Class 12-NCERT Solutions-Chapter-1-Relations and Functions-Mis -

NCERT Solutions Class-12-Chapter-1-Relations and Functions

Miscellaneous Exercise on Chapter 1

Note: These problems cover advanced concepts of functions (bijectivity) and relations (equivalence classes). Recall the definitions of one-one, onto, and the properties of relations.

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

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Solution:

One-one:

Case 1: Let \( x, y \) be positive. \( f(x) = f(y) \Rightarrow \frac{x}{1+x} = \frac{y}{1+y} \Rightarrow x + xy = y + xy \Rightarrow x = y \).

Case 2: Let \( x, y \) be negative. \( f(x) = f(y) \Rightarrow \frac{x}{1-x} = \frac{y}{1-y} \Rightarrow x - xy = y - xy \Rightarrow x = y \).

Case 3: If one is positive and one negative, their images will have different signs (positive and negative), so they cannot be equal.

Thus, \( f \) is one-one.

Onto:

Let \( y \in (-1, 1) \).

If \( y \geq 0 \), let \( x = \frac{y}{1-y} \). Since \( 0 \leq y < 1 \), \( x \) is a defined real number and \( x \geq 0 \). Then \( f(x) = \frac{\frac{y}{1-y}}{1 + \frac{y}{1-y}} = \frac{y}{1-y+y} = y \).

If \( y < 0 \), let \( x = \frac{y}{1+y} \). Since \( -1 < y < 0 \), \( x \) is a defined real number and \( x < 0 \). Then \( f(x) = \frac{\frac{y}{1+y}}{1 - \frac{y}{1+y}} = \frac{y}{1+y-y} = y \).

Thus, every element in the codomain has a pre-image. \( f \) is onto.

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

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Solution:

Let \( x, y \in \mathbf{R} \) such that \( f(x) = f(y) \).

\[ x^3 = y^3 \]

Taking the cube root on both sides (since the cube root is unique for real numbers):

\[ x = y \]

Therefore, \( f \) is injective (one-one).

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), ARB if and only if \( A \subset B \). Is R an equivalence relation on P(X)? Justify your answer.

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Solution:

Reflexive: For any set A, \( A \subset A \) is always true. So, R is reflexive.
Symmetric: Let \( A = \{1\} \) and \( B = \{1, 2\} \). \( A \subset B \) is true. However, \( B \subset A \) is false. Thus, R is not symmetric.
Transitive: If \( A \subset B \) and \( B \subset C \), then \( A \subset C \). Thus, R is transitive.

Therefore, R is not an equivalence relation because it is not symmetric.

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

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Solution:

The domain and codomain have the same number of elements (\( n \)).

For a function from a finite set to itself to be onto, it must also be one-one. Essentially, we are looking for the number of bijections (permutations) of \( n \) elements.

The number of such functions is \( n! \) (n factorial).

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 \( g(x) = 2\left|x - \frac{1}{2}\right| - 1, x \in A \). Are \( f \) and \( g \) equal? Justify your answer.

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Solution:

Two functions are equal if their domains are the same and \( f(x) = g(x) \) for all \( x \) in the domain. The domain A is common.

We check the values for each element in A:

For \( x = -1 \):
\( f(-1) = (-1)^2 - (-1) = 1 + 1 = 2 \)
\( g(-1) = 2|-1 - 0.5| - 1 = 2(1.5) - 1 = 3 - 1 = 2 \)
For \( x = 0 \):
\( f(0) = 0^2 - 0 = 0 \)
\( g(0) = 2|0 - 0.5| - 1 = 2(0.5) - 1 = 1 - 1 = 0 \)
For \( x = 1 \):
\( f(1) = 1^2 - 1 = 0 \)
\( g(1) = 2|1 - 0.5| - 1 = 2(0.5) - 1 = 1 - 1 = 0 \)
For \( x = 2 \):
\( f(2) = 2^2 - 2 = 4 - 2 = 2 \)
\( g(2) = 2|2 - 0.5| - 1 = 2(1.5) - 1 = 3 - 1 = 2 \)

Since \( f(x) = g(x) \) for all \( x \in A \), the functions are equal.

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

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Answer: (A)

Solution:

The relation must be reflexive, so it must contain \( \{(1, 1), (2, 2), (3, 3)\} \).

It must contain (1, 2) and (1, 3). Since it is symmetric, it must also contain (2, 1) and (3, 1).

Current set: \( R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\} \).

Check transitivity: We have (2, 1) and (1, 3), but (2, 3) is not in R. So this relation is NOT transitive. This is a valid relation.

If we add (2, 3) to make it transitive (or just to add elements), we must also add (3, 2) for symmetry. The relation then becomes the universal relation (containing all pairs), which IS transitive.

Thus, there is only 1 such relation.

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

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Answer: (B)

Solution:

An equivalence relation must be Reflexive, Symmetric, and Transitive.

Relation 1: Smallest equivalence relation containing (1, 2).
Must have reflexive pairs: \( \{(1, 1), (2, 2), (3, 3)\} \).
Must have symmetry for (1, 2): Add (2, 1).
\( R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} \).
Is this transitive? Yes.

Relation 2: Can we add more elements?
If we add (1, 3), we must add (3, 1) (Symmetry).
Now we have (2, 1) and (1, 3), so we must add (2, 3) (Transitivity).
If we have (2, 3), we must add (3, 2) (Symmetry).
This results in the Universal Relation \( A \times A \), which is an equivalence relation.

Therefore, there are exactly 2 such equivalence relations.