CBSE 2025, Set 3

Question

Let $A=\{1,2,3\}$ and $B=\{4,5,6\}$. A relation $R$ from $A$ to $B$ is defined as $R=\{(x,y):x+y=6, x\in A, y\in B\}$.
(i) Write all elements of $R$.
(ii) Is $R$ a function? Justify.
(iii) Determine domain and range of $R$.

Edvaya · Accepted Answer

The correct answer is Subjective. **Given:** 
 $A = \{1, 2, 3\}$
 $B = \{4, 5, 6\}$
 $R = \{(x,y) : x+y=6, x \in A, y \in B\}$

**(i) Elements of R:**
 We check which combinations of $x \in A$ and $y \in B$ satisfy $x + y = 6$:
 If $x = 1$, then $y = 6 - 1 = 5$. Since $5 \in B$, $(1, 5) \in R$.
 If $x = 2$, then $y = 6 - 2 = 4$. Since $4 \in B$, $(2, 4) \in R$.
 If $x = 3$, then $y = 6 - 3 = 3$. But $3 
otin B$, so there is no pair for $x = 3$.
 Therefore, the elements are:
 $$R = \{(1, 5), (2, 4)\}$$

**(ii) Is R a function?**
 A relation from set $A$ to set $B$ is a function if every element of set $A$ has one and only one image in set $B$.
 In the relation $R$, the element $3 \in A$ does not have any corresponding image in $B$ because there is no ordered pair starting with $3$.
 Therefore, $R$ is **not a function**.

**(iii) Domain and Range of R:**
 The domain is the set of all first elements of the ordered pairs in $R$.
 $$	ext{Domain}(R) = \{1, 2\}$$
 The range is the set of all second elements of the ordered pairs in $R$.
 $$	ext{Range}(R) = \{4, 5\}$$

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