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NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

Edited By Komal Miglani | Updated on Mar 31, 2025 03:06 AM IST | #CBSE Class 12th

Throughout our everyday encounters, we relate objects, people, or ideas together. For example, a student relates to his or her school, a country relates to its capital, or a person relates to the date he or she was born. Those are what we call relations in math. A relation is a way of relating elements of one set to elements of another and is a subset of the Cartesian product of two sets. A function is a relation in which each element of the first set (domain) relates to one element of the second set (codomain). Functions have various applications in real life, thereby allowing us to forecast weather patterns, design computer programs, and many other applications.

This Story also Contains
  1. NCERT Exemplar Class 12 Maths Solutions Chapter 1
  2. Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1
  3. NCERT Exemplar Class 12 Maths Solutions
  4. Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 1
  5. NCERT solutions of class 12 - Subject-wise
  6. NCERT Notes of class 12 - Subject Wise
  7. NCERT Books and NCERT Syllabus
  8. NCERT Exemplar Class 12 Solutions - Subject Wise

It is easy to understand relations and functions through NCERT Class 12 solutions. The solutions provide precise step-by-step solutions for various kinds of relations such as reflexive, symmetric, transitive, and equivalence, and functions, function composition, and inverse functions. Practicing NCERT problems on a daily basis will not just improve your concept clarity but will also increase your logical reasoning and problem-solving ability. Practice problems in NCERT textbooks form a great base for board and competitive examinations. So, step into the world of relations and functions through NCERT solutions and learn how mathematical relationships are involved in numerous aspects of life and education.

NCERT Exemplar Class 12 Maths Solutions Chapter 1

Class 12 Maths Chapter 1 exemplar solutions Exercise: 1.3
Page number: 11-17
Total questions: 62
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Question:1

Let A = {a, b, c} and the relation R be defined on A as follows:

R = {(a, a), (b, c), (a, b)}.

Then, write the minimum number of ordered pairs to be added in R to make R reflexive and transitive.

Answer:

Here, R = {(a, a), (b, c), (a, b)}

The minimum number of ordered pairs to be added to make R as reflexive is (b, b) and (c, c) to R. Whereas, to make R transitive, the minimum number of ordered pairs to be added is (a, c) to R.

Therefore, we need 3 ordered pairs to add to R to make it reflexive and transitive.

Question:2

Let D be the domain of the real-valued function f defined by f(x)=25x2 . Then, write D.

Answer:

f(x)=25x2
Therefore, the function can be defined as:
25x20

or,x2255x5

Therefore, the given function has a domain of [-5, 5].

Question:3

Let f, g:RR be defined by f(x) = 2x + 1 and g(x)=x22,xR, respectively. Then, find gof.

Answer:

Here,

f(x) = 2x + 1
g(x)=x22,xR
Therefore,

gof=g(f(x))=g(2x+1)

=(2x+1)22

=4x2+4x+12

=4x2+4x1

Question:4

. Let f: RR be the function defined by f(x)=2x3,xR. write f1.

Answer:

Here, f(x)=2x3,xR
Let’s say,
y=2x3

or,x=(y+3)/2

Therefore,
f1(x)=(x+3)/2

Question:5

If A={a,b,c,d} and the function f={(a,b),(b,d),(c,a),(d,c)} , write f1.

Answer:
Here, A={a,b,c,d} and the function f={(a,b),(b,d),(c,a),(d,c)}
Therefore, f1={(b,a),(d,b),(a,c),(c,d)}

Question:6

If f:RR is defined by f(x)=x23x+2, write f(f(x)).

Answer:

Here, f(x)=x23x+2

Therefore,
f(f(x))=f(x23x+2)=(x23x+2)23(x23x+2)+2,

=x4+9x2+46x3+4x212x3x2+9x6+2

=x46x3+10x23x
Similarly,
f(f(x))=x46x3+10x23x

Question:7

Is g={(1,1),(2,3),(3,5),(4,7)} a function? If g is described by g(x)=αx+β , then what value should be assigned to αandβ .

Answer:

Here,g={(1,1),(2,3),(3,5),(4,7)}
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also, g(x)=αx+β
Therefore,
g(1)=α(1)+β=1α+β=1..(1)
Similarly,
g(2)=α(2)+β=2α+β=3..(2)
By solving (1) and (2), we get
α=2andβ=1
Therefore, g(x)=2x1

Question:8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) {(x,y):x is a person, y is the mother of x}.
(ii) {(a,b):a is a person, b is an ancestor of a}.

Answer:
(i)Here, {(x,y):x is a person, y is the mother of x}.
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs makes a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here, {(a,b):a is a person, b is an ancestor of a}.
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.

Question:9

If the mappings f and g are given by f={(1,2),(3,5),(4,1)} and g={(2,3),(5,1),(1,3)} , write fog.

Answer:
Here,

f={(1,2),(3,5),(4,1)} and g={(2,3),(5,1),(1,3)}
Therefore,
fog(2)=f(g(2))=f(3)=5

fog(5)=f(g(5))=f(1)=2

fog(1)=f(g(1))=f(3)=5
So, we can write that:
fog={(2,5),(5,2),(1,5)}

Question:10

Let C be the set of complex numbers. Prove that the mapping f:CR is given by f(z)=|z|,zC, is neither one-one nor onto.

Answer:

Here, f:CR is given by f(z)=|z|,zC
If we assume z=4+3i
Then,
f(4+3i)=|4+3i|=(42+32)=25=5
Similarly, for z=43i
f(43i)=|43i|=(42+32)=25=5
Therefore, it is clear that f(z) is many-one.
So, |z|0,zC,
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Question:11

Let the function f:RR be defined by f(x)=cosx,xR. Show that f is neither one-one nor onto.

Answer:
It is given that, f:RR,f(x)=cosx,xR
So, we can write:
f(x1)=f(x2)or,cosx1=cosx2
Hence, x1=2nπ±x2,wherenZ
It is understandable that for any value of x1andx2, the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.

Question:12

Let X={1,2,3}andY={4,5} . Find whether the following subsets of X×Y are functions from X to Y or not.
(i) f={(1,4),(1,5),(2,4),(3,5)}
(ii) g={(1,4),(2,4),(3,4)}
(iii) h={(1,4),(2,5),(3,5)}
(iv) k={(1,4),(2,5)}.

Answer:

Here, X={1,2,3}andY={4,5}
Therefore, X×Y={(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)}
(i) f={(1,4),(1,5),(2,4),(3,5)}
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii) g={(1,4),(2,4),(3,4)}
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii) h={(1,4),(2,5),(3,5)}
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv) k={(1,4),(2,5)}
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

Question:13

If functions f:ABandg:BA satisfies gof=IA, then show that f is one-one and g is onto.

Answer:

Here, it is given:
f:ABandg:BA satisfies gof=IA
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.

Question:14

Let f:RR be the function defined by f(x)=1/(2cosx)xR. Then, find the range off.

Answer:

Here, f(x)=1/(2cosx)xR
Let’s say,
y=1/(2cosx)2yycosx=1Or,cosx=(2y1)/yOr,cosx=21/y
As the range of cos x is, 1cosx1
Therefore,
121/y131/y111/y31/3y1
Hence, the range of the given function is [1/3, 1].

Question:15

Let n be a fixed positive integer. Define a relation R in Z as follows: a,bZ, aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Answer:

Here, we have to a relation R in Z as follows: a,bZ, aRb if and only if a - b is divisible by n.
Here, aRa (aa) is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, aRb(ab) is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.

Question:16

If A = {1, 2, 3 }, define relations on A which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric, and transitive.

Answer:

Here, A={1,2,3}.
(i) Assume R1={(1,1),(1,2),(1,3),(2,3),(2,2),(1,3),(3,3)}
Here, (1, 1), (2, 2) and (3, 3)  R1.R1 is reflexive.
(1,2)R1,(2,3)R1(1,3)R1
. Hence,R1 is transitive.
Now, (1,2)R1(2,1)R1.
Therefore, R1 is not symmetric.
(ii) Let say, R2={(1,2),(2,1)}
So, (1,2)R2,(2,1)R2
Therefore, R2 is symmetric,
(1,1)R2. Therefore, R2 is not reflexive.
(1,2)R2, (2,1)R2 but (1,1)R2. Hence, R2 is not transitive.
(iii) Let R3 = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}
R3 is reflexive as (1,1)(2,2)and(3,3)R3
R3 is symmetric as (1,2),(1,3),(2,3)R3 (2,1),(3,1),(3,2)R3
Therefore, R3 is reflexive, symmetric and transitive.

Question:17

Let R be relation defined on the set of natural number N as follows:
R={(x,y):xN,yN,2x+y=41} . Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric, and transitive.

Answer:

Here, R={(x,y):xN,yN,2x+y=41}
So, the domain D={1,2,3,..,20}
And the Range ={1,3,5,..,39}
Here, (2,2)Ras2×2+241. Therefore, R is not reflexive.
Again, (1,39)Rbut(39,1)R. So, R is not symmetric.
Again, (11,19)R,(19,3)R;but(11,3)R. So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.

Question:18

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

(b) a mapping from A to B which is not injective

(c) a mapping from B to A.

Answer:

Here, A={2,3,4},B={2,5,6,7}
(i) Assume, f:AB denote a mapping
Then, f={(x,y):y=x+3}or
f={(2,5),(3,6),(4,7)} , this is an injective mapping.
(ii) Again assume, g:AB denotes a mapping: g={(2,2),(3,2),(4,5)} . Hence, it is not an injective mapping.
(iii) Again assume, h: BA denotes a mapping: h={(2,2),(5,3),(6,4),(7,4)} . Hence, it is a mapping from B to A.

Question:19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.
Answer:
(i) Say, f:NN, be a mapping defined by f(x)=x2
If, f(x1)=f(x2)
Then, x12=x22
So,x1=x2(as,x1+x1cannot be 0)
f(x1)=f(x2), hence, f(x) is one-one.
However, ‘f’ is not onto, as for 1N, therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let f:R[0,), is a mapping that is defined by f(x)=|x|
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But |x|0, so the range is [0,].
Therefore, f(x) is onto.
(iii) Assume, f:RR , be a mapping which is defined by f(x)=x2
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is [0,).
Therefore, f (x) is neither one-one nor onto.

Question:20

Let A=R{3},B=R{1} . Let f:AB be defined by f(x)=x2/x3xA . Then show that f is bijective.

Answer:

Here,

A=R{3},B=R{1}
f:AB be defined by f(x)=x2/x3xA
Hence, f(x)=(x3+1)/(x3)=1+1/(x3)
Say f(x1)=f(x2)
1+1x13=1+1x23
1x13=1x23x1=x2
So, f (x) is an injective function.
If,y=(x2)/(x3)

Or,x2=xy3y

Or,x(1y)=23y

Or,x=(3y2)/(y1)

Or,yR{1}=B
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.

Question:21

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:
(i) f(x)=x/2 (ii) g(x)=|x|
(iii) h(x)=x|x| (iv) k(x)=x2

Answer:

Here, A = [-1, 1]
(i) f:[1,1][1,1],

f(x)=x/2
If, f(x1)=f(x2)
x1/2=x2/2
So, f(x) is one-one.
Also, x[1,1]
x/2=f(x)=[1/2,1/2]
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
g(x)=|x|

Letg(x1)=g(x2)

|x1|=|x2|

x1=±x2
So, g(x) is not one-one.
Also, g(x)=|x|0,for all real x
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
h(x)=x|x|

Let h(x1)=h(x2)x1|x1|=x2|x2|

If x1,x2>0x12=x22x12x22=0(x1x2)(x1+x2)=0x1=x2(asx1+x20)

Again ,x1,x2<0,andx1=x2

Therefore ,x1andx2of opposite sign,x1x2.
Hence, f(x) is one-one.
For x[0,1],f(x)=x2[0,1]

For x<0,f(x)=x2[1,0)
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
k(x)=x2

Let k(x1)=k(x2)

x12=x22

x1=±x2
Therefore, k(x) is not one-one.

Question:22

Each of the following defines a relation on N:
(i) x is greater than y, x,yN
(ii) x + y = 10, x,yN
(iii) x y is square of an integer x,yN
(iv) x + 4y = 10 x,yN
Determine which of the above relations are reflexive, symmetric, and transitive.

Answer:

(i)Here, x is greater than y; x,yN
If (x,x)R, then x > x, that does not satisfy for any xN.\
Therefore, R is not reflexive.
Say,(x,y)R
Or,xRy

Or,x>y
y>x For any x,yN, the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
Or,x>yandy>z

Or,x>z

Or,xRz
Hence, R is transitive.
(ii) x + y = 10;x,yN
Thus,
R={(x,y);x+y=10,x,yN}

R={(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)}
Therefore, (1,1)R
So, R is not reflexive.
Again, (x,y)R

(y,x)R
Therefore, R is symmetric.
And, (1,9)R,(9,1)R,but(1,1)R
Therefore, R is not transitive.
(iii)Here, xy is square of an integer x,yN
R={(x,y):xy is a square of an integerx,yN}
So, (x,x)R,xN
For any xN,x2 is an integer.
Thus, R is reflexive.
If (x,y)R

(y,x)R
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then, xy=m2andyz=n2for some m, nZ
x=m2/yandz=n2/yxz=m2n2/y2, this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10; x,yN
R={(x,y):x+4y=10;x,yN}

R={(2,2),(6,1)}Here,(1,1)R
Hence, R is not symmetric.
(x,y)R

x+4y=10And,(y,z)R

y+4z=10

Or, x16z=30

Or, (x,z)R
Therefore, R is not transitive.

Question:23

Let A = {1, 2, 3, … 9} and R be the relation in A×A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A×A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer:

Here,A={1,2,3,9}and(a,b)R(c,d)

ifa+d=b+cfor(a,b),(c,d)A×A.Say(a,b)R(a,b)

Therefore,a+b=b+a,a,bA.
This must be true for any a,bA.
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
a+d=b+cc+b=d+a(c,d)R(a,b)
Therefore, R is symmetric.
Let (a,b)R(c,d),and,(c,d)R(e,f)
a+d=b+c,and,c+f=d+ea+d=b+c,and,d+e=c+f

Or,(a+d)(d+e=(b+c)(c+f)

Or,ae=bf

Or,a+f=b+e

Or,(a,b)R(e,f)
So, R is transitive.
Therefore, R is an equivalence relation.

Question:24

Using the definition, prove that the function f:AB is invertible if and only if f is both one-one and onto.

Answer:

Say, f:AB be many-one functions.
If, f(a)=p and f(b)=p
Then f1(p)=aandf1(p)=b
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, f:AB is not onto function.
B={p,q,r}and{p,q} is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then f=XY is invertible.

Question:25

Functions f,g:RR are defined, respectively, by f(x)=x2+3x+1,g(x)=2x3, find
(i) fog (ii) gof (iii) fof (iv) gog

Answer:

Here,
f(x)=x2+3x+1,g(x)=2x3

(i)fog=f(g(x))=f(2x3)=(2x3)2+3(2x3)+1=4x2 +912x+6x9+1=4x2 6x+1

(ii)gof=g(f(x))=g(x2 +3x+1)=2(x2 +3x+1)3

=2x2 +6x1

(iii)fof=f(f(x))=f(x2 +3x+1)=(x2 +3x+1)2+3(x2+3x+1)+1

=x4+9x2 +1+6x3+6x+2x2 +3x2 +9x+3+1

=x4+6x3+14x2 +15x+5(iv)gog=g(g(x))

=g(2x3)=2(2x3)3=4x63=4x9

Question:26

Let be the binary operation defined on Q. Find which of the following binary operations are commutative
(i)ab=aba,bQ(ii)ab=a2+b2a,bQ(iii)ab=a+aba,bQ(iv)ab=(ab)2a,bQ

Answer:

Here, is a binary operation defined on Q.
(i)
ab=ab,a,bQandba=baSo,abba
Hence, is not commutative.
(ii)
ab=a2+b2ba=b2+a2
Therefore, is commutative.
(iii)
ab=a+abba=b+abHence,a+abb+ab
Therefore, is not commutative.
(iv)
ab=(ab)2,a,bQba=(ba)2As,(ab)2=(ba)2
Therefore, is commutative.

Question:27

Let be binary operation defined on R by ab=1+ab,a,bR. Then the operation is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative

Answer:

(i) Here,

is a binary operation defined on R by ab=1+ab,a,bR
So, ab=ab+1=ba
Therefore, is a commutative binary operation.
Now,a(bc)=a(1+bc)=1+a(1+bc)=1+a+abc

Again,(ab)c=(1+ab)c=1+(1+ab)c=1+c+abc

Therefore,a(bc)(ab)c
Hence, is not associative.
Therefore, is commutative but not associative.

Question:28

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to ba,bT. Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these

Answer:

(C) equivalence
Here aRb, if a is congruent to b, a,bT.
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say, aRbab
babRa
Therefore, R is symmetric.
Say, aRb and bRc
abandbcacaRc
Therefore, R is transitive.
Therefore, R is an equivalence relation.

Question:29

Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is

(A) symmetric but not transitive (B) transitive but not symmetric

(C) neither symmetric nor transitive (D) both symmetric and transitive

Answer:

(B) transitive but not symmetric

If, aRb means a is brother of b.

Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.

Therefore, R is not symmetric.

If, aRb implies that a is the brother of b.

and bRc implies that b is the brother of c.

Therefore, a must be the brother of c.

Hence, R is transitive.

Question:30

The maximum number of equivalence relations on the set A = {1, 2, 3} are

(A) 1 (B) 2

(C) 3 (D) 5

Answer:

(D) 5

Given, set A={1,2,3}
Now, the number of equivalence relations as follows
R1={(1,1),(2,2),(3,3)}

R2={(1,1),(2,2),(3,3),(1,2),(2,1)}

R3={(1,1),(2,2),(3,3),(1,3),(3,1)}

R4={(1,1),(2,2),(3,3),(2,3),(3,2)}

R5={(1,2,3)A×A=A2}
Thus, the maximum number of equivalence relations is ‘5’.

Question:31

If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is

(A) reflexive (B) transitive

(C) symmetric (D) none of these

Answer:

(D) none of these

If, R be defined on the set {1, 2, 3} by R = {(1, 2)}

Then, we can say that R is not reflexive, transitive, and symmetric.

Question:32

Let us define a relation R in R as aRb if ab. Then R is

(a) an equivalence relation (b) reflexive, transitive but not symmetric (c) symmetric, transitive but not reflexive (d) neither transitive nor reflexive but symmetric.

Answer:

(b).
The defined relation R in R as aRb if ab.
Similarly, aRa implies aa which is true.
Therefore, it is reflexive.
Let aRbab,but,ba.
Therefore, we can’t write it as Rba
Hence, R is not symmetric.
Now, ab,bc. So, ac, and this is true.
Therefore, R is transitive.

Question:33

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is

(a) reflexive but not symmetric (b) reflexive but not transitive (c) symmetric and transitive (d) neither symmetric nor transitive.

Answer:

(a)

Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Therefore, it can be written as: 1R1, 2R2 and 3R3.

Therefore, R is reflexive.

Here, 1R2 is not the same as 2R1 and 2R3 is not the same as 3R2.

Therefore, R is not symmetric.

Again, 1R1 and 1R21R3

Therefore, R is transitive.

Question:34

The identity element for the binary operation * defined on Q ~ {0} as a * b = 2 ab a, b Q ~ {0} is

(a) 1 (b) 0 (c) 2 (d) None of these

Answer:

(c)
Here: ab=ab/2a,bQ{0}
Assume ‘e’ as the identity element.
Therefore, ae=ae/2=ae=2

Question:35

If the set A contains 5 elements and set B contains 6 elements, then the number of one-one and onto mapping from A to B is

(a) 720 (b) 120 (c) 0 (d) None of these

Answer:

Let, the number of elements in A and B set are m and n respectively. Therefore, one-one and onto mapping

from A to B is n! when m = n

and, 0 if m ≠ n

It is given that, m = 5 and n = 6.

As,5 ≠ 6 So, from A to B mapping = 0

Question:36

Let A={1,2,3,...,n} and B={a,b} . Then the number of surjections from A to B is
(a) nP2
(b) 2n2
(c)2n1
(d) None of these

Answer:

(d)
It is given that, A={1,2,3,...,n} and B={a,b}
Say, the number of elements in set A and B are m and n respectively.
Therefore, nCm×m! is the number of surjections from A to B, for nm
Given, m=2.
Therefore, the number of surjections from A to B
=nC2×2!=n!/[2!×(n2)!]=n(n1)=n2n

Question:37

Let f:RR be defined by f(x)=1/xxR then f is
(a) one-one (b) onto (c) bijective (d) f is not defined

Answer:

(d)
Here, f(x) = 1/x
Say, x = 0. Then f(x) = 1/0 = undefined
Hence, f(x) is not defined.

Question:38

Let f:RR be defined by f(x)=3x25 and g:RR by g(x)=x/(x2+1), then gof is
(a)(3x25)/(9x430x2+26)(b)(3x25)/(9x46x2+26)(c)3x2/(x4+2x24)(d)(3x2)/(9x4302+2)

Answer:

(a)

Given, f(x)=3x25 and g(x)=x/(x2+1)
Therefore,
gof=gof(x)=g(3x25)

=(3x25)/[(3x25)2+1]

=(3x25)/(9x430x2+26)

Question:39

Which of the following functions from Z to Z are bijections?
(a)f(x)=x3(b)f(x)=x+2(c)f(x)=2x+1(d)f(x)=x2+1

Answer:

(b)
Given that f:ZZ
Say, x1,x2f(x)f(x1)=x1+2;f(x2)=x2+2;
For, f(x1)=f(x2) x1+2=x2+ 2
or, x1=x2
Therefore, the function f(x) is one-one.
Say, y = x + 2
or, x=y2yZ
Hence, f(x) is onto.
Therefore, the function f(x) is bijective.

Question:40

Let f:RR be the functions defined by f(x)=x3+5. Then f1(x)is
(a)(x+5)1/3(b)(x5)1/3(c)(5x)1/3(d)5x

Answer:

(b)Here,

f(x)=x3+5

Say,y=x3+5

or,x3=y5

or,x=(y5)1/3 f11(x)=(x5)1/3

Question:41

Let f:AB and g:BC be the bijective functions. Then(gof)1 is
(a)f1og1(b)fog(c)g1of1(d)gof
Answer:

(a)
Given that, f:AB and g:BC
Then, (gof)1=f1og1

Question:42

Let f:{R3/5}R be defined by f(x)=(3x+2)/(5x3) then
(a)f(x)=f(x);(b)f1(x)=f(x)(c)(fof)x=x(d)f1(x)=f(x)/19

Answer:

(a)

Here,f(x)=(3x+2)/(5x3)x3/5
So,y=(3x+2)/(5x3)

or,y(5x3)=(3x+2)

or,5xy3y=3x+2

or,x=(3y+2)/(5y3)

Therefore,f1(x)=(3x+2)/(5x3)

Hence,f1(x)=f(x)

Question:43

Let f:[0,1][0,1] be defined by

f(x) = { x, if is rational

{ 1-x , if is irrational
Then (fof)x is

(a) constant

(b) 1 + x

(c) x

(d) None of these

Answer:

(c)
Here, f:[0,1][0,1]
f=f1Therefore,(fof)x=x

Question:44

Let f:[2,) and R be the function defined by f(x)=x24x+5, then the range of f is
(a)R(b)[1,)(c)[4,)(d)[5,)
Answer:

(b)

Here,f(x)=x24x+5

Assume,y=x2 4x+5

or,x2 4x+5y=0

or,x=[4±{(4)24(4)(5y)}]/2x=2±(y1)

When x is real,y10

Therefore,y1Hence the range is[1,)

Question:45

Let f:NR be the function defined by f(x)=(2x1)/2 and g:QR be another function defined by g(x)=x+2 then, gof(3/2) is
(a)1(b)1(c)7/2(d)Noneofthese

Answer:

(d)
Given that, f(x)=(2x1)/2andg(x)=x+2
Therefore, gof(x)=g[(f(x)]=f(x)+2=(2x1)/2=(2x+3)/2
Hence, gof(3/2)=3

Question:46

Let f : RR be defined by

\\f(x) = \{2x, when x> 3 \\ \ \ \ \: \: \: \: \: \: \: \: \ \ \ \ \ \{ x\textsuperscript{2}, when 1< x \leq 3\\ \ \ \ \ \ \ \ \ \ \ \ \{ 3x, when x \leq 1\\ then f(- 1) + f(2) + f(4) is\\

(a)9(b)14(c)5(d)Noneofthese

Answer:

(a)
Here,f(1)+f(2)+f(4)=3(1)+(2)2+2(4)=3+4+8=9

Question:47

If f:RR be given by f(x)=tanx, then f-1 is
(a)π4(b)(nπ+π/4)where nZ(c) does not exist(d)None of these
Answer:

(a)

Here,f(x)=tanx

Assume,f(x)=y=tanx

Then,  x=tan1(y)

or,f1(x)=tan1 (x)

or,f1(1)=tan1(1)

or, f1(1)=tan1tan(π/4)=π4

Question:48

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ............

Answer:

Here, aRb : 2a + 3b = 30

Or, 3b = 30 – 2a

Or, b = (30 - 2a)/ 3

for a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

Therefore, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Question:49

Let the relation R be defined on the set A={1,2,3,4,5} by R={(a,b):|a2b2|<8}. Then R is given by ............

Answer:

Here
A={1,2,3,4,5}andR={(a,b):|a2b2|<8}

Therefore, we can say,

R={(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(4,3)(3,4),(4,4),(5,5)}

Question:50

Let f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}. Then gof = ................ and fog = ...................

Answer:

Here,f={(1,2),(3,5),(4,1)}

and,g={(2,3),(5,1),(1,3)}

Then,gof(1)=g[f(1)]=g(2)=3

Again,gof(3)=g[f(3)]=g(5)=1

Again,gof(4)=g[f(4)]=g(1)=3

Therefore,gof={(1,3),(3,1),(4,3)}

Again,fog(2)=f[g(2)]=f(3)=5

Again,fog(5)=f[g(5)]=f(1)=2

Again,fog(1)=f[g(1)]=f(3)=5

Hence,fog={(2,5),(5,2),(1,5)}

Question:51

Let f:RR be defined by f(x)=x/(1+x2) then (fofof)(x)=.................

Answer:

( fofof )(x)=f[f(x))]=f[f(x1+x2)]

=f(x1+x21+x21+x2)

=f(x1+2x2)
 fofof (x)=x1+2x21+x21+2x2=x1+3x2

Question:52

If, f(x)=[4(x7)3], then f1(x) = .............

Answer:

Here,f(x)=[4(x7)3]

Say,y=[4(x7)3]

Or,(x7)3=(4y)

Or,(x7)=(4y)1/3

Or,x=7+(4y)1/3

Therefore,f1(x)=7+(4x)1/3

Question:53

Let R = {(3,1), (1,3), (3,3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive and not reflexive.

Answer:

False Here in the question, R = {(3,1), (1,3), (3,3)} which is defined on the set A = {1, 2, 3}

As (1 ,1) R, R is not a reflexive one.

As (3, 1) R and (1, 3) belongs to R, then R is symmetric.

Again, (1, 3) R, (3, 1) R. However (1, 1) does not belong to R. Then R is not transitive.

Question:54

Let f: R→R be the function defined by f(x)=sin(3x+2), x R. Then f is invertible.

Answer:

False
Here, f(x)=sin(3x+2), x R. This function is not a one-one onto function for all xR.
Again, f(x)=sin(3x+2)=0
or,3x+2=nπ,n Z.
Hence, the function is not invertible.

Question:55

Every relation which is symmetric and transitive is also reflexive.

Answer:

False

Assume, a relation R : R= {(1,2), (2,1), (2,2)} on the set A = {1,2}

Therefore, it is clear that (1,1) ∉ R. Hence, it is not reflexive.

Question:56

An integer m is said to be related to another integer n, if m is an integral multiple of n. This relation Z is reflexive, symmetric, and transitive.

Answer:

False.

Here, the given relation in the question is reflexive and transitive. However, it is not symmetric

Question:57

Let A={0,1} and N be the set of natural numbers. Then the mapping f:NA defined by f(2n1)=0,f(2n)=1,nN,is onto.

Answer:

True
Here, A={0,1},andf(2n1)=0,f(2n)=1,AnN
Therefore, the range of f is {0,1}
Hence, f:NA mapping is onto.

Question:58

The relation R on the set A = {1, 2, 3} defined as R = {(1,1), (1,2), (2,1), (3,3)}

Answer:

False

As, R={(1,1),(1,2),(2,1),(3,3)}

Therefore, (2,2)R

Hence, R is not reflexive.

Question:60

The composition of function is associative.

Answer:

True.

fo(goh)(x) = (fog)oh

Question:61

Every function is invertible.

Answer:

False.

Only bijective functions are invertible.

Question:62

A binary operation on a set always has an identity element.

Answer:

False

‘+’ is a binary operation on the set N but it has no identity element.

Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1

This chapter is divided into various sub-topics that are related to relations and functions and its operations. They are mentioned below:

  • Introduction
  • Types of relations
  • Types of functions
  • Composition of functions & invertible functions
  • Binary operations
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NCERT Exemplar Class 12 Maths Solutions

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 1

NCERT Exemplar solutions for Class 12 Maths chapter 1, students will learn about the basic concept of functions and relations.

  • The function is a relation, but is used for obtaining the output from a set of inputs. Learning the relations and functions at the very start of 12th Class will help in setting one's roots in calculus mathematics for cracking entrance exams and higher education.
  • A variety of questions with solutions are available to develop a strong base. Practicing these questions will help you to gain confidence and to score good marks in your exams.
JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

NCERT solutions of class 12 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Notes of class 12 - Subject Wise

Given below are the subject-wise NCERT Notes of class 12 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

NCERT Exemplar Class 12 Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:


Frequently Asked Questions (FAQs)

1. Are these solutions helpful for board examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 1  are designed well to prepare you for board examinations.

2. What are the important topics of this chapter?

The important topics of this NCERT Exemplar Class 12 Maths solutions chapter 1 include Types of Relations, Types of Functions, Binary Operations and Composition of Functions and Invertible Function.

3. How to remember the formulae well?

The best ways to remember formulae is to keep revising them and solving related problems. Other methods include reading them on a daily basis. You can also prepare charts to stick near your study table or bed to go through in your leisure time.

4. How many questions are there in this chapter?

There are a total 70 questions in NCERT Exemplar Class 12 Maths solutions chapter 1 based on different concepts.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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