RD Sharma Class 12 Exercise 2.1 Function Solutions Maths - Download PDF Free Online

Functions Excercise:2.1

Functions exercise 2.1 question1(i)

$f(x)=2x$
Given:
The example for a function which is one-one but not onto.
Hint:
One-one function means every element in the domain has a distinct image in the co-domain
.$f(x_1)=f(x_2)$, then$x_1=x_2$
Where, $X_1,X_2$ domain of$f(x)$
Onto function defined as every element in the co-domain has at least one pre image in the domain of function.
Solution:
Here, we need to give an example of a function for one-one but not onto.
Let we consider the function,
$f:N\to N$, given by$f(x)=2N$
Let us consider two elements$x_1$ and $x_2$ in the domain of f, so we get
$f(x_1)=2x_1$and$f(x_2)=x_2$
Now we know the condition for finding a one-one function.
$2x_1=2x_2$
$x_1=x_2$
Hence, the function is one-one.
Now, we need to prove$f(x)=2x$is not onto
Let $f(x)=y$, such that x,$y\in N$
We get,
$2x=y$
$x=\frac{y}{2}$
If we put $y=1,$
$x=\frac{1}{2}=0.5$, which is cannot be true as $x\in N$ supposed in solution.
Hence, the given function is not onto.
So, $f(x)=2x$ is an example of one-one but not onto function.

Functions exercise 2.1 question 1 (ii)

$f(x)=y$
Given:
The example for a function which is not one-one but onto.
Hint:
One-one function means every element in the domain has a distinct image in the co-domain. If one-one is given for any function.$f(x_1)=f(x_2)$, then$x_1=x_2$
Where, $x_1,x_2\in$ domain of $f(x)$
Onto function defined as every element in the co-domain has at least one pre image in the domain of function.
Solution:
Here, we need to give an example of a function for a not one-one but onto function.
Let the function$f:N\to N$, given by$f(1)=f(2)=1$
Here, $f(x)=f(2)=1$
$f(x)=f(1)=1$
Since different elements $1,2$ have same image 1,
$\therefore f$ is not one-one.
Let$f(x)=y$, such that $y\in N$
Here,y is a natural number and for every y there is a value of x which is a natural number.
Hence,f is onto.
So, the function $f:N\to N$, given by$f(1)=f(2)=1$ is not one-one but onto.

Functions exercise 2.1 question 1 (iii)

The example for a function which is neither one-one nor onto.
Hint:
One-one function means every element in the domain has a distinct image in the co-domain. If one-one is given for any function$f(x)$ as if$f(x_1)=f(x_2)$, then$x_1=x_2$
Where, $x_1,x_2\in$ domain of $f(x)$
Onto function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Let function$f:N\to N$ , given by$f(x)=x^2$
Calculate $f(x_1)$and$f(x_2)$
$\begin{array}{rl}& f\left(x_1\right)=x_1^2\\ & f\left(x_2\right)=x_2^2\\ & {\left(x_1\right)}^2={\left(x_2\right)}^2\\ & x_1=x_2\text{ or }{x}_1=-x_2\end{array}$
Since,$x_1$ doesn’t have unique image,
Now,
$f(x)=x^2$
Let $f(x)=y$ such that $y\in R$
$x^2=y$
$x=\pm \sqrt{y}$
Since y is real number, then it can be negative also.
If $y=-2,x=\pm \sqrt{-2}$ which is not possible as the root of a negative number is not real.
Hence,x is not real, so f is not onto.
$\therefore$ Function $f:N\to N$ given by$f(x)=x^2$is neither one-one nor onto.

Functions exercise 2.1 question 2 (i)

$f_1$ is one-one and onto function.
Given:
$f_1=\{(1,3),(2,5),(3,7)\};A=\{1,2,3\};B=\{3,5,7\}$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the codomain has at least one pre image in the domain of function.
Solution:
We need to find the one-one function
Injectivity: (one-one)
$\begin{array}{rl}& f_1(1)=3\\ & f_1(2)=5\\ & f_1(3)=7\end{array}$
Every element of A has different images in B.
$\therefore$ Function $f_1$ is one-one.
Surjectivity: (onto)
Co-domain of$f_1=\{3,5,7\}$
Range of $f_1$ set of image$\{3,5,7\}$
Co-domain=Range
So, $f_1$ is onto.

Functions exercise 2.1 question 2 (ii)

$f_2$ is one-one and onto function.
Given:
$f_2=\{(2,a),(3,b),(4,c)\};A=\{2,3,4\};B=\{a,b,c\}$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the codomain has at least one pre image in the domain of function.
Solution:
We need to find the one-one function
Injectivity:
$\begin{array}{rl}& f_2(2)=a\\ & f_2(3)=b\\ & f_2(4)=c\end{array}$
Every element of A has different images in B.
$\therefore$ Function $f_2$ is one-one.
Surjectivity:
Co-domain of $f_2=\{a,b,c\}$
Range of $f_2=$ set of image$\{a,b,c\}$
Codomain = Range
So, $f_2$ is onto

Functions exercise 2.1 question 2 (iii)

$f_3$ is not one-one and onto function.
Given:
$f_3=\{(a,x),(b,x),(c,z),(d,z)\};A=\{a,b,c,d\};B=\{x,y,z\}$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the codomain has at least one pre image in the domain of function.
Solution:
We need to find the one-one function
Injectivity: (one-one)
$\begin{array}{rl}& f_3(a)=x,f_3(c)=z\\ & f_3(b)=x,f_3(d)=z\end{array}$
Here $f_3(a),f_3(b)$ have the same image x
Also $f_3(c),f_3(d)$ have the same image z
Therefore,$f_3$ is not one-one.

Functions exercise 2.1 question 3

$f:N\to N$, given by$f(x)=x^2+x+1$is one-one but not onto.
Hint:
Let x and y be any two elements in the domain (N), such that $f(x)=f(y)$
So,f is one-one.
Given:
$f:N\to N$, defined by$f(x)=x^2+x+1$
Let us prove that the given function is one-one.
Solution:
We have$f(x)=x^2+x+1$ … (i)
Calculate $f(y)=y^2+x+1$ … (ii)
Now equate equation (i) and (ii)
$\begin{array}{rl}& f(x)=f(y)\\ & x^2+x+1=y^2+y+1\\ & x^2+x-y^2-y=0\end{array}$
$\begin{array}{r}{x}^2-y^2+x-y=0\\ [\because a^2-b^2=(a+b)(a-b)]\\ (x+y)(x-y)+x-y=0\\ (x-y)[(x+y+1)]=0\end{array}$
From (iii) equation, we can write
$x-y=0,x+y+1\ne 0$
$x+y+1\ne 0$ , because if we substitute$x=1,y=1$
$1+1+1=3$
Hence, $x+y+1\ne 0$ for any$x\in N$
Where as,
$x-y=0$
$x=y$
So,f is a one-one function.
But, $f(x)=x^3+x+1\ge 3$ for all $x\in N$
So,$f(x)$ doesn’t assume values of1 and 2
$\therefore f$ is not an onto function.
Hence, we proved that, $f(x)=x^2+x+1$ is one-one function not an onto function.

Functions exercise 2.1 question 4

$f:A\to A$ is neither one-one nor onto.
Given:
$A=\{-1,0,1\}\text{ and }f=\{(x,x):x\in A\}$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
We need to prove that$f:A\to A$ is neither one-one nor onto.
First of all we check the function with one-one,
$\begin{array}{rl}& \begin{array}{c}f(x)=x^2\\ \text{ If }x=1,f(1)=1^2=1\\ \text{ If }x=-1,f(-1)=(-1{)}^2=1\end{array}\\ & \text{ Here }f(1)\text{ and }f(-1)\text{ have the same image }1\text{ . }\end{array}$
So,f is not one-one.
Now, we go for onto function,
Co-domain of $f=\{0,1\}$
$\begin{array}{rl}& f(x)=x^2\\ & f(1)=1^2=1\\ & f(-1)=(-1{)}^2=1\\ & f(0)=0\end{array}$
Range of $f=\{-1,0.1\}$
Codomain of$f\ne$ Range of $f$ . (both are not same)
Hence, f is not onto

Functions exercise 2.1 question 5 (i)

One-one but not onto.
Given:
$f:N\to N$ given by$f(x)=x^2$ .
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
For any function to be a bijective, the given function be one-one and onto.
Solution:
Let's start with the injection test.
Let x and y be any two elements in the domain(N) such that$f(x)=f(y)$
$\begin{array}{rl}& f(x)=x^2,f(y)=y^2\\ & f(x)=f(y)\\ & x^2=y^2\\ & x=y\end{array}$
Where x and y are in N , so we don’t get 1 value.
So,f is an injection.
Surjection test:
Let y be any element in the codomain (N) , such that$f(x)=y$ for some element x in N .
$f(x)=y$
$x^2=y$
$x=\pm \sqrt{y}$ which may not be in N
Example:
If $y=2,x=\pm \sqrt{2}$ which is not in N .
So,f is not a surjection.
The condition of bijection is as we know the function should be one-one and onto.
Hence for this, f is not bijection

Functions exercise 2.1 question 5 (ii)

Neither one-one nor onto.
Given:
$f:Z\to Z$ given by $f(x)=x^2$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
For any function to be a bijective, the given function be one-one and onto.
Solution:
Let's start with the injection test.
Let x and y be any two elements in the domain (Z) such that $f(x)=f(y)$
$\begin{array}{rl}& f(x)=x^2,f(y)=y^2\\ & f(x)=f(y)\\ & x^2=y^2\\ & x=\pm y\end{array}$
Therefore,f is not an injection
Surjection test:
Let y be any element in the codomain (Z) , such that$f(x)=y$ for some element x in Z .
$f(x)=y$
$x^2=y$
$x=\pm \sqrt{y}$ which may not be in N
Example:
If $y=2,x=\pm \sqrt{2}$ which is not in N .
So,f is not a surjection.
The condition of bijection is as we know the function should be one-one and onto.
Hence for this, f is not bijection.

Functions exercise 2.1 question 5 (iii)

Injective but not surjective.
Given:
$f:N\to N$ given by$f(x)=x^3$ .
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
For any function to be bijective, the given function be one-one and onto.
Solution:
Let us check for the given function is injection, surjection and bijection.
Let's start with the injection test.
Let x and y be any two elements in the domain (N) such that$f(x)=f(y)$
$\begin{array}{rl}& f(x)=x^3,f(y)=y^3\\ & f(x)=f(y)\\ & x^3=y^3\\ & x=y\end{array}$
So,f is an injection
Surjection test:
Let y be any element in the codomain (N) , such that$f(x)=y$ for some element x in N .
$f(x)=y$
$x^3=y$
$x=\sqrt[3]{y}$ which may not be in N
So,f is not a surjection.
Here f is injective but not surjective, so the f is not a bijective.
The condition of bijection is as we know the function should be one-one and onto.
Hence for this, f is not bijection.

Functions exercise 2.1 question 5 (iv)

Injective but not surjective.
Given:
$f:Z\to Z$ given by$f(x)=x^3$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
For any function to be a bijective, the given function be one-one and onto.
Solution:
Let us check for the given function is injection, surjection and bijection.
Let's start with the injection test.
Let x and y be any two elements in the domain (N) such that$f(x)=f(y)$
$\begin{array}{rl}& f(x)=x^3,f(y)=y^3\\ & f(x)=f(y)\\ & x^3=y^3\\ & x=y\end{array}$
So,f is an injection
Surjection test:
Let y be any element in the codomain (N) , such that$f(x)=y$ for some element x in N .
$f(x)=y$
$x^3=y$
$x=\sqrt[3]{y}$ which may not be in N
So,f is not a surjection.
Here f is injective but not surjective, so the f is not a bijection.
The condition of bijection is as we know the function should be one-one and onto.
Hence for this, f is not bijection.

Functions exercise 2.1 question 5 (v)

Neither an injection nor a surjection.
Given:
$f:R\to R$, defined by $f(x)=\left|x\right|$.
Hint:
Injective function means every element in the domain has a distinct image in the co-domain.
Surjective
function means every element in the co-domain has at least one pre image in the domain of function.
Bijection ⇒ Function should fulfill the injection, surjection condition.
Solution:
Let us check if the given function is injective, surjective, or bijective.
Injection:
Let x and y be any two elements in the domain (R) such that$f(x)=f(y)$
$\begin{array}{rl}& f(x)=|x|,f(y)=|y|\\ & x=\pm y\end{array}$
Therefore f is not an injection.
Surjection test:
Let y be any element in the codomain (R) , such that$f(x)=y$ for some element x in R .
$\begin{array}{rl}& f(x)=y\\ & |x|=y\\ & x=\pm y\in Z\end{array}$
Therefore, f is surjective.
Since f is not an injective , but surjective. So the f is not bijective.

Functions exercise 2.1 question 5 (vi)

Neither injective nor surjective.
Given:
$f:Z\to Z$ defined by $f(x)=x^2+x$.
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
Bijective ⇒ Function should fulfill the injective, surjective condition.
Solution:
Let us check if the given function is an injective, surjective, bijective.
Injection:
Let x,y be any two elements in domain $f(x)=f(y)$.
$f(x)=x^2+x,f(y)=y^2+y$
Here we cannot say that x=y.
For example,
$x=2$ and $y=-3$
Then $x^2+x=2^2+2=6$
$y^2+y=(-3{)}^2-3=6$
Therefore, we have two number 2 and -3 in the domain z whose image is same as 6.
So f is not an injective.
Surjection:
Let y be any element in co-domain (R) such that $f(x)=(y)$ for some element x in (R)
$f(x)=(y)$
$x^2+x=y$
Here we can’t say $x\in Z$.
$\begin{array}{rl}& y=-4\\ & x^2+x=-4\\ & x^2+x+4=0\text{ , on solving this }\\ & x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\text{ (formula) }\\ & a=1,b=1,c=4\\ & x=\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot 4}}{2\cdot 1}\\ & x=\frac{-1\pm \sqrt{-15}}{2\cdot 1}\\ & x=\frac{-1\pm i\sqrt{15}}{2\cdot 1}\text{ which is not in }Z\end{array}$
So, f is not surjective and f is not a bijective.

Functions exercise 2.1 question 5 (vii)

Bijective.
Given:
As $f:Z\to Z$ defined by $f(x)=x-5$
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
Bijection ⇒ Function should fulfill the injection, surjection condition.
Solution:
Injection test:
Let $x,y$ be any two elements in domain(Z) such that $f(x)=(y),f(x)=x-5,f(y)=y-5$
$f(x)=f(y)$
$x-5=y-5$
$x=y$
So, f is an injection.
Surjection test:
Let y be any element in the co-domain(Z) such that $f(x)=y$ for some element x in Z( domain)
$f(x)=y$
$x-5=y$
$x=y+5$ which is in Z.
So, Z is a surjection and f is a bijection.

Functions exercise 2.1 question 5 (viii)

Neither injective nor surjective.
Given:
$f:R\to R$ defined by $f(x)=\sin x$.
Hint:
One-one function means every element in the domain has a distinct image in the co-domain.
Onto function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Injection test
Let x and y be any two elements in the domain (R) such that$f(x)=f(y)$
$\sin x=\sin y$
Here x may not be equal to y because $\sin 0=\sin \pi =0$
So, f is not an injection.
Surjection test:
Range of $f=[-1,1]$
Co-domain of $f=R$
Both are not the same.
So f is not a surjection and f is not a bijection

Functions exercise 2.1 question 5 (ix)

Bijective
Given:
$f:R\to R$ defined by $f(x)=x^3+1$.
Hint:
Bijection: function should be one-one and onto function.
Solution:
Injection test:
Let x and y be any two elements in the domain (R) such that$f(x)=f(y)$
$x^3+1=y^3+1$
$x^3=y^3$
$x=y$
So, f is an injection.
Surjection test:
Let y be any element in the codomain (N) , such that$f(x)=y$ for some element x in R .
$f(x)=y$
$x^3+1=y$
$x^3=y-1$
$x=\sqrt[3]{(y-1)}\in R$
So, f is a surjection.
So, f is a bijection.

Functions exercise 2.1 question 5 (x)

Surjective but not injective
Given:
$f:R\to R$, defined by $f(x)=x^3-x$
Hint:
Injective function means every element in the domain has a distinct image in the co-domain.
surjective function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Injective test:
Let x and y be any two elements in the domain (R), such that
$f(x)=f(y)$
$x^3-x=y^3-y$
Here we cannot say $x=y$
For example $x=1,y=1$
$x^3-x=1-1=0$
$y^3-y=(-1{)}^3-(-1)=-1+1=0$
So, -1 and 1 have the same image 0.
Thus f is not an injection.
Surjection test:
Let x and y be any two elements in the domain (R) such that$f(x)=f(y)$ for some element (R) in R.
$f(x)=f(y)$
$x^3-x=y$
In order to identity surjection,
$\begin{array}{rl}& f(x)=x^3-x\\ & f^{\prime }(x)=3x^2-1=0\\ & x=\pm \frac{1}{\sqrt{3}}\end{array}$
Therefore, f is a surjective.
Therefore, f is not a bijective.

Functions exercise 2.1 question 5 (xi)

Neither injective nor surjective.
Given:
$f:R\to R$ defined by $f(x)={\sin }^{2}x+{\cos }^{2}x$.
Hint:
Injective (one-one): function means every element in the domain has a distinct image in the co-domain.
Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Injection condition:
$f(x)={\sin }^{2}x+{\cos }^{2}x$
${\sin }^{2}x+{\cos }^{2}x=1$ ( by formula)
So,$f(x)=1$ for every x in R.
For all elements in the domain the image is 1. So,f is not an injective.
Surjection:
Range of $f=\left\{1\right\}$, co-domain of f=R
Both are not the same.
So, f is not a surjection and f is not a bijection.

Functions exercise 2.1 question 5 (xii)

Injective but not surjective
Given:
$f:Q-\{3\}\to Q,\text{ defined by }f(x)=\frac{2x+3}{x-3}$
Hint:
Injective (one-one): function means every element in the domain has a distinct image in the co-domain.
Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Injection:
let x,y be any two elements in the domain$(Q-\{3\})$, such that $f(x)=f(y)$.
$\begin{array}{rl}& f(x)=\frac{(2x+3)}{(x-3)},f(y)=\frac{(2y+3)}{(y-3)}\\ & f(x)=f(y)\\ & \frac{(2x+3)}{(x-3)}=\frac{(2y+3)}{(y-3)}\\ & \begin{array}{c}(2x+3)(y-3)=(2y+3)(x-3)\\ 2xy-6x+3y-9=2xy-6y+3x-9\\ 9x=9y\\ x=y\end{array}\end{array}$
So, f is an injection.
Subjection test:
Let y be any element in the co-domain Q, such that $f(x)=y$ for some element x in ($(Q-\{3\})$ (domain).
$\begin{array}{rl}& f(x)=y\\ & \frac{(2x+3)}{(x-3)}=y\\ & 2x+3=xy-3y\\ & 2x-xy=-3y-3\\ & x(2-y)=-3(y+1)\\ & x=-3(y+1)/(2-y),\text{ which is not defined at }y=2\end{array}$
So, f is not surjective.
Therefore, f is not bijective

Functions exercise 2.1 question 5 (xiii)

Injective
Given:
$f:Q\to Q$ defined by $f(x)=x^3+1$
Hint:
Injective (one-one): function means every element in the domain has a distinct image in the co-domain.
Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Injection: Let x,y be any two element in the domain Q,
such that $f(x)=f(y)$.
$\begin{array}{rl}& x^3+1=y^3+1\\ & x^3=y^3\\ & x=y\end{array}$
Therefore, f is an injection.
Surjection: Let y be any two element in the co-domain Q, such that $f(x)=y$ for some element x in Q.
$\begin{array}{rl}& f(x)=y\\ & x^3+1=y\\ & x^3=y-1\\ & x=\sqrt[3]{(y-1)}\end{array}$ which may not be Q.
So, f is not surjective.
Therefore, f is not bijective.

Functions exercise 2.1 question 5 (xvi)

Bijective
Given:
$f:R\to R$, defined by $f(x)=5x^3+4$
Hint:
Bijective function should be one-one and onto function.
Solution:
Let x and y be any two elements in the domain(R), such that $f(x)=f(y)$.
$\begin{array}{rl}& \Rightarrow 5x^3+4=5y^3+4\\ & \Rightarrow 5x^3=5y^3\\ & \Rightarrow x^3=y^3\\ & \Rightarrow x=y\end{array}$
So, f is an injection.
Subjection test:
Let y be any element in the co-domain(R), such that $f(x)=y$.
$\begin{array}{rl}& 5x^3+4=y\\ & x^3=\frac{(y-4)}{5}\in R\end{array}$
So, f is a surjective.
Hence, f is a bijective.

Functions exercise 2.1 question 5 (xv)

Bijective
Hint:
$f:R\to R$, defined by $f(x)=3-4x$.
Hint:
Bijective function should be one-one and onto function.
Solution:
Let
$\begin{array}{rl}& x_1,x_2\in R\\ & f\left(x_1\right)=3-4x_1,f\left(x_2\right)=3-4x_2\\ & f\left(x_1\right)=f\left(x_2\right)\\ & 3-4x_1=3-4x_2\\ & x_1=x_2\end{array}$
Hence f is an injective.
Subjection test:
Let
$\begin{array}{rl}& y=3-4x\\ & x=\frac{3-y}{4}\\ & \therefore \\ & \mathrm{\forall }y\in R,\mathrm{\exists }\frac{3-y}{4}\in R\\ & f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-4}{4}\right)=y\end{array}$
$\therefore f$ is onto
Hence, f is bijection.

Functions exercise 2.1 question 5 (xvi)

Neither injective nor surjective.
Given:
$f:R\to R$ defined by $f(x)=1+x^2$
Hint:
Injective (one-one): function means every element in the domain has a distinct image in the co-domain.
Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Injection condition:
Let x,y be any two elements in domain R such that
$\begin{array}{rl}& f(x)=f(y),f(x)=1+x^2,f(y)=1+y^2\\ & 1+x^2=1+y^2\\ & x^2=y^2\\ & x=\pm y\end{array}$
Therefore, f is not an injection.
Surjection test:
Let y be any two elements in domain (R) such that $f(x)=y$ for some element x in R.
$\begin{array}{rl}& f(x)=y\\ & 1+x^2=y\\ & x^2=y-1\\ & x=\sqrt[2]{(y-1)}\\ & \text{ For, }y=O\in R\\ & x=\pm \sqrt{-1}=\pm i\text{ is not in }R\text{ . }\end{array}$
So, f is not a surjection and f is not a bijection

Functions exercise 2.1 question 5 (xvii)

Neither injective nor surjective.
Given:
$f:R\to R$ defined by $f(x)=\frac{x}{1+x^2}$
Hint:
Injective (one-one): function means every element in the domain has a distinct image in the co-domain.
Surjective(Onto): function means every element in the co-domain has at least one pre image in the domain of function.
Solution:
Let x and y be any two elements in domain (R), such that f(x) = f(y).
$\frac{x}{1+x^2}=\frac{y}{1+y^2}$
xy²+ x = x²y + y
xy² − x²y + x − y = 0
−xy(−y + x) + 1(x − y) = 0
(x − y) (1 – xy) = 0
x = y or x = $\frac{1}{y}$
So, f is not an injection.
Surjection test:
Let y be any element in co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
$\frac{x}{1+x^2}=y$
yx² – x + y = 0
x((-1) ± √(1 - 4x²))/(2y) if y ≠ 0
= (1 ± √(1 - 4y²))/(2y), which may not be in R
For example, if y = 1, then (1 ± √(1-4))/(2y) = (1 ± i√3)/2, which is not in R Therefore, f is not surjection and f is not bijection.

Functions exercise 2.1 question 6

One
Given:
$F:A\to B$ is an injection such that the range of$f=\left\{a\right\}$
Hint:
For an injection there will be exactly one image for each element.
Solution:
Here $f=\left\{a\right\}$
⇒ Since F is an injection, there will be exactly one image for each element of f.
⇒ So, the number of images of f is one.
$\therefore$ Number of element in A is one.

Functions exercise 2.1 question 8 (i)

f is not bijective.
Given:
$f:A\to A$, given by$f(x)=\frac{x}{2}$
To prove:
Here we have to show that the given function is one-one and onto.
Solution:
For one-one:
Let $x_1$ and $x_2$ be any two elements in the domain A, such that
$\begin{array}{rl}& f\left(x_1\right)=f\left(x_2\right)\\ \Rightarrow & \frac{x_1}{2}=\frac{x_2}{2}\\ \Rightarrow & x_1=x_2\end{array}$
So,f is one-one.
For onto:
Let y be any element in the codomain A, such that $f(x)=(y)$ for some element x in domain A
Then, $f(x)=(y)$
$\Rightarrow \frac{x}{2}=y$ $[f(x)=\frac{x}{2}]$
$\Rightarrow x=2y$, which may not be in A=[-1, 1]
So f is not onto.
Hence,f is not bijective.

Functions exercise 2.1 question 8 (ii)

f is not bijective
Given:
$g:A\to A$, given by$g(x)=\left|x\right|$
To prove:
Function f is one-one, onto or bijective.
Hint:
For any function to be a bijective, the given function should be one-one and onto.
Solution:
First we will check whether the given function is one-one or not.
We know for one-one function
$\begin{array}{rl}& \Rightarrow g\left(x_1\right)=g\left(x_2\right)\\ & \Rightarrow \left|x_1\right|=\left|x_2\right|\\ & \Rightarrow x_1=\pm x_2\end{array}$
So, g is not one-one.
$g(x)=\left|x\right|$
Let $g(x)=y$such that$y\in A$
$y=|x|$
Hence, the value of y is defined only if y is positive,
But y is a real number
Hence, if y is negative, there is not corresponding element of x
i.e; $y=-x$, there is no value of x inA
Hence, f is not onto.
So, g is not onto.
Hence g is not bijective.

Functions exercise 2.1 question 8 (iii)

f is not bijective.
Given:
$h:A\to A$, given by$h(x)=(x{)}^2$
To prove:
Function h is one-one, onto or bijective.
Hint:
For any function to be bijective, the given function should be one-one and onto.
Solution:
First we will check whether the given function is one-one or not.
We know for function h to be one-one
$\begin{array}{rl}& \begin{array}{rl}& h\left(x_1\right)=h\left(x_2\right)\\ \Rightarrow & x_1^2=x_2^2\\ \Rightarrow & x_1=\pm x_2\end{array}\\ & \text{ So, }h\text{ is not one-one. }\end{array}$
$h(x)=x^2$
Let $h(x)=y$ such that$y\in A$
$y=x^2$
Hence, the value of y is defined only if y is positive,
But y is a real number
Hence, if y is negative, there is not corresponding element of x
i.e; for $y=-1$ , there is no value of x in A.
Hence, h is not onto.
Hence, h is not bijective.

Functions exercise 2.1 question 9 (i)

f is surjective.
Let$f=\left\{(x,y)\right\}:x$ is a person, y is the mother of x
Hint:
As for each element x in the domain set, there is a unique element y in the co-domain set.
So, f is a function.
Solution:
First we check for injection or one-one.
We know: For one-one,for each element x in the domain set, there should be a unique element y in the co-domain set.
But here,y can be the mother of two or more persons.
So, f is not injective.
Again for surjective,
For every mother y defined by (x,y) , there exists a person x for whom y is mother.
So, f is surjective.
Hence, f is surjective function.

Functions exercise 2.1 question 9 (ii)

Let$g:\left\{(a,b)\right\}:a$ is person and b is an ancestor of a
To prove:
The mapping is injective, surjective or not.
Hint:
As for each element x in the domain set, there is a unique related element y in the co-domain set.
So, f is a function.
Solution:
Since the ordered map (a,b) is not a map because "a" can have many ancestors. So the codomain set is not unique.
So, g is not a function.
Note:
If the function does not exist then we can’t show that the function is an injective or surjective.

Functions exercise 2.1 question 10

Given $A=\{1,2,3\}$
Number of element in A =3
We know that,
Number of one-one functionnumber of ways of arranging 3 elements.
=3! = 6
Hence, the number of one-one function as following:
$$\begin{gathered} (i)\{(1,1),(2,2),(3,3)\} \\ (ii)\{(1,1),(2,3),(3,2)\} \\ (iii)\{(1,2),(2,1),(3,3)\} \\ (iv)\{(1,2),(2,3),(3,1)\} \\ (v)\{(1,3),(2,2),(3,1)\} \\ (vi)\{(1,3),(2,1),(3,2)\} \end{gathered}$$

Functions exercise 2.1 question 11

f is bijection.
Given:
$f:R\to R$ is a function defined by $f(x)=4x^3+7$.
To prove:
f is bijection.
Hint:
Any function to be bijection. The given function should be one-one and onto.
Solution:
First we will check whether the function is one-one or not.
Let x and y be any two element in the domain R such that
$\begin{array}{rl}& f(x)=f(y)\\ \Rightarrow & 4x^3+7=4y^3+7\\ \Rightarrow & 4x^3=4y^3\\ \Rightarrow & x^3=y^3\\ \Rightarrow & x=y\end{array}$
So, f is one-one.
Now, we will check if the given function is onto or not.
Let y be any element in the co-domain R such that $f(x)=y$ for some element x in R (domain).
$f(x)=y$
$\begin{array}{rl}& \Rightarrow 4x^3+7=y\\ & \Rightarrow 4x^3=y-7\\ & \Rightarrow x=\sqrt[3]{y-7}\end{array}$
So, for every element in the co-domain there exists some pre-image in the domain.
So, f is onto.
Hence f is bijection.

Functions exercise 2.1 question 12

Given:
$f:R\to R$, given by$f(x)=e^x$.
To prove:
f is one-one but not onto.
Hint:
If different element in A have distinct images in B then the function is one-one and if every element in B has at least one pre-image in A then the function is onto.
Solution:
First we will check whether the function is one-one or not.
Here, let $x,y\in R$ such that
$f(x)=f(y)$
$\begin{array}{ll}\Rightarrow & e^x=e^y\\ \Rightarrow & e^{x-y}=1=e^0\\ \Rightarrow & x-y=0\\ \Rightarrow & x=y\end{array}$
$\therefore f$ is one-one
Now, we check onto
$f(x)=e^x$
Let $f(x)=y$ such that $y\in R$
$y=e^x$
clearly the range of$e^x\text{ is }(0,\mathrm{\infty })=R^{+}$ but the given range isR
co-domain≠ Range
$\therefore f$ is not onto.
Note:
When co-domain is replaced by$R_0^{+}$, i.e. $(0,\mathrm{\infty })$ then f become onto function.

Functions exercise 2.1 question 13

Given:
$f:R_0^{+}\to R$ given by $f(x){\log }_{a}x:a>0$.
To prove:
$f(x){\log }_{a}x:a$ is bijection.
Hint:
For any function to be bijective, the function should be one-one and onto.
Solution:
Let $x,y\in R_0^{+}$ such that
$f(x)=f(y)$
$\begin{array}{ll}\Rightarrow & {\log }_{a}x={\log }_{a}y\\ \Rightarrow & {\log }_a\frac{x}{y}=0\\ \Rightarrow & \frac{x}{y}=1\\ \Rightarrow & x=y\end{array}$
$\therefore f$ is one-one.
Now, let $y\in R$ be arbitrary then
$f(x)=y$
$\begin{array}{rl}& \Rightarrow {\log }_{a}x=y\\ & \Rightarrow x=a^y\in R_0^{+}\\ & [\because a>0\Rightarrow a^y>0]\end{array}$
Thus for all $y\in R$, there exist $x=a^y$ such that $f(x)=y$.
$\therefore f$ is onto
$\therefore f$ is one-one and onto.
Hence f is bijective.

Functions exercise 2.1 question 14

Given:
$A=\{1,2,3\}$
To prove:
One-one function must be onto.
Solution:
Here $f:A\to A$ and $A=\{1,2,3\}$
We know that, since f is one-one, so the three elements of $\{1,2,3\}$ must be taken to different element of the co-domain $\{1,2,3\}$ under f.
Hence f must be onto.

Functions exercise 2.1 question 15

Given:
$f:A\to A$ , given by$A=\{1,2,3\}$.
To prove:
Onto function must be one-one.
Solution:
Let f is not one-one then there exists two elements say 1 and 2 in the domain having same image in the co-domain.
Also, the image of 3 under f can be only one element.
$\therefore$ The range set will have at most two element of the co-domain, but we know that in the range set we have three elements $\{1,2,3\}$.
i.e. There arises a contradiction that f is not onto function.
Hence f must be one-one.

Functions exercise 2.1 question 16

Given:
Here given that $\{1,2,3....n\}$ is the set.
Here we have find all onto function from the given set.
Hint:
Since if f is onto then all element have a unique pre image.
Solution:
Taking the set $\{1,2,3\}$
$\therefore$ Total number of one-one function $3\times 2\times 1=6$
As example since f is onto, all elements of $\{1,2,3\}$ have unique pre-image



So as the diagram, we getSo as the diagram, we get $n\times (n-1)\times (n-2)\times \dots \times 2\times 1=n!$