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Q11:
Let $f$ be the subset of $\mathbb{Z} \times \mathbb{Z}$ defined by $f = \{(ab, a + b) : a, b \in \mathbb{Z}\}$. Is $f$ a function from $\mathbb{Z}$ to $\mathbb{Z}$? Justify your answer.
Solution :
1. Definitions and Criterion for a Function
A relation $f: A \to B$ is classified as a function if and only if it satisfies two fundamental criteria:
- Existence: Every element $x \in A$ must have at least one image in $B$.
- Uniqueness: For every $x \in A$, there must exist exactly one image $y \in B$. That is, if $(x, y_1) \in f$ and $(x, y_2) \in f$, then $y_1 = y_2$.
In the given problem, the relation $f$ is defined as $f = \{(ab, a + b) : a, b \in \mathbb{Z}\}$. Here, the "input" corresponds to the first component $x = ab$ and the "output" corresponds to the second component $y = a + b$.
2. Testing the Uniqueness Criterion
To determine if $f$ is a function, we must test if the same input $x$ can produce different outputs $y$ for different pairs $(a, b)$.
Consider the case where $x = 6$. We search for integer pairs $(a, b)$ such that $ab = 6$:
| Pair $(a, b)$ | Input $x = ab$ | Output $y = a + b$ |
|---|---|---|
| $(2, 3)$ | $2 \times 3 = 6$ | $2 + 3 = 5$ |
| $(1, 6)$ | $1 \times 6 = 6$ | $1 + 6 = 7$ |
3. Logical Analysis
From the table above, we observe the following mapping within the relation $f$:
- $(2, 3) \in f \implies (6, 5) \in f$ [By definition of the set $f$]
- $(1, 6) \in f \implies (6, 7) \in f$ [By definition of the set $f$]
Since the same input $x = 6$ is associated with two distinct outputs $y = 5$ and $y = 7$, the condition of uniqueness is violated. Specifically, $f(6) = 5$ and $f(6) = 7$, which implies $5 \neq 7$.
4. Visual Representation of the Non-Functional Mapping
5. Conclusion
A relation is a function if and only if each input maps to exactly one unique output. Because there exists at least one input ($x=6$) that maps to multiple distinct outputs ($y=5, 7$), the relation fails the vertical line test/uniqueness condition.
Final Solution: No, $f$ is not a function from $\mathbb{Z}$ to $\mathbb{Z}$ because the relation is not well-defined; the same input $ab$ can result in different values of $a+b$, violating the uniqueness requirement of a function.
More Questions from Class 11 Mathematics Relations and Functions Miscellaneous Exercise on Chapter 2
- Q1: The relation $f$ is defined by $f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases}$. The relation $g$ is defined by $g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases}$. Show that $f$ is a function and $g$ is not a function.
- Q2: If $f(x) = x^2$, find $\frac{f(1.1) - f(1)}{1.1 - 1}$.
- Q3: Find the domain of the function $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.
- Q4: Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x - 1}$.
- Q5: Find the domain and the range of the real function $f$ defined by $f(x) = |x – 1|$.
- Q6: Let $f = \{\left(x, \frac{x^2}{1+x^2}\right) : x \in \mathbb{R}\}$ be a function from $\mathbb{R}$ into $\mathbb{R}$. Determine the range of $f$.
- Q7: Let $f, g : \mathbb{R} \to \mathbb{R}$ be defined, respectively by $f(x) = x + 1$, $g(x) = 2x – 3$. Find $f + g$, $f – g$ and $\frac{f}{g}$.
- Q8: Let $f = \{(1,1), (2,3), (0,–1), (–1, –3)\}$ be a function from $\mathbb{Z}$ to $\mathbb{Z}$ defined by $f(x) = ax + b$, for some integers $a, b$. Determine $a, b$.
- Q9(i): Let $R$ be a relation from $\mathbb{N}$ to $\mathbb{N}$ defined by $R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$. Are the following true? (i) $(a,a) \in R$, for all $a \in \mathbb{N}$. Justify your answer in each case.
- Q9(ii): Let $R$ be a relation from $\mathbb{N}$ to $\mathbb{N}$ defined by $R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$. Are the following true? (ii) $(a,b) \in R$, implies $(b,a) \in R$. Justify your answer in each case.
- Q9(iii): Let $R$ be a relation from $\mathbb{N}$ to $\mathbb{N}$ defined by $R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$. Are the following true? (iii) $(a,b) \in R, (b,c) \in R$ implies $(a,c) \in R$. Justify your answer in each case.
- Q10(i): Let $A =\{1,2,3,4\}$, $B = \{1,5,9,11,15,16\}$ and $f = \{(1,5), (2,9), (3,1), (4,5), (2,11)\}$. Are the following true? (i) $f$ is a relation from $A$ to $B$. Justify your answer in each case.
- Q10(ii): Let $A =\{1,2,3,4\}$, $B = \{1,5,9,11,15,16\}$ and $f = \{(1,5), (2,9), (3,1), (4,5), (2,11)\}$. Are the following true? (ii) $f$ is a function from $A$ to $B$. Justify your answer in each case.
- Q12: Let $A = \{9,10,11,12,13\}$ and let $f : A \to \mathbb{N}$ be defined by $f(n) = \text{the highest prime factor of } n$. Find the range of $f$.
CBSE Solutions for Class 11 Mathematics Relations and Functions
Chapters in CBSE - Class 11 Mathematics
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