MATHEMATICS CLASS- 12
CHAPTER-1
(RELATIONS AND FUNCTIONS)
Miscellaneous Exercise
Question 1
Show that the function
$$ f:\mathbb{R}\rightarrow\{x\in\mathbb{R}:-1is one-one and onto.
Solution
To show that the function is one-one:
Let
$$ f(x_1)=f(x_2). $$Then
$$ \frac{x_1}{1+|x_1|} = \frac{x_2}{1+|x_2|}. $$We consider the following cases.
Case 1: If
$$ x_1,\ x_2\ge0, $$then
$$ |x_1|=x_1,\qquad |x_2|=x_2. $$Hence,
$$ \frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}. $$Cross-multiplying, we get
$$ x_1(1+x_2)=x_2(1+x_1). $$Therefore,
$$ x_1+x_1x_2=x_2+x_1x_2, $$which gives
$$ x_1=x_2. $$Case 2: If
$$ x_1,\ x_2<0, $$then
$$ |x_1|=-x_1,\qquad |x_2|=-x_2. $$Hence,
$$ \frac{x_1}{1-x_1} = \frac{x_2}{1-x_2}. $$Cross-multiplying,
$$ x_1(1-x_2)=x_2(1-x_1). $$Therefore,
$$ x_1-x_1x_2=x_2-x_1x_2, $$which gives
$$ x_1=x_2. $$Case 3: One of
$$ x_1,\ x_2 $$is non-negative and the other is negative.
Then
$$ f(x_1)\ge0 $$and
$$ f(x_2)<0, $$which is impossible because
$$ f(x_1)=f(x_2). $$Hence, this case cannot occur.
Therefore, in every case,
$$ x_1=x_2. $$Thus, the function is one-one (injective).
To show that the function is onto:
Let
$$ y\in(-1,1). $$We find
$$ x $$such that
$$ \frac{x}{1+|x|}=y. $$If
$$ 0\le y<1, $$take
$$ x=\frac{y}{1-y}. $$Then
$$ x\ge0 $$and
$$ f(x)=\frac{x}{1+x} =\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}} =y. $$If
$$ -1Since
$$ y<0, $$we have
$$ x<0. $$Now,
$$ f(x)=\frac{x}{1-x} =\frac{\frac{y}{1+y}}{1-\frac{y}{1+y}} =y. $$Thus, every element of the co-domain has a pre-image in
$$ \mathbb{R}. $$Hence, the function is onto (surjective).
Answer:
The function is both one-one and onto.
$$ \boxed{\text{Hence, }f(x)=\frac{x}{1+|x|}\text{ is a bijective function.}} $$Question 2
Show that the function
$$ f:\mathbb{R}\rightarrow\mathbb{R} $$defined by
$$ f(x)=x^3 $$is injective.
Solution
To prove that the function is injective, let
$$ f(x_1)=f(x_2). $$Then
$$ x_1^3=x_2^3. $$Taking cube roots on both sides, we obtain
$$ x_1=x_2. $$Hence, different elements of the domain cannot have the same image.
Therefore, the function is one-one (injective).
Alternatively, observe that the function
$$ f(x)=x^3 $$is strictly increasing on
$$ \mathbb{R}. $$A strictly increasing function is always one-one.
Answer:
$$ \boxed{\text{The function }f(x)=x^3\text{ is injective.}} $$Question 3
Given a non-empty set
$$ X, $$consider
$$ P(X), $$the set of all subsets of
$$ X. $$Define the relation
$$ R $$on
$$ P(X) $$by
$$ ARB \iff A\subseteq B. $$Is
$$ R $$an equivalence relation on
$$ P(X)? $$Justify your answer.
Solution
To determine whether the relation is an equivalence relation, we check the three properties: reflexivity, symmetry and transitivity.
Reflexivity:
For every subset
$$ A\in P(X), $$we always have
$$ A\subseteq A. $$Therefore,
$$ (A,A)\in R. $$Hence, the relation is reflexive.
Symmetry:
Suppose
$$ A\subseteq B. $$For the relation to be symmetric, it must follow that
$$ B\subseteq A. $$This is not always true.
For example, let
$$ X=\{1,2\}, $$and take
$$ A=\{1\},\qquad B=\{1,2\}. $$Then
$$ A\subseteq B, $$but
$$ B\nsubseteq A. $$Hence, the relation is not symmetric.
Transitivity:
Suppose
$$ A\subseteq B $$and
$$ B\subseteq C. $$Then every element of
$$ A $$belongs to
$$ B, $$and every element of
$$ B $$belongs to
$$ C. $$Therefore, every element of
$$ A $$belongs to
$$ C. $$Hence,
$$ A\subseteq C. $$Thus,
$$ (A,C)\in R, $$and the relation is transitive.
Answer:
The relation is reflexive and transitive, but it is not symmetric.
Therefore, it is not an equivalence relation.
$$ \boxed{\text{The relation }R\text{ is not an equivalence relation on }P(X).} $$Question 4
Find the number of all onto functions from the set
$$ A=\{1,2,3,\ldots,n\} $$to itself.
Solution
The set
$$ A=\{1,2,3,\ldots,n\} $$contains
$$ n $$elements.
An onto function from
$$ A $$to itself must map every element of the co-domain to at least one element of the domain.
Since the domain and co-domain have the same finite number of elements, namely
$$ n, $$every onto function is also one-one.
Thus, every onto function is a bijection.
Now, the number of bijections from a set containing
$$ n $$elements onto itself is equal to the number of permutations of its elements.
The first element of the domain can be mapped to any one of the
$$ n $$elements of the co-domain.
After choosing the image of the first element, the second element can be mapped in
$$ n-1 $$ways.
Similarly, the third element can be mapped in
$$ n-2 $$ways, and so on.
Therefore, the total number of onto functions is
$$ n(n-1)(n-2)\cdots2\cdot1=n!. $$Answer:
$$ \boxed{n!} $$Hence, the number of all onto functions from the set
$$ \{1,2,3,\ldots,n\} $$to itself is
$$ \boxed{n!}. $$Question 5
Let
$$ A=\{-1,0,1,2\},\qquad B=\{-4,-2,0,2\} $$and let
$$ f,g:A\rightarrow B $$be defined by
$$ f(x)=x^2-x $$and
$$ g(x)=2\left|x-\frac12\right|-1,\qquad x\in A. $$Are
$$ f $$and
$$ g $$equal? Justify your answer.
Solution
Two functions are equal if they have the same domain, the same co-domain and assign the same image to every element of the domain.
Here, both functions have the same domain
$$ A=\{-1,0,1,2\} $$and the same co-domain
$$ B=\{-4,-2,0,2\}. $$Now, compute the values of both functions for each element of
$$ A. $$| \(x\) | \(f(x)=x^2-x\) | \(g(x)=2\left|x-\frac12\right|-1\) |
|---|---|---|
| \(-1\) | \((-1)^2-(-1)=2\) | \(2\left|-\frac32\right|-1=3-1=2\) |
| \(0\) | \(0^2-0=0\) | \(2\left|-\frac12\right|-1=1-1=0\) |
| \(1\) | \(1^2-1=0\) | \(2\left|\frac12\right|-1=1-1=0\) |
| \(2\) | \(2^2-2=2\) | \(2\left|\frac32\right|-1=3-1=2\) |
Thus, for every element of
$$ A, $$we have
$$ f(x)=g(x). $$Therefore, the two functions assign the same image to every element of the domain.
Answer:
Since
$$ f(x)=g(x) $$for every
$$ x\in A, $$the two functions are equal.
$$ \boxed{f=g.} $$Question 6
Let
$$ A=\{1,2,3\}. $$Find the number of relations containing
$$ (1,2)\ \text{and}\ (1,3) $$which are reflexive and symmetric but not transitive.
Solution
Since the relation is reflexive, it must contain the pairs
$$ (1,1),\ (2,2),\ (3,3). $$Since it is symmetric and contains
$$ (1,2)\ \text{and}\ (1,3), $$it must also contain their symmetric pairs
$$ (2,1)\ \text{and}\ (3,1). $$Thus, the following seven pairs are compulsory:
$$ (1,1),\ (2,2),\ (3,3),\ (1,2),\ (2,1),\ (1,3),\ (3,1). $$The only remaining pairs are
$$ (2,3)\ \text{and}\ (3,2). $$Since the relation is symmetric, these two pairs must either be included together or excluded together.
Case 1: Neither
$$ (2,3) $$nor
$$ (3,2) $$is included.
Then
$$ (2,1)\in R \quad\text{and}\quad (1,3)\in R, $$but
$$ (2,3)\notin R. $$Therefore, the relation is not transitive.
This gives one relation.
Case 2: Both
$$ (2,3) $$and
$$ (3,2) $$are included.
Now every transitivity condition is satisfied, so the relation becomes transitive.
Hence, this case is not required.
Answer:
Only one relation satisfies all the given conditions.
$$ \boxed{\text{Option (A): }1} $$Question 7
Let
$$ A=\{1,2,3\}. $$Find the number of equivalence relations containing
$$ (1,2). $$Solution
An equivalence relation corresponds to a partition of the set.
Since
$$ (1,2) $$belongs to the relation, the elements
$$ 1 $$and
$$ 2 $$must belong to the same equivalence class.
There are only two possible partitions of
$$ A=\{1,2,3\} $$having
$$ 1 $$and
$$ 2 $$in the same class.
Partition 1:
$$ \{\{1,2\},\{3\}\}. $$Partition 2:
$$ \{\{1,2,3\}\}. $$Each partition determines exactly one equivalence relation.
Hence, there are exactly
$$ 2 $$such equivalence relations.
Answer:
$$ \boxed{\text{Option (B): }2} $$