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### Course: Class 12 math (India)>Unit 1

Lesson 8: Solutions to select NCERT problems

# Select problems from exercise 1.2

solutions ncert selected problems.
In this article we will look at solutions of a few selected problems from exercise 1.2 of NCERT.
Problem 1:
Show that the function $f:{\mathbb{R}}_{\ast }\to {\mathbb{R}}_{\ast }$ defined by $f\left(x\right)=\frac{1}{x}$ is one-one and onto, where ${\mathbb{R}}_{\ast }$ is the set of all non-zero real numbers. Is the result true, if the domain ${\mathbb{R}}_{\ast }$ is replaced by $\mathbb{N}$ with co-domain being same as ${\mathbb{R}}_{\ast }$?
Solution:
A function is one-one if no two different inputs give the same output.
Let's check some of the function outputs.
$f\left(x\right)$$\frac{1}{2}$$\frac{1}{7}$$\frac{-1}{3}$
$x$$2$$7$$-3$
In the second row, try to look for a different input which gives the same output in the above row. See that it is impossible to do so.
Let us now generally show that $f$ is one-one. Let ${x}_{1}$ and ${x}_{2}$ be two inputs having the same output.
So ${x}_{1}={x}_{2}$, meaning inputs are the same. Thus for the same output, we cannot have two different inputs. $f$ is one-one.
Now, a function is onto if its range is equal to its co-domain. In this case $f$ is onto if the function output takes all non-zero real numbers.
Let's check the graph of $f\left(x\right)=\frac{1}{x}$.
$f$ takes all real numbers except zero. On both sides, the graph tends to zero, but never actually touches zero.
To show $f$ is onto algebraically, take any $y\in {\mathbb{R}}_{\ast }$. Pick $x=\frac{1}{y}$. Because $y$ is a non-zero real number, $x$ is also a non-zero real number, or $x\in {\mathbb{R}}_{\ast }$.
$f\left(x\right)=f\left(\frac{1}{y}\right)=y$
Hence for every $y\in {\mathbb{R}}_{\ast }$, there exists some $x\in {\mathbb{R}}_{\ast }$ such that $f\left(x\right)=y$. Therefore, $f$ is onto.
Now, for the second part of the question, let us consider $f:\mathbb{N}\to {\mathbb{R}}_{\ast }$, $f\left(x\right)=\frac{1}{x}$.
As far as one-one nature is concerned, nothing changes here. Earlier no two different real number inputs gave the same output. And now, no two different natural number inputs will give the same output. So, $f$ remains one-one.
The main question is whether $f$ remains onto as well. Because $x\in \mathbb{N}$, $x=1,2,3,\dots$. So $f\left(x\right)=1,\frac{1}{2},\frac{1}{3},\dots$.
See that $f\left(x\right)$ does not take any negative real number. Also, many positive real numbers like those between $\frac{1}{2}$ and $1$ are left out. So, $f$ is not onto in this case.
Problem 2:
Let $A$ and $B$ be sets. Show that $f:A×B\to B×A$ such that $f\left(a,b\right)=\left(b,a\right)$ is a bijective function.
Solution:
A function is bijective if it is both one-one and onto.
Let us first see an example. Let $A=\left\{1,2\right\}$ and $B=\left\{p,q,r\right\}$.
$\begin{array}{rl}A×B& =\left\{\left(1,p\right),\left(1,q\right),\left(1,r\right),\left(2,p\right),\left(2,q\right),\left(2,r\right)\right\}\\ \\ B×A& =\left\{\left(p,1\right),\left(q,1\right),\left(r,1\right),\left(p,2\right),\left(q,2\right),\left(r,2\right)\right\}\end{array}$
See that $f$ will map each element in the first line above to the element exactly below it. No two different inputs have the same output and all elements of $B×A$ are covered. For this example, the function is one-one and onto.
Let us now generally show that $f$ is one-one. Let $\left({a}_{1},{b}_{1}\right)$ and $\left({a}_{2},{b}_{2}\right)$ be two different inputs having the same output.
$\begin{array}{rl}& f\left({a}_{1},{b}_{1}\right)=f\left({a}_{2},{b}_{2}\right)\\ \\ & \left({b}_{1},{a}_{1}\right)=\left({b}_{2},{a}_{2}\right)\end{array}$
Now, recall two ordered pairs are equal if and only if their first and second elements are respectively equal. So, ${b}_{1}={b}_{2}$ and ${a}_{1}={a}_{2}$.
This means the two inputs are the same and our initial assumption is false. For the same output, we cannot have two different inputs. $f$ is one-one.
To show $f$ is onto, consider the number of elements in $A×B$ and $B×A$.
$n\left(A×B\right)=n\left(B×A\right)=n\left(A\right)×n\left(B\right)$
Now, $f:A×B\to B×A$ is a one-one function where the domain and co-domain have the same number of elements. So, $f$ must be onto.
To conclude, $f$ is a bijective function.
Problem 3:
Let $f:\mathbb{N}\to \mathbb{N}$ be defined by for all $n\in \mathbb{N}$.
State whether the function $f$ is bijective. Justify your answer.
Solution:
For $f$ to be bijective, $f$ should be both one-one and onto.
$f$ will be one-one if no two different inputs give the same output. Also $f$ will be onto if its range covers the entire set of natural numbers.
Let's check the behaviour of the function.
$n$$1$$2$$3$$4$$\dots$
oddevenoddeven$\dots$
$f\left(n\right)$$1$$1$$2$$2$$\dots$
Clearly, $f$ is not one-one. $f\left(1\right)=f\left(2\right)=1$.
But see that $f$ is onto. If we see the pattern in the second row, we find that all natural numbers will be covered by $f$.
$f$ is onto but not one-one. So, $f$ is not bijective.
Problem 4:
Let $A=\mathbb{R}-\left\{3\right\}$ and $B=\mathbb{R}-\left\{1\right\}$. Consider the function $f:A\to B$ defined by $f\left(x\right)=\frac{x-2}{x-3}$. Is $f$ one-one and onto? Justify your answer.
Solution:
$f$ will be one-one if no two different inputs give the same output. Also $f$ will be onto if its range is equal to $\mathbb{R}-\left\{1\right\}$, i.e. the function output takes all real numbers except $1$.
It is a little difficult to understand the behaviour of the function the way it is written. However, if we manipulate the expression a little bit, we can write $f$ in a much simpler form.
$\begin{array}{rl}f\left(x\right)& =\frac{x-2}{x-3}\\ \\ & =\frac{x-3+1}{x-3}\\ \\ & =\frac{x-3}{x-3}+\frac{1}{x-3}\\ \\ & =1+\frac{1}{x-3}\end{array}$
Now $\frac{1}{x-3}$ is one-one for $x\in \mathbb{R}-\left\{3\right\}$.
So, $f\left(x\right)$ is one-one.
Also, $\frac{1}{x-3}$ takes all real values except $0$. Its behaviour is similar to $\frac{1}{x}$. Here's the graph.
So, $f\left(x\right)=1+\frac{1}{x-3}$ takes all real values except $1$.
Range of $f=\mathbb{R}-\left\{1\right\}$. Thus $f$ is onto.

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