ab})$


Since for any value of$x,yϵR1+x^2{y}^2\le 1+x^2+y^2+x^2{y}^2$

$\Rightarrow \sqrt{1+x^2{y}^2}\le \sqrt{1+x^2+y^2+x^2{y}^2}$

$\therefore$ by using (i) and (ii) we get

$f\left(xy\right)\le f\left(x\right)\cdot f\left(y\right)$

Hence, the correct option is (C)