CBSE Class 12 Mathematics: Relations and Functions NCERT Solutions

The given set is $A\; =\; \backslash \{1,\; 2,\; 3,\; \backslash dots,\; 13,\; 14\backslash \}$ and the relation is $R\; =\; \backslash \{(x,\; y):\; 3x\; -\; y\; =\; 0\backslash \}$. This can be rewritten as $y\; =\; 3x$. Let's find the pairs in R:

• If $x\; =\; 1$, $y\; =\; 3(1)\; =\; 3$. So, $(1,\; 3)\; \backslash in\; R$.
• If $x\; =\; 2$, $y\; =\; 3(2)\; =\; 6$. So, $(2,\; 6)\; \backslash in\; R$.
• If $x\; =\; 3$, $y\; =\; 3(3)\; =\; 9$. So, $(3,\; 9)\; \backslash in\; R$.
• If $x\; =\; 4$, $y\; =\; 3(4)\; =\; 12$. So, $(4,\; 12)\; \backslash in\; R$.
• If $x\; =\; 5$, $y\; =\; 3(5)\; =\; 15$. Since $15\; otin\; A$, we stop here.

Thus, R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}.

Reflexive: For R to be reflexive, (x, x) \in R for all x \in A. However, pairs like (1, 1), (2, 2), ..., (14, 14) are not in R (e.g., for x=1, 3(1) - 1 = 2

eq 0). Therefore, R is not reflexive.

Symmetric: For R to be symmetric, if (x, y) \in R, then (y, x) \in R. We see that (1, 3) \in R, but (3, 1)

otin R because 3(3) - 1 = 9 - 1 = 8

eq 0. Therefore, R is not symmetric.

Transitive: For R to be transitive, if (x, y) \in R and (y, z) \in R, then (x, z) \in R. We have (1, 3) \in R (since 3(1) - 3 = 0) and (3, 9) \in R (since 3(3) - 9 = 0). However, (1, 9)

otin R because 3(1) - 9 = 3 - 9 = -6

eq 0. Therefore, R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

The set is (natural numbers). The relation is . The condition means $x$ can be 1, 2, or 3.

• If $x\; =\; 1$, $y\; =\; 1\; +\; 5\; =\; 6$. So, $(1,\; 6)\; \backslash in\; R$.
• If $x\; =\; 2$, $y\; =\; 2\; +\; 5\; =\; 7$. So, $(2,\; 7)\; \backslash in\; R$.
• If $x\; =\; 3$, $y\; =\; 3\; +\; 5\; =\; 8$. So, $(3,\; 8)\; \backslash in\; R$.

Thus, R = \{(1, 6), (2, 7), (3, 8)\}.

Reflexive: For R to be reflexive, (x, x) \in R for all x \in {f N}. However, (1, 1), (2, 2), etc., are not in R. For example, (1, 1) is not in R because y = x + 5 is not satisfied (1

eq 1 + 5). Therefore, R is not reflexive.

Symmetric: For R to be symmetric, if (x, y) \in R, then (y, x) \in R. We see that (1, 6) \in R. However, (6, 1)

otin R because the condition x < 4 is not met for x = 6 (and also 1

eq 6 + 5). Therefore, R is not symmetric.

Transitive: For R to be transitive, if (x, y) \in R and (y, z) \in R, then (x, z) \in R. In the set R = {(1, 6), (2, 7), (3, 8)}, there are no pairs (x, y) and (y, z) such that the first element of the second pair is the second element of the first pair. For instance, we cannot find a pair starting with 6, 7, or 8. Since the condition for transitivity requires checking pairs like (x, y) and (y, z), and no such chain exists within the given pairs, the condition is vacuously true. However, the standard interpretation in such cases where no such chain can be formed is that the relation is transitive. Let's re-examine the definition. If we take (x, y) \in R and (y, z) \in R, then y = x + 5 and z = y + 5. This implies z = (x + 5) + 5 = x + 10. For (x, z) to be in R, we need z = x + 5 and x < 4. Since z = x + 10, the condition z = x + 5 is not met. Therefore, R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

The set is A = \{1, 2, 3, 4, 5, 6\} and the relation is R = \{(x, y): y \text{ is divisible by } x\}.

Reflexive: For R to be reflexive, (x, x) \in R for all x \in A. This means x must be divisible by x. Since every integer is divisible by itself, (x, x) \in R for all x \in A. For example, (1, 1), (2, 2), ..., (6, 6) are all in R. Therefore, R is reflexive.

Symmetric: For R to be symmetric, if (x, y) \in R, then (y, x) \in R. Let's take an example. (2, 4) \in R because 4 is divisible by 2. However, (4, 2)

otin R because 2 is not divisible by 4. Therefore, R is not symmetric.

Transitive: For R to be transitive, if (x, y) \in R and (y, z) \in R, then (x, z) \in R. This means if y is divisible by x, and z is divisible by y, then z must be divisible by x. Let y = kx for some integer k, and z = ly for some integer l. Substituting the first equation into the second, we get z = l(kx) = (lk)x. Since lk is an integer, z is divisible by x. Thus, (x, z) \in R. For example, (2, 4) \in R (4 is divisible by 2) and (4, 8) \in R (8 is divisible by 4). Then (2, 8) \in R (8 is divisible by 2). Note that 8 is not in the set A, but the property holds for any integers x, y, z. If we restrict to A, consider (2, 4) \in R and (4, 8) is not in R. Let's take (2, 4) \in R and (4, 4) \in R. Then (2, 4) \in R, which is true. Let's take (3, 6) \in R and (6, ?). No pair starts with 6. Let's take (2, 6) \in R and (6, ?). No pair starts with 6. Let's take (1, 2) \in R and (2, 4) \in R. Then (1, 4) \in R (4 is divisible by 1). This holds. Therefore, R is transitive.

Conclusion: R is reflexive and transitive, but not symmetric.

The set is $\backslash mathbf\{Z\}$ (all integers) and the relation is $R\; =\; \backslash \{(x,\; y):\; x\; -\; y\; \backslash text\{\; is\; an\; integer\}\backslash \}$.

Reflexive: For R to be reflexive, $(x,\; x)\; \backslash in\; R$ for all $x\; \backslash in\; \backslash mathbf\{Z\}$. This means $x\; -\; x$ must be an integer. Since $x\; -\; x\; =\; 0$, and 0 is an integer, the condition is satisfied for all $x\; \backslash in\; \backslash mathbf\{Z\}$. Therefore, R is reflexive.

Symmetric: For R to be symmetric, if $(x,\; y)\; \backslash in\; R$, then $(y,\; x)\; \backslash in\; R$. If $(x,\; y)\; \backslash in\; R$, then $x\; -\; y$ is an integer. Let $x\; -\; y\; =\; k$, where $k\; \backslash in\; \backslash mathbf\{Z\}$. Then $y\; -\; x\; =\; -(x\; -\; y)\; =\; -k$. Since $k$ is an integer, $-k$ is also an integer. Thus, $y\; -\; x$ is an integer, which means $(y,\; x)\; \backslash in\; R$. Therefore, R is symmetric.

Transitive: For R to be transitive, if $(x,\; y)\; \backslash in\; R$ and $(y,\; z)\; \backslash in\; R$, then $(x,\; z)\; \backslash in\; R$. If $(x,\; y)\; \backslash in\; R$, then $x\; -\; y$ is an integer. If $(y,\; z)\; \backslash in\; R$, then $y\; -\; z$ is an integer. Let $x\; -\; y\; =\; k\_1$ and $y\; -\; z\; =\; k\_2$, where $k\_1,\; k\_2\; \backslash in\; \backslash mathbf\{Z\}$. We need to check if $x\; -\; z$ is an integer. We can write $x\; -\; z\; =\; (x\; -\; y)\; +\; (y\; -\; z)\; =\; k\_1\; +\; k\_2$. Since the sum of two integers is always an integer, $k\_1\; +\; k\_2$ is an integer. Thus, $x\; -\; z$ is an integer, which means $(x,\; z)\; \backslash in\; R$. Therefore, R is transitive.

Conclusion: R is reflexive, symmetric, and transitive. (This means R is an equivalence relation).

Let A be the set of human beings in a town at a particular time.

1. Relation: $R\; =\; \backslash \{(x,\; y):\; x\; \backslash text\{\; and\; \}\; y\; \backslash text\{\; work\; at\; the\; same\; place\}\backslash \}$

• Reflexive: Since every person works at their own place, $(x,\; x)\; \backslash in\; R$ for all $x\; \backslash in\; A$. Thus, R is reflexive.
• Symmetric: If $x$ works at the same place as $y$, then $y$ also works at the same place as $x$. So, if $(x,\; y)\; \backslash in\; R$, then $(y,\; x)\; \backslash in\; R$. Thus, R is symmetric.
• Transitive: If $x$ works at the same place as $y$, and $y$ works at the same place as $z$, then $x$, $y$, and $z$ all work at the same place. Therefore, $x$ works at the same place as $z$. So, if $(x,\; y)\; \backslash in\; R$ and $(y,\; z)\; \backslash in\; R$, then $(x,\; z)\; \backslash in\; R$. Thus, R is transitive.

Conclusion: R is reflexive, symmetric, and transitive (an equivalence relation).

1. Relation: $R\; =\; \backslash \{(x,\; y):\; x\; \backslash text\{\; and\; \}\; y\; \backslash text\{\; live\; in\; the\; same\; locality\}\backslash \}$

• Reflexive: Every person lives in their own locality. So, $(x,\; x)\; \backslash in\; R$ for all $x\; \backslash in\; A$. Thus, R is reflexive.
• Symmetric: If $x$ lives in the same locality as $y$, then $y$ lives in the same locality as $x$. So, if $(x,\; y)\; \backslash in\; R$, then $(y,\; x)\; \backslash in\; R$. Thus, R is symmetric.
• Transitive: If $x$ lives in the same locality as $y$, and $y$ lives in the same locality as $z$, then $x$, $y$, and $z$ all live in the same locality. Therefore, $x$ lives in the same locality as $z$. So, if $(x,\; y)\; \backslash in\; R$ and $(y,\; z)\; \backslash in\; R$, then $(x,\; z)\; \backslash in\; R$. Thus, R is transitive.

Conclusion: R is reflexive, symmetric, and transitive (an equivalence relation).

1. Relation: $R\; =\; \backslash \{(x,\; y):\; x\; \backslash text\{\; is\; exactly\; 7\; cm\; taller\; than\; \}\; y\backslash \}$

• Reflexive: A person cannot be exactly 7 cm taller than themselves. So, $(x,\; x)\; otin\; R$ for any $x\; \backslash in\; A$. Thus, R is not reflexive.
• Symmetric: If $x$ is exactly 7 cm taller than $y$, then $y$ is exactly 7 cm shorter than $x$. $y$ cannot be taller than $x$. So, if $(x,\; y)\; \backslash in\; R$, then $(y,\; x)\; otin\; R$. Thus, R is not symmetric.
• Transitive: If $x$ is exactly 7 cm taller than $y$ (i.e., $x\; =\; y\; +\; 7$), and $y$ is exactly 7 cm taller than $z$ (i.e., $y\; =\; z\; +\; 7$), then $x\; =\; (z\; +\; 7)\; +\; 7\; =\; z\; +\; 14$. This means $x$ is 14 cm taller than $z$, not exactly 7 cm taller. So, if $(x,\; y)\; \backslash in\; R$ and $(y,\; z)\; \backslash in\; R$, then $(x,\; z)\; otin\; R$. Thus, R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

1. Relation: $R\; =\; \backslash \{(x,\; y):\; x\; \backslash text\{\; is\; wife\; of\; \}\; y\backslash \}$

• Reflexive: A person cannot be their own wife. So, $(x,\; x)\; otin\; R$ for any $x\; \backslash in\; A$. Thus, R is not reflexive.
• Symmetric: If $x$ is the wife of $y$, then $y$ must be the husband of $x$. The relation is defined as 'wife of', not 'husband of'. So, if $(x,\; y)\; \backslash in\; R$, then $(y,\; x)\; otin\; R$ (unless $y$ is also a wife, which is not possible in a standard definition). Thus, R is not symmetric.
• Transitive: If $x$ is the wife of $y$, and $y$ is the wife of $z$, this implies $y$ is married to $x$ and $y$ is married to $z$. This is impossible in a monogamous society. If we consider polygamy, $x$ is a wife of $y$, and $y$ is a wife of $z$ implies $y$ is married to two people. If $x$ is wife of $y$, and $y$ is wife of $z$, then $x$ cannot be the wife of $z$. Thus, R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

1. Relation: $R\; =\; \backslash \{(x,\; y):\; x\; \backslash text\{\; is\; father\; of\; \}\; y\backslash \}$

• Reflexive: A person cannot be their own father. So, $(x,\; x)\; otin\; R$ for any $x\; \backslash in\; A$. Thus, R is not reflexive.
• Symmetric: If $x$ is the father of $y$, then $y$ is the son or daughter of $x$. $y$ cannot be the father of $x$. So, if $(x,\; y)\; \backslash in\; R$, then $(y,\; x)\; otin\; R$. Thus, R is not symmetric.
• Transitive: If $x$ is the father of $y$, and $y$ is the father of $z$, then $x$ is the grandfather of $z$. $x$ is not the father of $z$. So, if $(x,\; y)\; \backslash in\; R$ and $(y,\; z)\; \backslash in\; R$, then $(x,\; z)\; otin\; R$. Thus, R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.