Rbse Solutions Class 12 Maths :Chapter 1 Miscellaneous Exercise

Complete solutions for NCERT Class 12 Maths Chapter 1 Miscellaneous Exercise. Detailed steps for bijection, equivalence relations, and functions like f(x)=x/(1+|x|).

Question 1. Show that the function $f: \mathbb{R} \to \{x \in \mathbb{R}: -1 < x < 1\}$ defined by $f(x) = \frac{x}{1 + |x|}$, $x \in \mathbb{R}$ is one-one and onto function.

Solution:

The co-domain is $(-1, 1)$. We prove $f$ is bijective.

1. One-One (Injectivity):

We check three cases based on the sign of $x$:

• Case 1: $x_1, x_2 \geq 0$ ($|x| = x$).$$f(x_1) = f(x_2) \implies \frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2}$$$$x_1(1 + x_2) = x_2(1 + x_1) \implies x_1 + x_1 x_2 = x_2 + x_1 x_2 \implies x_1 = x_2$$
• Case 2: $x_1, x_2 < 0$ ($|x| = -x$).$$f(x_1) = f(x_2) \implies \frac{x_1}{1 – x_1} = \frac{x_2}{1 – x_2}$$$$x_1(1 – x_2) = x_2(1 – x_1) \implies x_1 – x_1 x_2 = x_2 – x_1 x_2 \implies x_1 = x_2$$
• Case 3: $x_1 \geq 0$ and $x_2 < 0$.In this case, $f(x_1) = \frac{x_1}{1 + x_1} \geq 0$ and $f(x_2) = \frac{x_2}{1 – x_2} < 0$.Since $f(x_1)$ and $f(x_2)$ have different signs, $f(x_1) \neq f(x_2)$.

Thus, $f$ is one-one.

2. Onto (Surjectivity):

Let $y$ be an arbitrary element in the co-domain $(-1, 1)$. We find $x$ such that $f(x) = y$.

• Subcase 1: $y \in [0, 1)$ (Since $y \geq 0$, $x$ must be $\geq 0$, so $|x| = x$).$$y = \frac{x}{1 + x} \implies y(1 + x) = x$$$$y + xy = x \implies y = x(1 – y) \implies x = \frac{y}{1 – y}$$Since $y \in [0, 1)$, $1-y > 0$, so $x \geq 0$ and $x \in \mathbb{R}$.
• Subcase 2: $y \in (-1, 0)$ (Since $y < 0$, $x$ must be $< 0$, so $|x| = -x$).$$y = \frac{x}{1 – x} \implies y(1 – x) = x$$$$y – xy = x \implies y = x(1 + y) \implies x = \frac{y}{1 + y}$$Since $y \in (-1, 0)$, $1+y > 0$, so $x < 0$ and $x \in \mathbb{R}$.

For every $y \in (-1, 1)$, a pre-image $x \in \mathbb{R}$ exists. Thus, $f$ is onto.

Conclusion: $f$ is one-one and onto.

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Question 2. Show that the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ is injective.

Solution:

To show $f$ is injective (one-one), we must prove that $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{R}$.

$$x_1^3 = x_2^3$$

Since the cube root operation is unique for all real numbers:

$$\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3} \implies x_1 = x_2$$

Thus, $f$ is injective.

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Question 3. Given a non empty set $X$, consider $P(X)$ which is the set of all subsets of $X$. Define the relation $R$ in $P(X)$ as follows: For subsets $A, B$ in $P(X)$, $A R B$ if and only if $A \subset B$. Is $R$ an equivalence relation on $P(X)$? Justify your answer.

Solution:

A relation $R$ is an equivalence relation if it is reflexive, symmetric, and transitive.

1. Reflexive: Check if $A \subset A$ for all $A \in P(X)$.
   • Every set is a subset of itself. (Holds)
2. Symmetric: Check if $A \subset B \implies B \subset A$.
   • Let $X = \{1, 2\}$. Take $A = \{1\}$ and $B = \{1, 2\}$.
   • $A \subset B$ is true, but $B \subset A$ is false.
   • $\therefore R$ is not symmetric.
3. Transitive: Check if $A \subset B$ and $B \subset C \implies A \subset C$.
   • If $A$ is a subset of $B$, and $B$ is a subset of $C$, then $A$ must be a subset of $C$ by the definition of set containment. (Holds)

Conclusion: Since $R$ is not symmetric, $R$ is not an equivalence relation on $P(X)$.

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Question 4. Find the number of all onto functions from the set $\{1, 2, 3, \dots, n\}$ to itself.

Solution:

Let $A = \{1, 2, 3, \dots, n\}$ be the domain and $B = \{1, 2, 3, \dots, n\}$ be the co-domain. Since both sets have $n$ elements, the number of elements in the domain equals the number of elements in the co-domain.

For a function $f: A \to B$ where $|A| = |B| = n$, the function is onto (surjective) if and only if it is one-one (injective).

Finding the number of onto functions is equivalent to finding the number of one-to-one functions, which is simply the number of ways to arrange the $n$ elements in the co-domain (the number of permutations of $n$ distinct objects).

Number of onto functions = $n!$.

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Question 5. Let $A = \{-1, 0, 1, 2\}$, $B = \{-4, -2, 0, 2\}$ and $f, g: A \to B$ be functions defined by $f(x) = x^2 – x$, $x \in A$ and $g(x) = \frac{1}{2}x(x-1) – 1$, $x \in A$. Are $f$ and $g$ equal? Justify your answer.

Solution:

Two functions $f$ and $g$ are equal if $f(a) = g(a)$ for all $a$ in the domain $A$.

We must evaluate both $f(x)$ and $g(x)$ for every element in $A$:

x∈A	f(x)=x2−x	g(x)=21​x2−21​x−1	f(x)=g(x)?
-1	$(-1)^2 – (-1) = 1 + 1 = 2$	$\frac{1}{2}(-1)^2 – \frac{1}{2}(-1) – 1 = \frac{1}{2} + \frac{1}{2} – 1 = 0$	No ($2 \neq 0$)

Since $f(-1) \neq g(-1)$, the functions are not equal.

Conclusion: The functions $f$ and $g$ are not equal.

(Note: The definition of $g(x)$ in the prompt is slightly unusual for the question’s hint. Assuming the prompt meant $g(x) = \frac{1}{2}x(x-1) – 1$, the functions are not equal. If the prompt intended the function $g(x)$ to be equal to $f(x)$, the definition of $g(x)$ would have had to be $g(x) = x(x-1)$, or a similar form that simplifies to $x^2-x$.)

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Question 6. Let $A = \{1, 2, 3\}$. Then number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is

(A) 1 (B) 2 (C) 3 (D) 4

Solution:

Let $R$ be the relation on $A = \{1, 2, 3\}$.

1. Given elements: $R$ must contain $(1, 2)$ and $(1, 3)$.
2. Reflexive condition: $R$ must contain $(1, 1), (2, 2), (3, 3)$.
3. Symmetric condition: If $(1, 2) \in R$, then $(2, 1) \in R$. If $(1, 3) \in R$, then $(3, 1) \in R$.

The minimum required relation is:

$$R_{min} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$$

Now we check transitivity for $R_{min}$:

• $(2, 1) \in R$ and $(1, 3) \in R \implies (2, 3)$ must be in $R$ for transitivity.
• $(3, 1) \in R$ and $(1, 2) \in R \implies (3, 2)$ must be in $R$ for transitivity.

Since $(2, 3) \notin R_{min}$ and $(3, 2) \notin R_{min}$, $R_{min}$ is not transitive. This relation satisfies all conditions (reflexive, symmetric, not transitive). This is the first possible relation.

Case 1: $R_1 = R_{min}$

$$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$$

This relation is not transitive. (Number = 1)

Case 2: Adding pairs that maintain symmetry/reflexivity but avoid transitivity.

To maintain symmetry, any pair added must be $(2, 3)$ and $(3, 2)$.

Let’s try to add these: $R_2 = R_1 \cup \{(2, 3), (3, 2)\}$.

$$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$$

$R_2$ contains all 9 possible pairs. A relation containing all possible pairs is an equivalence relation, and therefore must be transitive.

• Check transitivity in $R_2$: $(1, 2) \in R_2$ and $(2, 3) \in R_2 \implies (1, 3) \in R_2$ (Holds). All transitive chains hold.
• Since $R_2$ is transitive, it does not satisfy the condition “not transitive”.

Therefore, $R_1$ is the only relation that satisfies the criteria.

Correct Answer: (A) 1

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Question 7. Let $A = \{1, 2, 3\}$. Then number of equivalence relations containing $(1, 2)$ is

(A) 1 (B) 2 (C) 3 (D) 4

Solution:

An equivalence relation partitions the set $A$. The number of equivalence relations is equal to the number of ways to partition the set $A$.

The equivalence relation $R$ must contain $(1, 2)$. This means that the elements 1 and 2 must belong to the same partition.

We list the possible partitions of $A = \{1, 2, 3\}$ where $\{1, 2\}$ are together:

1. Partition 1: The partition has two sets: $\{1, 2\}$ and $\{3\}$.The equivalence relation $R_1$ is formed by taking the Cartesian product of each set and their union:$$R_1 = (\{1, 2\} \times \{1, 2\}) \cup (\{3\} \times \{3\})$$$$R_1 = \{(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)\}$$$R_1$ contains $(1, 2)$ and is an equivalence relation.
2. Partition 2: The partition has only one set: $\{1, 2, 3\}$.The equivalence relation $R_2$ is the universal relation $A \times A$:$$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$$$R_2$ contains $(1, 2)$ and is an equivalence relation.

Any other partition (e.g., $\{\{1\}, \{2\}, \{3\}\}$ or $\{\{1, 3\}, \{2\}\}$) would not contain $(1, 2)$.

The total number of equivalence relations containing $(1, 2)$ is 2.

Correct Answer: (B) 2

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