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In a vending machine, if you press a button, you will get a specific snack every time. This is what a function is- One output for one input. But if there is a button, whenever it is pressed, it gives random snacks each time. That's called a relation, as the output is inconsistent every time. NCERT Solutions for Exercise 1.2 Class 12 Maths Chapter 1 Relations and Functions hold a significant importance in a student's journey as it clarifies the types of functions to students and gives them the chance to check their progression by doing the exercise.
The 12th class Maths exercise 1.2 solutions of the NCERT are prepared by Careers360 subject matter experts following the latest CBSE guidelines. All the problems are well explained, and necessary formulas and graphs are provided with the solutions.
Answer:
Given, $f: R_* \longrightarrow R_{*}$ is defined by $f(x) = \frac{1}{x}$ .
One - One :
$f(x)=f(y)$
$\frac{1}{x}=\frac{1}{y}$
$x=y$
$\therefore$ f is one-one.
Onto:
We have $y \in R_*$ , then there exists $x=\frac{1}{y} \in R_*$ ( Here $y\neq 0$ ) such that
$f(x)= \frac{1}{(\frac{1}{y})} = y$
$\therefore f is \, \, onto$ .
Hence, the function is one-one and onto.
If the domain R ∗ is replaced by N with co-domain being same as R ∗ i.e. $g: N \longrightarrow R_{*}$ defined by
$g(x)=\frac{1}{x}$
$g(x_1)=g(x_2)$
$\frac{1}{x_1}=\frac{1}{x_2}$
$x_1=x_2$
$\therefore$ g is one-one.
For $1.5 \in R_*$ ,
$g(x) = \frac{1}{1.5}$ but there does not exists any x in N.
Hence, function g is one-one but not onto.
Question:2(i) Check the injectivity and surjectivity of the following functions:
(i) $f : N\rightarrow N$ given by $f(x) = x^2$
Answer:
$f : N\rightarrow N$
$f(x) = x^2$
One- one:
$x,y \in N$ then $f(x)=f(y)$
$x^{2}=y^{2}$
$x=y$
$\therefore$ f is one- one i.e. injective.
For $3 \in N$ there is no x in N such that $f(x)=x^{2}=3$
$\therefore$ f is not onto i.e. not surjective.
Hence, f is injective but not surjective.
Question:2(ii) Check the injectivity and surjectivity of the following functions:
(ii) $f : Z \rightarrow Z$ given by $f(x) = x^2$
Answer:
$f : Z \rightarrow Z$
$f(x) = x^2$
One- one:
For $-1,1 \in Z$ then $f(x) = x^2$
$f(-1)= (-1)^{2}$
$f(-1)= 1$ but $-1 \neq 1$
$\therefore$ f is not one- one i.e. not injective.
For $-3 \in Z$ there is no x in Z such that $f(x)=x^{2}= -3$
$\therefore$ f is not onto i.e. not surjective.
Hence, f is neither injective nor surjective.
Question:2(iii) Check the injectivity and surjectivity of the following functions:
(iii) $f: R \rightarrow R$ given by $f(x) = x^2$
Answer:
$f: R \rightarrow R$
$f(x) = x^2$
One- one:
For $-1,1 \in R$ then $f(x) = x^2$
$f(-1)= (-1)^{2}$
$f(-1)= 1$ but $-1 \neq 1$
$\therefore$ f is not one- one i.e. not injective.
For $-3 \in R$ there is no x in R such that $f(x)=x^{2}= -3$
$\therefore$ f is not onto i.e. not surjective.
Hence, f is not injective and not surjective.
Question:2(iv) Check the injectivity and surjectivity of the following functions:
(iv) $f: N \rightarrow N$ given by $f(x) = x^3$
Answer:
$f : N\rightarrow N$
$f(x) = x^3$
One- one:
$x,y \in N$ then $f(x)=f(y)$
$x^{3}=y^{3}$
$x=y$
$\therefore$ f is one- one i.e. injective.
For $3 \in N$ there is no x in N such that $f(x)=x^{3}=3$
$\therefore$ f is not onto i.e. not surjective.
Hence, f is injective but not surjective.
Question:2(v) Check the injectivity and surjectivity of the following functions:
(v) $f : Z \rightarrow Z$ given by $f(x) = x^3$
Answer:
$f : Z \rightarrow Z$
$f(x) = x^3$
One- one:
For $(x,y) \in Z$ then $f(x) = f(y)$
$x^{3}=y^{3}$
$x=y$
$\therefore$ f is one- one i.e. injective.
For $3 \in Z$ there is no x in Z such that $f(x)=x^{3}= 3$
$\therefore$ f is not onto i.e. not surjective.
Hence, f is injective but not surjective.
Answer:
$f : R\longrightarrow R$
$f (x) = [x]$
One- one:
For $1.5,1.7 \in R$ then $f(1.5)=\left [ 1.5 \right ] = 1$ and $f(1.7)=\left [ 1.7 \right ] = 1$
but $1.5\neq 1.7$
$\therefore$ f is not one- one i.e. not injective.
For $0.6 \in R$ there is no x in R such that $f(x)=\left [ 0.6 \right ]$
$\therefore$ f is not onto i.e. not surjective.
Hence, f is not injective but not surjective.
Answer:
$f : R \rightarrow R$
$f (x) = | x |$
$f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0$
One- one:
For $-1,1 \in R$ then $f (-1) = | -1 |= 1$
$f (1) = | 1 |= 1$
$-1\neq 1$
$\therefore$ f is not one- one i.e. not injective.
For $-2 \in R$ ,
We know $f (x) = | x |$ is always positive there is no x in R such that $f (x) = | x |=-2$
$\therefore$ f is not onto i.e. not surjective.
Hence, $f (x) = | x |$ , is neither one-one nor onto.
Question:5 Show that the Signum Function $f : R \rightarrow R$ , given by
$f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.$ is neither one-one nor onto.
Answer:
$f : R \rightarrow R$ is given by
$f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.$
As we can see $f(1)=f(2)=1$ , but $1\neq 2$
So it is not one-one.
Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain $R$ ,there does not exists x in domain $R$ such that $f(x)= -3$ .
So it is not onto.
Hence, signum function is neither one-one nor onto.
Answer:
$A = \{1, 2, 3\}$
$B = \{4, 5, 6, 7\}$
$f = \{(1, 4), (2, 5), (3, 6)\}$
$f : A \rightarrow B$
$\therefore$ $f(1)=4,f(2)=5,f(3)=6$
Every element of A has a distant value in f.
Hence, it is one-one.
Question:7(i) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) $f: R\rightarrow R$ defined by $f(x) = 3 -4x$
Answer:
$f: R\rightarrow R$
$f(x) = 3 -4x$
Let there be $(a,b) \in R$ such that $f(a)=f(b)$
$3-4a = 3 -4b$
$-4a = -4b$
$a = b$
$\therefore$ f is one-one.
Let there be $y \in R$ , $y = 3 -4x$
$x = \frac{(3-y)}{4}$
$f(x) = 3 -4x$
Puting value of x, $f(\frac{3-y}{4}) = 3 - 4(\frac{3-y}{4})$
$f(\frac{3-y}{4}) = y$
$\therefore$ f is onto.
f is both one-one and onto hence, f is bijective.
Question:7(ii) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(ii) $f : R\rightarrow R$ defined by $f(x) = 1 + x^2$
Answer:
$f : R\rightarrow R$
$f(x) = 1 + x^2$
Let there be $(a,b) \in R$ such that $f(a)=f(b)$
$1+a^{2} = 1 +b^{2}$
$a^{2}=b^{2}$
$a = \pm b$
For $f(1)=f(-1)=2$ and $1\neq -1$
$\therefore$ f is not one-one.
Let there be $-2 \in R$ (-2 in codomain of R)
$f(x) = 1 + x^2 = -2$
There does not exists any x in domain R such that $f(x) = -2$
$\therefore$ f is not onto.
Hence, f is neither one-one nor onto.
Answer:
$f : A \times B \rightarrow B \times A$
$f (a, b) = (b, a)$
Let $(a_1,b_1),(a_2,b_2) \in A\times B$
such that $f (a_1, b_1) = f(a_2, b_2)$
$(b_1,a_1)=(b_2,a_2)$
$\Rightarrow$ $b_1= b_2$ and $a_1= a_2$
$\Rightarrow$ $(a_1,b_1) = (a_2,b_2)$
$\therefore$ f is one- one
Let, $(b,a) \in B\times A$
then there exists $(a,b) \in A\times B$ such that $f (a, b) = (b, a)$
$\therefore$ f is onto.
Hence, it is bijective.
Question:9 Let $f : N \rightarrow N$ be defined by $f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;even \end{matrix}\right.$ for all $n\in N$ . State whether the function f is bijective. Justify your answer.
Answer:
$f : N \rightarrow N$ , $n\in N$
$f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;evem \end{matrix}\right.$
Here we can observe,
$f(2)=\frac{2}{2}=1$ and $f(1)=\frac{1+1}{2}=1$
As we can see $f(1)=f(2)=1$ but $1\neq 2$
$\therefore$ f is not one-one.
Let, $n\in N$ (N=co-domain)
case1 n be even
For $r \in N$ , $n=2r$
then there is $4r \in N$ such that $f(4r)=\frac{4r}{2}=2r$
case2 n be odd
For $r \in N$ , $n=2r+1$
then there is $4r+1 \in N$ such that $f(4r+1)=\frac{4r+1+1}{2}=2r +1$
$\therefore$ f is onto.
f is not one-one but onto
hence, the function f is not bijective.
Answer:
$A = R - \{3\}$
$B = R - \{1\}$
$f : A\rightarrow B$
$f(x) = \left (\frac{x-2}{x-3} \right )$
Let $a,b \in A$ such that $f(a)=f(b)$
$\left (\frac{a-2}{a-3} \right ) = \left ( \frac{b-2}{b-3} \right )$
$(a-2)(b-3)=(b-2)(a-3)$
$ab-3a-2b+6=ab-2a-3b+6$
$-3a-2b=-2a-3b$
$3a+2b= 2a+3b$
$3a-2a= 3b-2b$
$a=b$
$\therefore$ f is one-one.
Let, $b \in B = R - \{1\}$ then $b\neq 1$
$a \in A$ such that $f(a)=b$
$\left (\frac{a-2}{a-3} \right ) =b$
$(a-2)=(a-3)b$
$a-2 = ab -3b$
$a-ab = 2 -3b$
$a(1-b) = 2 -3b$
$a= \frac{2-3b}{1-b}\, \, \, \, \in A$
For any $b \in B$ there exists $a= \frac{2-3b}{1-b}\, \, \, \, \in A$ such that
$f(\frac{2-3b}{1-b}) = \frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}$
$=\frac{2-3b-2+2b}{2-3b-3+3b}$
$=\frac{-3b+2b}{2-3}$
$= b$
$\therefore$ f is onto
Hence, the function is one-one and onto.
Question:11 Let $f : R \rightarrow R$ be defined as $f(x) = x^4$ . Choose the correct answer.
(D) f is neither one-one nor onto.
Answer:
$f : R \rightarrow R$
$f(x) = x^4$
One- one:
For $a,b \in R$ then $f(a) = f(b)$
$a^{4}=b^{4}$
$a=\pm b$
$\therefore f(a)=f(b)$ does not imply that $a=b$
example: and $2\neq -2$
$\therefore$ f is not one- one
For $2\in R$ there is no x in R such that $f(x)=x^{4}= 2$
$\therefore$ f is not onto.
Hence, f is neither one-one nor onto.
Option D is correct.
Question:12 Let $f : R\rightarrow R$ be defined as $f(x) = 3x$ . Choose the correct answer.
(D) f is neither one-one nor onto.
Answer:
$f : R\rightarrow R$
$f(x) = 3x$
One - One :
Let $\left ( x,y \right ) \in R$
$f(x)=f(y)$
$3x=3y$
$x=y$
$\therefore$ f is one-one.
Onto:
We have $y \in R$ , then there exists $x=\frac{y}{3} \in R$ such that
$f(\frac{y}{3})= 3\times \frac{y}{3} = y$
$\therefore f is \, \, onto$ .
Hence, the function is one-one and onto.
The correct answer is A .
Also read,
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Type of Function | Definition | Example |
One-One (Injective) | Each input has a unique output. | f(x) = 2x (e.g., f(1)=2, f(2)=4 — all outputs are different) |
Onto (Surjective) | Every element in the codomain is covered by some input. | f(x) = x³ over all real numbers |
One-One & Onto (Bijective) | A function that is both injective and surjective. | f(x) = x + 5 (every input gives a unique output, all outputs are used) |
Many-One | Multiple inputs give the same output. | f(x) = x² |
Also, read,
These are the subject-wise links for the NCERT solutions of other subjects.
NCERT Exemplar are a good resource for practice. Students can use the links below for that cause.
Concepts related to one to one functions, reflexive functions etc, are discussed in the Exercise 1.2 Class 12 Maths
In Mathematics, A set is a collection of distinct or well-defined numbers or elements
Weightage of the chapters 'relation and function' is around 5 % weightage in the CBSE final board exam.
There are 3 ways to represent a set:
a. Statement form.
b. Roaster form .
c. Set Builder form.
A set with no elements is called an empty set. Also known by Null set or void set.
A relation is the set of ordered pair numbers.
12 questions are there in Exercise 1.2 Class 12 Maths
5 exercises are there including a miscellaneous exercise in the NCERT class 12 maths chapter 1.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
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I hope this information helps you.
Hi,
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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
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