MCQ 11 Mark
Assertion $(A):$ The function $f : R ^* \rightarrow R ^*$ defined by $f(x)=\frac{1}{x}$ is one$-$one and onto, where $R*$ is the set of all non$-$zero real numbers.
Reason $(R):$ The function $g : N \rightarrow R ^*$ defined by $f(x)=\frac{1}{x}$ is one$-$one and onto.
Reason $(R):$ The function $g : N \rightarrow R ^*$ defined by $f(x)=\frac{1}{x}$ is one$-$one and onto.
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- ✓$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer
View full question & answer→Correct option: C.
$A$ is true but $R$ is false.
Assertion: It is given that$f: R^* \rightarrow R^*$ is defined by
$f(x)=\frac{1}{x}$
For one$-$one, $f(x)=f(y)$
$\Rightarrow \frac{1}{x}=\frac{1}{y}$
$\Rightarrow x=y$
Therefore, $f$ is one$-$one.
For onto, it is clear that for $y \in R^*$,
there exists $x=\frac{1}{y} \in R^* ($exists as $y \neq 0 )$ such that $f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$
Therefore, $f$ is onto.
Thus, the given function $( f )$ is one$-$one and onto.
Reason: Now, consider function $g : N \rightarrow R ^*$ defined by $g(x)=\frac{1}{x}$.
We have, $g\left(x_1\right)=g\left(x_2\right) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$
$f(x)=\frac{1}{x}$
For one$-$one, $f(x)=f(y)$
$\Rightarrow \frac{1}{x}=\frac{1}{y}$
$\Rightarrow x=y$
Therefore, $f$ is one$-$one.
For onto, it is clear that for $y \in R^*$,
there exists $x=\frac{1}{y} \in R^* ($exists as $y \neq 0 )$ such that $f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$
Therefore, $f$ is onto.
Thus, the given function $( f )$ is one$-$one and onto.
Reason: Now, consider function $g : N \rightarrow R ^*$ defined by $g(x)=\frac{1}{x}$.
We have, $g\left(x_1\right)=g\left(x_2\right) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$