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Relations and Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsRelations and Functions1 Mark
MCQ
Choose the correct answers: 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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