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Relations and Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRelations and Functions1 Mark
MCQ
Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and let $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}.$ Then,
  • ✓
    $S$ defines a function from $A$ to $B.$
  • B
    $S^o$  defines a function from $A$ to $C.$
  • C
    $S^o$ defines a function from $A$ to $B.$
  • D
    $S$ defines a function from $A$ to $C.$
✓

Answer

Correct option: A.
$S$ defines a function from $A$ to $B.$
Given that $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}$
$\text{x}^2+\text{y}^2=1$
$\Rightarrow\ \text{y}^2=1-\text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{1-\text{x}^2}$
$\text{y}\in\text{B}$
Hence$, S$ defines a function from $A$ to $B.$

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