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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 1 · Set 1 · Q26
(a) A relation R on set A ={1,2,3,4,5} is defined as R ={( x, y ):| x^2- y^2|<8}. Check whether the relation R is reflexive, symmetric and transitive.
A relation R on set $\displaystyle \mathrm{A}=\{1,2,3,4,5\}$ is defined as $\displaystyle \mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}):\left|\mathrm{x}^{2}-\mathrm{y}^{2}\right|<8\right\}$. Check whether the relation R is reflexive, symmetric and transitive.
A function $\displaystyle \mathrm{f}$ is defined from $\displaystyle \mathrm{R} \rightarrow \mathrm{R}$ as $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{a} \mathrm{x}+\mathrm{b}$, such that $\displaystyle \mathrm{f}(1)=1$ and $\displaystyle \mathrm{f}(2)=3$. Find function $\displaystyle \mathrm{f}(\mathrm{x})$. Hence, check whether function $\displaystyle \mathrm{f}(\mathrm{x})$ is one-one and onto or not.
Marking-scheme solution
(a)
Reflexive:
$\displaystyle \because\left|\mathrm{x}^{2}-\mathrm{x}^{2}\right|<\mathbf{8} \forall \mathrm{x} \in \mathrm{A} \Rightarrow(\mathrm{x}, \mathrm{x}) \in \mathrm{R} \quad \therefore \mathrm{R}$ is reflexive.
(b)
Symmetric:
Let $\displaystyle (\mathrm{x}, \mathrm{y}) \in \mathrm{R}$ for some $\displaystyle \mathrm{x}, \mathrm{y} \in \mathrm{A}$
$$\therefore\left|\mathrm{x}^{$\displaystyle 2$}-\mathrm{y}^{$\displaystyle 2$}\right|<$\displaystyle 8$ \Rightarrow\left|\mathrm{y}^{$\displaystyle 2$}-\mathrm{x}^{$\displaystyle 2$}\right|<$\displaystyle 8$ \Rightarrow(\mathrm{y}, \mathrm{x}) \in \mathrm{R}
$$Hence $\displaystyle \mathbf{R}$ is symmetric.
(c)
Transitive:
$\displaystyle (1,2),(2,3) \in \mathrm{R}$ as $\displaystyle \left|1^{2}-2^{2}\right|<8,\left|2^{2}-3^{2}\right|<8$ respectively
But $\displaystyle \left|1^{2}-3^{2}\right| \nless 8 \Rightarrow(1,3) \notin \mathrm{R}$
Hence $\displaystyle \mathrm{R}$ is not transitive.
$\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$
Solving $\displaystyle \mathrm{a}+\mathrm{b}=1$ and $\displaystyle 2 \mathrm{a}+\mathrm{b}=3$ to get $\displaystyle \mathrm{a}=2, \mathrm{~b}=-1$
$$\mathrm{f}(\mathrm{x})=$\displaystyle 2$ \mathrm{x}-$\displaystyle 1$
$$Let $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)$ for some $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}$
$$$2$\displaystyle \mathrm{x}_{$1$\displaystyle }-$1$\displaystyle =$2$\displaystyle \mathrm{x}_{$2$\displaystyle }-$1$\displaystyle \Rightarrow \mathrm{x}_{$1$\displaystyle }=\mathrm{x}_{$2$}
$$Hence f is one - one.
Let $\displaystyle \mathrm{y}=2 \mathrm{x}-1, \mathrm{y} \in \mathrm{R}$ (Codomain)
$\displaystyle \Rightarrow \mathrm{x}=\frac{\mathrm{y}+1}{2} \in \mathrm{R}$ (domain)
Also, $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{f}\left(\frac{\mathrm{y}+1}{2}\right)=\mathrm{y}$
∴ f is onto.
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