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Relations and Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsRelations and Functions1 Mark
MCQ
Let $\text{f(x)}=\sqrt{\text{x}^2+1}$ Then which of the following is correct?
  • 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 os these.
✓

Answer

Correct option: C.
$\text{f(xy)}\leq\text{f(x)}\text{f(y)}$
Given, $\text{f(x)}=\sqrt{\text{x}^2+1}\ ...(\text{i})$
Replacing $x$ by $y$ in $(i)$, we get
$\text{f(y)}=\sqrt{\text{y}^2+1}$
$\therefore\ \text{f(x)}\text{f(y)}=\sqrt{\text{x}^2+1}\sqrt{\text{y}^2+1}$
$=\sqrt{(\text{x}^2+1)(\text{y}^2+1)}$
$=\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
Also, replacing $x$ by $xy$ in $(i),$ we get
$\text{f(xy)}=\sqrt{\text{x}^2\text{y}^2+1}$
Now,
$\text{x}^2\text{y}^2+1\leq\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1$
$\Rightarrow\sqrt{\text{x}^2\text{y}^2+1}\leq\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
$\Rightarrow\text{f}(\text{xy})\leq\text{f(x)}\text{f(y)}$

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