## NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 - FREE PDF Download

The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 Relation and Functions provides complete solutions to the problems in the Exercise. These NCERT Solutions are intended to assist students with the CBSE Class 10 board examination.

Students should thoroughly study this NCERT solution in order to solve all types of questions based on Relation and Functions. By completing these practice questions with the NCERT Maths Solutions Chapter 1 Exercise 1.2 Class 12, you will be better prepared to understand all of the different types of questions that may be asked in the class 12 maths ex 1.2 board exams.

## Glance on NCERT Solutions Maths Chapter 1 Exercise 1.2 Class 12 | Vedantu

Chapter 1, Exercise 1.2 of Class 12 Maths textbook likely deals with advanced topics related to Relations and Functions, building on the concepts introduced earlier in the chapter.

Explore different types of relations beyond basic ones, like reflexive, symmetric, and transitive relations. Concepts like equivalence relations might be introduced.

The focus will shift towards different properties of functions. Solving problems to determine if a function is one-one (injective), onto (surjective), or both (bijective).

Composition of Functions involves problems where you combine two functions and analyze the resulting composite function.

Questions related to identifying the domain (input values) and range (output values) for functions and composite functions.

Class 12 ex 1.2 Maths NCERT Solutions has over all 12 Questions.

## Access PDF for Maths NCERT Chapter 1 Relation and Functions Exercise 1.2 Class 12

Exercise 1.2

1. Show that the function ${\text{f}}:{{\mathbf{R}}_*} \to {{\mathbf{R}}_*}$ defined by ${\text{f(}}x) = \dfrac{1}{x}$ is one-one and onto, where ${{\mathbf{R}}_*}$ is the set of all non-zero real numbers. Is the result true, if the domain ${{\mathbf{R}}_*}$ is replaced by ${\text{N}}$ with co-domain being same as ${{\mathbf{R}}_*}$?

Ans: Given that, f: ${{\text{R}}^*} \to {R_*}$ is defined by ${\text{f}}(x) = \dfrac{1}{x}$.

Consider $x,\,\,y \in R*$ such that ${\text{f}}(x) = {\text{f}}(y)$

$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{y}$

$ \Rightarrow x = y$

Thus, ${\text{f}}$ is one-one.

It is clear that for $y \in R*$, there exists $x = \dfrac{1}{y} \in R*[$ as $y \ne 0]$ such that

${\text{f}}(x) = \dfrac{1}{{\left( {\dfrac{1}{y}} \right)}} = {\text{y}}$

Thus, ${\text{f}}$ is onto.

Therefore, the given function ${\text{f}}$ is one-one and onto.

Now, consider function ${\text{g}}:{\text{N}} \to {{\text{R}}_*}$ defined by ${\text{g}}(x) = \dfrac{1}{x}$

We have, $g\left( {{x_1}} \right) = g\left( {{x_2}} \right)\quad \Rightarrow \, = \dfrac{1}{{{x_1}}} = \dfrac{1}{{{x_2}}}$

$ \Rightarrow {x_1} = {x_2}$

Thus, \[{\text{g}}\] is one-one.

It is clear that ${\text{g}}$ is not onto as for $1.2 \in = {{\text{R}}_*}$, there does not exist any $x$ in ${\text{N}}$ such that

${\text{g}}(x)$ $ = \dfrac{1}{{1.2}}$

Therefore, the function ${\text{g}}$ is one-one but not onto.

2. Check the injectivity and surjectivity of the following functions:

**(i) ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^2}$**

Ans: Here, ${\text{f}}\,{\text{:}}\,{\text{N}} \to {\text{N}}$ is given by ${\text{f}}(x) = {x^2}$

For $x,y \in N,\,\,$

${\text{f}}(x) = {\text{f}}(y) \Rightarrow {x^2} = {y^2} \Rightarrow x = y$

Thus, ${\text{f}}$ is injective.

Now, $2 \in {\text{N}}$. But, there does not exist any $x$ in ${\text{N}}$ such that ${\text{f}}(x) = {x^2} = 2$

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is injective but not surjective.

**(ii) ${\text{f}}:{\text{Z}} \to {\text{Z}}$ given by ${\text{f}}(x) = {x^2}$**

Ans: Here, ${\text{f}}:{\text{Z}} \to {\text{Z}}$ is given by ${\text{f}}(x) = {x^2}$

It is seen that ${\text{f}}( - 1) = {\text{f}}(1) = 1$, but $ - 1 \ne 1$.

Thus, ${\text{f}}$ is not injective.

Now, $ - 2 \in {\text{Z}}$. But, there does not exist any element ${\text{x}} \in {\text{Z}}$ such that

$f(x) = - 2$ or ${x^2} = - 2$

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is neither injective nor surjective.

**(iii) f: ${\text{R}} \to {\text{R}}$ is given by ${\text{f}}(x) = {x^2}$**

Ans: Here, f: ${\text{R}} \to {\text{R}}$ is given by ${\text{f}}(x) = {x^2}$

It is seen that ${\text{f}}( - 1) = {\text{f}}(1) = 1$, but $ - 1 \ne 1$.

Thus, ${\text{f}}$ is not injective.

Now, $ - 2 \in {\text{R}}$. But, there does not exist any element $x \in {\text{R}}$ such that ${\text{f}}(x) = - 2$

or ${x^2} = - 2$.

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is neither injective nor surjective.

**(iv) ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^3}$**

Ans: Here, ${\text{f}}:{\text{N}} \to {\text{N}}$ given by ${\text{f}}(x) = {x^3}$

For $x,y \in N,$

$f(x) = f(y) \Rightarrow {x^3} = {y^3} \Rightarrow x = y$

Thus, ${\text{f}}$ is injective.

Now, $2 \in {\text{N}}$. But, there does not exist any element $x \in {\text{N}}$ such that

$f(x) = 2$ or ${x^3} = 2$

Thus, ${\text{f}}$ is not surjective

Therefore, function ${\text{f}}$ is injective but not surjective.

**(v) ${\text{f}}\,{\text{:}}\,{\text{Z}} \to {\text{Z}}$ is given by $f(x) = {x^3}$**

Ans: Here, ${\text{f}}\,{\text{:}}\,{\text{Z}} \to {\text{Z}}$ is given by $f(x) = {x^3}$

For $x,y \in {\text{Z}},$

${\text{f}}(x) = {\text{f}}(y) \Rightarrow {x^3} = {y^3} \Rightarrow x = y$

Thus, ${\text{f}}$ is injective.

Now, $2 \in {\text{Z}}$. But, there does not exist any element $x \in {\text{Z}}$ such that

${\text{f}}(x) = 2$ or ${x^3} = 2$

Thus, ${\text{f}}$ is not surjective.

Therefore, the function ${\text{f}}$ is injective hut not surjective.

3. Prove that the Greatest Integer Function ${\text{f}}:{\text{R}} \to {\text{R}}$ given by ${\text{f}}(x) = [x]$, is neither one

one nor onto, where \[[x]\]denotes the greatest integer less than or equal to $x$.

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is given by, ${\text{f}}(x) = [x]$

It is seen that ${\text{f}}(1.2) = [1.2] = 1,{\text{f}}(1.9) = [1.9] = 1$.

Then, ${\text{f}}(1.2) = {\text{f}}(1.9)$, but $1.2 \ne 1.9$

Thus, ${\text{f}}$ is not one-one.

Consider $0.7 \in {\text{R}}$

It is known that ${\text{f}}(x) = [x]$ is always an integer.

Thus, there does not exist any element $x \in {\text{R}}$ such that ${\text{f}}(x) = 0.7$

Therefore, ${\text{f}}$ is not onto.

Hence, the greatest integer function is neither one-one nor onto.

4. Show that the Modulus Function ${\text{f}}:{\text{R}} \to {\text{R}}$ given by ${\text{f}}(x) = |x|$, is neither one-one nor

onto, where \[|x|\] is $x$, if ${\text{x}}$ is positive or 0 and $|x|$ is $ - x$, if $x$ is negative.

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is given by \[{\text{f}}(x) = |x| = \left\{ {\begin{array}{*{20}{l}} x&{{\text{ if }}x \geqslant 0} \\ { - x}&{{\text{ if }}x < 0} \end{array}} \right.\]

It is clear that ${\text{f}}( - 1) = | - 1| = 1$ and ${\text{f}}(1) = |1| = 1$

Now, ${\text{f}}( - 1) = {\text{f}}(1)$, but $ - 1 \ne 1$

Thus, ${\text{f}}$ is not one-one.

Now, consider $ - 1 \in {\text{R}}$.

It is known that ${\text{f}}(x) = |x|$ is always non-negative.

Thus, there does not exist any element $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = |x| = - 1$

Thus, ${\text{f}}$ is not onto.

Therefore, the modulus function is neither one-one nor onto.

5. Show that the Signum Function f: **${\text{R}} \to {\text{R}}$, given by ${\text{f}}(x) = \left\{ {\begin{array}{*{20}{l}}
1&{{\text{ if }}x > 0} \\
{0,}&{{\text{ if }}x = 0} \\
{ - 1,}&{{\text{ if }}x < 0}
\end{array}} \right.$**

is neither one-one nor onto.

Ans: Here, f: ${\text{R}} \to {\text{R}}$, given by ${\text{f}}(x) = \left\{ {\begin{array}{*{20}{l}} 1&{{\text{ if }}x > 0} \\ {0,}&{{\text{ if }}x = 0} \\ { - 1,}&{{\text{ if }}x < 0} \end{array}} \right.$

It is seen that ${\text{f}}(1) = {\text{f}}(2) = 1$, but $1 \ne 2$.

Thus, ${\text{f}}$ is not one-one.

Now, as ${\text{f}}(x)$ takes only \[3\] values $(1,\,\,0$, or $ - 1)$ for the element $ - 2$ in co-domain

${\text{R}}$, there does not exist any $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = - 2$.

Thus, ${\text{f}}$ is not onto

Therefore, the Signum function is neither one-one nor onto.

6. Let ${\text{A}} = \{ 1,2,3\} ,\,\,{\text{B}} = \{ 4,5,6,7\} $ and let ${\text{f}} = \{ (1,4),\,\,(2,5),\,\,(3,6)\} $ be a function from ${\text{A}}$ to ${\text{B}}$.

Show that ${\text{f}}$ is one-one.

Ans: Given that, ${\text{A}} = \{ 1,2,3\} $

${\text{B}} = \{ 4,5,6,7\} $

${\text{f}}\,{\text{:}}\,{\text{A}} \to {\text{B}}$ is defined as ${\text{f}} = \{ (1,4),\,\,(2,5),\,\,(3,6)\} $

Thus, ${\text{f}}(1) = 4,\,\,{\text{f}}(2) = 5,\,\,{\text{f}}(3) = 6$

It is seen that the images of distinct elements of ${\text{A}}$ under ${\text{f}}$ are distinct.

Therefore, the function ${\text{f}}$ is one-one.

**7. In each of the following cases, state whether the function is one-one,**

**onto or bijective. Justify your answer.**

(i) ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ defined by ${\text{f}}(x) = 3 - 4x$

**Ans: **Here, f: ${\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3 - 4x$.

Let ${x_1},\,\,{x_2} \in {\text{R}}$ such that ${\text{f}}\left( {{x_1}} \right) = {\text{f}}\left( {{x_2}} \right)$

$ \Rightarrow 3 - 4{x_1} = 3 - 4{x_2}$

$ \Rightarrow - 4{x_1} = - 4{x_2}$

$ \Rightarrow {x_1} = {x_2}$

Thus, ${\text{f}}$ is one-one.

For any real number \[\left( y \right)\] in ${\text{R}}$ such that ${\text{f}}\left( {\dfrac{{3 - y}}{4}} \right) = 3 - 4\left( {\dfrac{{3 - y}}{4}} \right) = y$

Thus, ${\text{f}}$ is onto.

Therefore, the function ${\text{f}}$ is bijective.

**(ii) ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 1 + {x^2}$**

**Ans:** Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ defined by ${\text{f}}(x) = 1 + {x^2}$

Let ${x_1},\,\,{x_2} \in {\text{R}}$ such that ${\text{f}}({x_1}) = {\text{f}}\left( {{x_2}} \right)$

$ \Rightarrow 1 + {({x_1})^2} = 1 + {({x_2})^2}$

$ \Rightarrow {({x_1})^2} = {({x_2})^2}$

$ \Rightarrow {x_1} = {x_2}$

Thus, ${\text{f}}\left( {{x_1}} \right) = {\text{f}}\left( {{x_2}} \right)$ does not imply that ${x_1} = {x_2}$.

For example, ${\text{f}}(1) = {\text{f}}( - 1) = 2$

Therefore, ${\text{f}}$ is not one-one.

Consider an element $ - 2$ in co-domain ${\text{R}}$.

It is seen that ${\text{f}}(x) = 1 + {x^2}$ is positive for all ${\text{x}} \in {\text{R}}$.

Thus, there does not exist any $x$ in domain ${\text{R}}$ such that ${\text{f}}(x) = - 2$.

Therefore, ${\text{f}}$ is not onto.

Hence, the function ${\text{f}}$ is neither one-one nor onto.

**8. Let ${\text{A}}$ and ${\text{B}}$ be sets. Show that ${\text{f}}:{\text{A}} \times {\text{B}} \to {\text{B}} \times {\text{A}}$ such that $(a,b) = (b,a)$ is**

**bijective function.**

**Ans:** ${\text{f}}\,{\text{:}}\,{\text{A}} \times {\text{B}} \to {\text{B}} \times {\text{A}}$ is defined as ${\text{f}}(a,b) = (b,a)$

Let $\left( {{a_1},{b_1}} \right),\,\,\left( {{a_2},{b_2}} \right) \in {\text{A \times B}}$ such that ${\text{f}}\left( {{a_1},{b_1}} \right) = {\text{f}}\left( {{a_2},{b_2}} \right)$

$ \Rightarrow \left( {{b_1},{a_1}} \right) = \left( {{b_2},{a_2}} \right)$

$ \Rightarrow {b_1} = {b_2}$ and \[{a_1} = {a_2}\]

$ \Rightarrow \left( {{a_1},\;{b_1}} \right) = \left( {{a_2},\;{b_2}} \right)$

Thus, ${\text{f}}$ is one-one.

Now, let $(b,a) \in {\text{B A}}$ be any element.

Then, there exists $(a,b) \in {\text{A B}}$ such that $f(a, b)=(b, a) .[$By definition of \[{\text{f}}]\]

Thus, ${\text{f}}$ is onto.

Therefore, the function ${\text{f}}$ is bijective.

**9. Let f: ****${\text{N}} \to {\text{N}}$ be defined by ${\text{f}}(n) = \left\{ {\begin{array}{*{20}{l}}
{\dfrac{{n + 1}}{2},}&{{\text{ if }}n{\text{ is odd }}} \\
{\dfrac{n}{2},}&{{\text{ if }}n{\text{ is even }}}
\end{array}} \right.$ for all $n \in {\text{N}}$**

**State whether the function ${\text{f}}$ is bijective. Justify your answer.**

Ans: ${\text{f}}:{\text{N}} \to {\text{N}}$ is defined as ${\text{f}}(n) = \left\{ {\begin{array}{*{20}{l}} {\dfrac{{n + 1}}{2},}&{{\text{ if }}n\,\,{\text{is}}\,{\text{odd }}} \\ {\dfrac{n}{2},}&{{\text{ if }}n{\text{ is even }}} \end{array}} \right.$ for all $n \in {\text{N}}$

It can be observed that

$f(1) = \dfrac{{1 + 1}}{2} = 1$ and $f(2) = \dfrac{2}{2} = 1$ (By definition of ${\text{f}}(n)$)

$f(1) = f(2)$, where $1 \ne 2$

Thus, ${\text{f}}$ is not one-one.

Consider a natural number \[\left( n \right)\] in co-domain \[{\text{N}}.\]

**Case 1:** $n$ is odd

Thus, $n = 2{\text{r}} + 1$ for some $r \in {\text{N}}$. Then, there exists $4r + 1 \in {\text{N}}$ such that

$f(4r + 1) = \dfrac{{4r + 1 + 1}}{2} = 2r + 1$

**Case 2:** $n$ is even

Thus, $n = 2r$ for some $r \in {\text{N}}$. Then, there exists $4r \in {\text{N}}$ such that

${\text{f}}(4r) = \dfrac{{4r}}{2} = 2r$

Therefore, ${\text{f}}$ is onto.

Hence, the function \[{\text{f}}\]is not a bijective function.

10. Let ${\text{A}} = {\text{R}} - \{ 3\} $ and ${\text{B}} = {\text{R}} - \{ 1\} .$ Consider the function f: ${\text{A}} \to {\text{B}}$ defined by ${\text{f}}(x)$

**$ = \left( {\dfrac{{x - 2}}{{x - 3}}} \right).$ Is f one-one and onto? Justify your answer.**

Ans: ${\text{A}} = {\text{R}} - \{ 3\} ,{\text{B}} = {\text{R}} - \{ 1\} $ and ${\text{f}}:{\text{A}} \to {\text{B}}$ defined by ${\text{f}}(x) = \left( {\dfrac{{x - 2}}{{x - 3}}} \right)$

Let $x,y \in $ A such that $f(x) = f(y)$

$ \Rightarrow \dfrac{{x - 2}}{{x - 3}} = \dfrac{{y - 2}}{{y - 3}}$

By cross multiplication,

$ \Rightarrow (x - 2)(y - 3) = (y - 2)(x - 3)$

Expand brackets,

$ \Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6$

$ \Rightarrow - 3x - 2y = - 2x - 3y \Rightarrow x = y$

Thus, ${\text{f}}$ is one-one.

Let $y \in {\text{B}} = {\text{R}} - \{ 1\} .$ Then, $y \ne 1$.

The function ${\text{f}}$ is onto if there exists $x \in {\text{A}}$ such that ${\text{f}}(x) = y$.

Now, $f(x) = y$

$ \Rightarrow \dfrac{{x - 2}}{{y - 3}} = y$

By cross multiplication,

$ \Rightarrow x - 2 = xy - 3y \Rightarrow x(1 - y) = - 3y + 2$

$ \Rightarrow x = \dfrac{{2 - 3y}}{{1 - y}} \in {\text{A}}$

$[y \ne 1]$

Thus, for any $y \in {\text{B}}$, there exists $\dfrac{{2 - 3y}}{{1 - y}} \in {\text{A}}$ such that

${\text{f}}\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) = \dfrac{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 2}}{{\left( {\dfrac{{2 - 3y}}{{1 - y}}} \right) - 3}}$

Take LCM,

$ = \dfrac{{2 - 3y - 2 + 2y}}{{2 - 3y - 3 + 3y}}$

$ = \dfrac{{ - y}}{{ - 1}} = y$

Thus, ${\text{f}}$ is onto.

Therefore, the function ${\text{f}}$ is one-one and onto.

11. Let ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ be defined as ${\text{f}}(x) = {x^4}$. Choose the correct answer.

**(A) ${\text{f}}$ is one-one onto**

**(B) ${\text{f}}$ is many-one onto**

**(C) ${\text{f}}$ is one-one but not onto**

**(D) ${\text{f}}$ is neither one-one nor onto**

Ans: Here, ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ is defined as ${\text{f}}(x) = {x^4}$.

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

$ \Rightarrow {x^4} = {y^4}$

$ \Rightarrow x = \pm y$

${\text{f}}(x) = {\text{f}}(y)$ does not imply that $x = y$

For example, ${\text{f}}(1) = {\text{f}}( - 1) = 1$

Thus, ${\text{f}}$ is not one-one.

Consider an element \[2\] in co-domain ${\mathbf{R}}$. It is clear that there does not exist any $x$ in

domain ${\text{R}}$ such that ${\text{f}}(x) = 2$

Thus, ${\text{f}}$ is not onto.

Hence, the function ${\text{f}}$ is neither one-one nor onto.

The correct answer is \[{\text{D}}.\]

12. Let ${\text{f}}\,{\text{:}}\,{\text{R}} \to {\text{R}}$ be defined as ${\text{f}}(x) = 3x$. Choose the correct answer**.**

**(A) f is one – one and onto **

**(B) f is many – one and onto **

**(C) f is one – one but not onto **

**(D) f is neither one – one nor onto**

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3x$.

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

$ \Rightarrow 3x = 3y$

$ \Rightarrow x = y$

Thus, ${\text{f}}$ is one-one.

Now, for any real number \[\left( y \right)\] in co-domain ${\text{R}}$, there exists \[\dfrac{y}{3}\] in ${\text{R}}$ such that ${\text{f}}\left( {\dfrac{y}{3}} \right) = 3\left( {\dfrac{y}{3}} \right) = y$

Thus, ${\text{f}}$ is onto.

Hence, the function ${\text{f}}$ is one-one and onto.

Therefore, the correct answer is \[{\text{A}}.\]

## Conclusion

Class 12 Exercise 1.2 of Maths Chapter 1 - Relation and Functions, is crucial for a solid foundation in math. Understanding the concept of the fundamental principles of relations and functions. Where students gain a deeper understanding of functions, a special type of relation where each input (domain element) has a unique output (range element). Regular practice of class 12 maths chapter 1 exercise 1.2 solutions NCERT solutions provided by Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward.

## NCERT Solutions for Class 12 Maths Chapter 1 Exercises

S.No. | Chapter 1 Relations and Functions All Exercises in PDF Format | |

1 | Class 12 Maths Chapter 1 Exercise 1.1 - 16 Questions & Solutions (3 Short Answers, 13 Long Answers) | |

2 | Class 12 Maths Chapter 1 Miscellaneous Exercise - 7 Questions & Solutions |

## CBSE Class 12 Maths Chapter 1 Other Study Materials

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## NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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## Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

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## FAQs on NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.2 - Relation and Functions

1. What is Exercise 1.2 based upon?

The problems in Exercise 1.2 are based on the concepts of one-one, onto, injective, surjective, bijective, etc. If the function can be inverted, then we must also discover its inverse. One-one refers to a situation where there is only one element in the co-domain for every element in the domain. Similar to this, if there is just one element in domain for every element in range, it is onto. A function is referred to as being invertible if it is one - one as well as onto.

2. How many days are required to finish Math exercise 1.2 for grade 12?

Students will need two to three days to do exercise 1.2 of class 12 math if they can devote two hours every day to it. This time is a rough estimate. Because no two students work at the same rate or with the same efficiency, this time may vary.

3. How can you simplify NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 (Ex 1.2)?

It will be simpler for the pupils to learn if they comprehend it and practise it. They will also receive full practise with the questions and the procedure from Vedantu ncert solutions for class 12th exercise 1.2. Students will benefit greatly from asking the same questions over.

4. What are the most crucial questions from Exercise 1.2 for the 12th-grade First Term Exam?

All of the problems and examples in exercise 1.2 of class 12 math are significant from the perspective of the exam, however, some of these issues are particularly significant for exams and might appear on the board exam. Examples 11, 12, 13, and 14 represent the issues at hand, along with issues 2, 3, 4, 5, 8, and 9.

5. Is the CBSE Grade 12 Maths exercise 1.2 the shortest one?

Yes, when compared to the other exercises in Chapter 1, exercise 1.2 in the grade 12 math textbook is the shortest. Eight examples (examples 7, 8, 9, 10, 11, 12, 13, 14) and 12 questions make up Exercise 1.2 of 12th Math. This could be easier with the Vedantu ncert solutions.