MATHEMATICS CLASS- 12
CHAPTER-1
(RELATIONS AND FUNCTIONS)
Exercise 1.2
Question 1
Show that the function
$$ f:\mathbb{R}^{*}\rightarrow\mathbb{R}^{*} $$defined by
$$ f(x)=\frac{1}{x}, $$where
$$ \mathbb{R}^{*}=\mathbb{R}-\{0\}, $$is one-one and onto. Is the result true if the domain
$$ \mathbb{R}^{*} $$is replaced by
$$ \mathbb{N} $$with co-domain remaining
$$ \mathbb{R}^{*}\ ? $$Solution
To show that the function is one-one (Injective):
Let
$$ f(x_1)=f(x_2). $$Then
$$ \frac{1}{x_1}=\frac{1}{x_2}. $$Multiplying both sides by
$$ x_1x_2\ (\neq 0), $$we get
$$ x_2=x_1. $$Hence,
$$ x_1=x_2. $$Therefore, distinct elements of the domain have distinct images. Hence, the function is one-one (injective).
To show that the function is onto (Surjective):
Let
$$ y\in\mathbb{R}^{*}. $$Choose
$$ x=\frac{1}{y}. $$Since
$$ y\neq0, $$we have
$$ x=\frac{1}{y}\in\mathbb{R}^{*}. $$Now,
$$ f(x)=\frac{1}{x} =\frac{1}{\frac{1}{y}} =y. $$Thus, every element of the co-domain has a pre-image in the domain. Hence, the function is onto (surjective).
Since the function is both one-one and onto, it is a bijection.
When the Domain is
$$ \mathbb{N} $$Now consider the function
$$ f:\mathbb{N}\rightarrow\mathbb{R}^{*}, \qquad f(x)=\frac{1}{x}. $$Injective: If
$$ f(x_1)=f(x_2), $$then
$$ \frac{1}{x_1}=\frac{1}{x_2}, $$which gives
$$ x_1=x_2. $$Hence, the function is still one-one.
Surjective: The function is not onto because many non-zero real numbers, such as
$$ 2,\quad -\frac12,\quad \pi, $$are not of the form
$$ \frac{1}{n}, \qquad n\in\mathbb{N}. $$Therefore, every element of
$$ \mathbb{R}^{*} $$does not have a pre-image in
$$ \mathbb{N}. $$Answer:
- The function $$ f(x)=\frac{1}{x} $$ from $$ \mathbb{R}^{*}\rightarrow\mathbb{R}^{*} $$ is one-one and onto.
- If the domain is changed to $$ \mathbb{N}, $$ the function remains one-one but is not onto.
Question 2 (i)
Check the injectivity and surjectivity of the function
$$ f:\mathbb{N}\rightarrow\mathbb{N} $$defined by
$$ f(x)=x^2. $$Solution
Injectivity: Let
$$ f(x_1)=f(x_2). $$Then
$$ x_1^2=x_2^2. $$Since
$$ x_1,x_2\in\mathbb{N}, $$both are positive integers. Therefore,
$$ x_1=x_2. $$Hence, the function is one-one (injective).
Surjectivity: For the function to be onto, every natural number must be the square of some natural number.
But the number
$$ 2 $$is a natural number and there is no natural number
$$ x $$such that
$$ x^2=2. $$Hence, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The function is one-one but not onto.}} $$Question 2 (ii)
Check the injectivity and surjectivity of the function
$$ f:\mathbb{Z}\rightarrow\mathbb{Z} $$defined by
$$ f(x)=x^2. $$Solution
Injectivity: Consider the integers
$$ 2\quad\text{and}\quad -2. $$Then
$$ f(2)=2^2=4 $$and
$$ f(-2)=(-2)^2=4. $$Since
$$ 2\neq-2 $$but
$$ f(2)=f(-2), $$the function is not one-one (not injective).
Surjectivity: Every integer is not a perfect square. For example,
$$ 2,\quad3,\quad5,\quad-1 $$are integers but are not equal to
$$ x^2 $$for any integer
$$ x. $$Hence, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The function is neither one-one nor onto.}} $$Question 2 (iii)
Check the injectivity and surjectivity of the function
$$ f:\mathbb{R}\rightarrow\mathbb{R} $$defined by
$$ f(x)=x^2. $$Solution
Injectivity: Consider the real numbers
$$ 2\quad\text{and}\quad -2. $$Then
$$ f(2)=4 $$and
$$ f(-2)=4. $$Since
$$ 2\neq-2 $$but
$$ f(2)=f(-2), $$the function is not one-one (not injective).
Surjectivity: Every real number is not the square of a real number. For example,
$$ -1 $$has no real number
$$ x $$such that
$$ x^2=-1. $$Hence, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The function is neither one-one nor onto.}} $$Question 2 (iv)
Check the injectivity and surjectivity of the function
$$ f:\mathbb{N}\rightarrow\mathbb{N} $$defined by
$$ f(x)=x^3. $$Solution
Injectivity: Let
$$ f(x_1)=f(x_2). $$Then
$$ x_1^3=x_2^3. $$Taking cube roots on both sides, we get
$$ x_1=x_2. $$Hence, the function is one-one (injective).
Surjectivity: For the function to be onto, every natural number must be a perfect cube.
But the number
$$ 2 $$is a natural number and there is no natural number
$$ x $$such that
$$ x^3=2. $$Hence, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The function is one-one but not onto.}} $$Question 2 (v)
Check the injectivity and surjectivity of the function
$$ f:\mathbb{Z}\rightarrow\mathbb{Z} $$defined by
$$ f(x)=x^3. $$Solution
Injectivity: 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, the function is one-one (injective).
Surjectivity: For the function to be onto, every integer must be a perfect cube.
But the integer
$$ 2 $$is not the cube of any integer.
Hence, there is no integer
$$ x $$such that
$$ x^3=2. $$Therefore, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The function is one-one but not onto.}} $$Question 3
Prove that the Greatest Integer Function
$$ f:\mathbb{R}\rightarrow\mathbb{R}, \qquad f(x)=[x], $$is neither one-one nor onto, where
$$ [x] $$denotes the greatest integer less than or equal to
$$ x. $$Solution
To show that the function is not one-one:
Consider the real numbers
$$ 1.2 \quad\text{and}\quad 1.8. $$Then
$$ f(1.2)=[1.2]=1 $$and
$$ f(1.8)=[1.8]=1. $$Since
$$ 1.2\neq1.8 $$but
$$ f(1.2)=f(1.8), $$different elements of the domain have the same image.
Hence, the function is not one-one (not injective).
To show that the function is not onto:
The greatest integer function always gives an integer as its output.
Therefore, numbers such as
$$ 1.5,\quad \sqrt2,\quad \pi $$belong to the co-domain
$$ \mathbb{R}, $$but none of them is the image of any real number.
Hence, every element of
$$ \mathbb{R} $$does not have a pre-image.
Therefore, the function is not onto (not surjective).
Answer:
The Greatest Integer Function is neither one-one nor onto.
$$ \boxed{\text{The function is neither one-one nor onto.}} $$Question 4
Show that the Modulus Function
$$ f:\mathbb{R}\rightarrow\mathbb{R}, \qquad f(x)=|x| $$is neither one-one nor onto, where
$$ |x|= \begin{cases} x, & \text{if } x\ge0,\\ -x, & \text{if } x<0. \end{cases} $$Solution
To show that the function is not one-one:
Consider the two distinct real numbers
$$ 2 \quad \text{and} \quad -2. $$Then
$$ f(2)=|2|=2 $$and
$$ f(-2)=|-2|=2. $$Since
$$ 2\neq-2 $$but
$$ f(2)=f(-2), $$different elements of the domain have the same image. Hence, the function is not one-one (not injective).
To show that the function is not onto:
The modulus of every real number is always non-negative. Therefore,
$$ |x|\ge0 $$for every
$$ x\in\mathbb{R}. $$Hence, no negative real number can be the image of any element of the domain. For example, there is no real number
$$ x $$such that
$$ |x|=-1. $$Therefore, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The modulus function is neither one-one nor onto.}} $$Question 5
Show that the Signum Function
$$ f:\mathbb{R}\rightarrow\mathbb{R} $$defined by
$$ f(x)= \begin{cases} 1, & \text{if } x>0,\\[4pt] 0, & \text{if } x=0,\\[4pt] -1, & \text{if } x<0 \end{cases} $$is neither one-one nor onto.
Solution
To show that the function is not one-one:
Consider the positive real numbers
$$ 2 \quad \text{and} \quad 5. $$Then
$$ f(2)=1 $$and
$$ f(5)=1. $$Since
$$ 2\neq5 $$but
$$ f(2)=f(5), $$different elements of the domain have the same image. Hence, the function is not one-one (not injective).
To show that the function is not onto:
The signum function can take only the three values
$$ -1,\ 0,\ 1. $$Since the co-domain is
$$ \mathbb{R}, $$many real numbers such as
$$ 2,\quad \frac12,\quad \pi $$are not images of any real number.
Therefore, every element of the co-domain does not have a pre-image. Hence, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The signum function is neither one-one nor onto.}} $$Question 6
Let
$$ A=\{1,2,3\}, \qquad B=\{4,5,6,7\} $$and let
$$ f=\{(1,4),(2,5),(3,6)\} $$be a function from
$$ A \text{ to } B. $$Show that
$$ f $$is one-one.
Solution
A function is one-one (injective) if distinct elements of the domain have distinct images.
From the given function, we have
$$ f(1)=4,\qquad f(2)=5,\qquad f(3)=6. $$The images of the elements
$$ 1,\ 2,\ 3 $$are
$$ 4,\ 5,\ 6, $$which are all distinct.
Thus, no two different elements of the domain have the same image. That is, whenever
$$ f(a)=f(b), $$it follows that
$$ a=b. $$Hence, the function satisfies the definition of an injective (one-one) function.
Also, note that the element
$$ 7\in B $$has no pre-image. Therefore, the function is not onto, but this does not affect the injective property.
Answer:
$$ \boxed{\text{The function }f\text{ is one-one (injective).}} $$Question 7 (i)
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
Given
$$ f:\mathbb{R}\rightarrow\mathbb{R}, \qquad f(x)=3-4x. $$Solution
To check whether the function is one-one:
Let
$$ f(x_1)=f(x_2). $$Then
$$ 3-4x_1=3-4x_2. $$Subtracting 3 from both sides, we get
$$ -4x_1=-4x_2. $$Dividing both sides by
$$ -4, $$we obtain
$$ x_1=x_2. $$Hence, distinct elements of the domain have distinct images. Therefore, the function is one-one (injective).
To check whether the function is onto:
Let
$$ y\in\mathbb{R}. $$We have
$$ y=3-4x. $$Solving for
$$ x, $$we get
$$ x=\frac{3-y}{4}. $$Since
$$ \frac{3-y}{4}\in\mathbb{R}, $$there exists a real number
$$ x $$for every
$$ y\in\mathbb{R}. $$Hence, every element of the co-domain has a pre-image. Therefore, the function is onto (surjective).
Answer:
The function is both one-one and onto.
$$ \boxed{\text{Hence, the function is bijective.}} $$Question 7 (ii)
Given
$$ f:\mathbb{R}\rightarrow\mathbb{R}, \qquad f(x)=1+x^2. $$Solution
To check whether the function is one-one:
Consider the real numbers
$$ 2 \quad\text{and}\quad -2. $$Then
$$ f(2)=1+2^2=5 $$and
$$ f(-2)=1+(-2)^2=5. $$Since
$$ 2\neq-2 $$but
$$ f(2)=f(-2), $$the function is not one-one (not injective).
To check whether the function is onto:
Since
$$ x^2\ge0 $$for every real number
$$ x, $$we have
$$ 1+x^2\ge1. $$Thus, the range of the function is
$$ [1,\infty). $$Since the co-domain is
$$ \mathbb{R}, $$numbers such as
$$ 0,\quad -2,\quad -5 $$are not images of any real number.
Hence, the function is not onto (not surjective).
Answer:
$$ \boxed{\text{The function is neither one-one nor onto.}} $$Question 8
Let
$$ A \text{ and } B $$be sets. Show that the function
$$ f:A\times B\rightarrow B\times A $$defined by
$$ f(a,b)=(b,a) $$is a bijective function.
Solution
To show that the function is one-one:
Let
$$ f(a_1,b_1)=f(a_2,b_2). $$Then
$$ (b_1,a_1)=(b_2,a_2). $$Since two ordered pairs are equal only when their corresponding components are equal, we get
$$ b_1=b_2 \quad\text{and}\quad a_1=a_2. $$Hence,
$$ (a_1,b_1)=(a_2,b_2). $$Therefore, the function is one-one (injective).
To show that the function is onto:
Let
$$ (b,a)\in B\times A. $$Choose the element
$$ (a,b)\in A\times B. $$Then
$$ f(a,b)=(b,a). $$Thus, every element of
$$ B\times A $$has a pre-image in
$$ A\times B. $$Hence, the function is onto (surjective).
Answer:
Since the function is both one-one and onto, it is a bijective function.
$$ \boxed{\text{The function }f(a,b)=(b,a)\text{ is bijective.}} $$Question 9
Let
$$ f:\mathbb{N}\rightarrow\mathbb{N} $$be defined by
$$ f(n)= \begin{cases} \dfrac{n+1}{2}, & \text{if } n \text{ is odd},\\[8pt] \dfrac{n}{2}, & \text{if } n \text{ is even}. \end{cases} $$State whether the function
$$ f $$is bijective. Justify your answer.
Solution
To check whether the function is one-one:
Consider the natural numbers
$$ 1 \quad\text{and}\quad 2. $$Then
$$ f(1)=\frac{1+1}{2}=1 $$and
$$ f(2)=\frac{2}{2}=1. $$Since
$$ 1\neq2 $$but
$$ f(1)=f(2), $$the function is not one-one (not injective).
To check whether the function is onto:
Let
$$ m\in\mathbb{N}. $$Choose
$$ n=2m. $$Since
$$ 2m $$is even,
$$ f(2m)=\frac{2m}{2}=m. $$Thus, every natural number has a pre-image.
Hence, the function is onto (surjective).
Answer:
The function is onto but not one-one. Therefore, it is not bijective.
$$ \boxed{\text{The function is not bijective.}} $$Question 10
Let
$$ A=\mathbb{R}-\{3\} \quad\text{and}\quad B=\mathbb{R}-\{1\}. $$Consider the function
$$ f:A\rightarrow B $$defined by
$$ f(x)=\frac{x-2}{x-3}. $$Is the function one-one and onto? Justify your answer.
Solution
To check whether the function is one-one:
Let
$$ f(x_1)=f(x_2). $$Then
$$ \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}. $$Cross-multiplying, we get
$$ (x_1-2)(x_2-3)=(x_2-2)(x_1-3). $$Expanding both sides,
$$ x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6. $$On simplifying,
$$ -x_1-x_2=-x_2-2x_1+x_1, $$which gives
$$ x_1=x_2. $$Hence, distinct elements of the domain have distinct images. Therefore, the function is one-one (injective).
To check whether the function is onto:
Let
$$ y\in B. $$Then
$$ y=\frac{x-2}{x-3}. $$Solving for
$$ x, $$we get
$$ yx-3y=x-2. $$Therefore,
$$ x(y-1)=3y-2, $$and hence
$$ x=\frac{3y-2}{y-1}. $$Since
$$ y\neq1, $$the denominator is never zero. Thus,
$$ x=\frac{3y-2}{y-1} $$belongs to
$$ A=\mathbb{R}-\{3\}. $$Also,
$$ f\!\left(\frac{3y-2}{y-1}\right)=y. $$Hence, every element of the co-domain has a pre-image in the domain. Therefore, the function is onto (surjective).
Answer:
The function is both one-one and onto.
$$ \boxed{\text{Hence, }f\text{ is a bijective function.}} $$Question 11
Let
$$ f:\mathbb{R}\rightarrow\mathbb{R} $$be defined by
$$ f(x)=x^4. $$Choose the correct answer.
(A) \(f\) is one-one onto
(B) \(f\) is many-one onto
(C) \(f\) is one-one but not onto
(D) \(f\) is neither one-one nor onto
Solution
To check whether the function is one-one:
Consider the real numbers
$$ 2 \quad\text{and}\quad -2. $$Then
$$ f(2)=2^4=16 $$and
$$ f(-2)=(-2)^4=16. $$Since
$$ 2\neq-2 $$but
$$ f(2)=f(-2), $$the function is not one-one. It is a many-one function.
To check whether the function is onto:
For every real number
$$ x, $$we have
$$ x^4\ge0. $$Hence, the range of the function is
$$ [0,\infty), $$which is not equal to the co-domain
$$ \mathbb{R}. $$For example, there is no real number
$$ x $$such that
$$ x^4=-1. $$Therefore, the function is not onto.
Answer:
$$ \boxed{\text{Option (D): The function is neither one-one nor onto.}} $$Question 12
Let
$$ f:\mathbb{R}\rightarrow\mathbb{R} $$be defined by
$$ f(x)=3x. $$Choose the correct answer.
(A) \(f\) is one-one onto
(B) \(f\) is many-one onto
(C) \(f\) is one-one but not onto
(D) \(f\) is neither one-one nor onto
Solution
To check whether the function is one-one:
Let
$$ f(x_1)=f(x_2). $$Then
$$ 3x_1=3x_2. $$Dividing both sides by
$$ 3, $$we obtain
$$ x_1=x_2. $$Hence, the function is one-one (injective).
To check whether the function is onto:
Let
$$ y\in\mathbb{R}. $$Choose
$$ x=\frac{y}{3}. $$Since
$$ \frac{y}{3}\in\mathbb{R}, $$we have
$$ f\left(\frac{y}{3}\right)=3\left(\frac{y}{3}\right)=y. $$Thus, every element of the co-domain has a pre-image in the domain.
Hence, the function is onto (surjective).
Answer:
$$ \boxed{\text{Option (A): The function is one-one and onto (bijective).}} $$