# Chapter 2 – Relations and Functions Questions and Answers: NCERT Solutions for Class 11 Maths

Class 11 Maths NCERT book solutions for Chapter 2 - Relations and Functions Questions and Answers

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Class 11 Maths NCERT book solutions for Chapter 2 - Relations and Functions Questions and Answers

These are ordered pairs which are equal with each other, then the corresponding elements should also be equal to each other.

Thus, we will have, x3+1=53×3+1=53

And also y−23=13y−23=13

Now, we will try to simplify the given equations and find our needed values.

x3+1=53×3+1=53

⇒x3=53−1⇒x3=53−1

Simplifying further,

⇒x3=5−33=23⇒x3=5−33=23

⇒x=2⇒x=2

So, we have the value of xx as 22 .

Again, for the second equation,

y−23=13y−23=13

⇒y=23+13⇒y=23+13

And, after more simplification,

⇒y=1+23=33⇒y=1+23=33

⇒y=1⇒y=1

So, we have the value of xx and yy as 22 and 11 respectively.

So, the number of elements in set BB is 33 .

Thus, the number of elements in (A×B)(A×B) will be,

= Number of elements in A×A× Number of elements in BB

=3×3=9=3×3=9

So, the number of elements in (A×B)(A×B) is 99 .

The Cartesian product of two non-empty sets AA and BB is defined as A×B={(a,b):a∈Aandb∈B}A×B={(a,b):a∈Aandb∈B}

So, the value of G×HG×H will be,

G×H={(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}G×H={(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}

And similarly the value of H×GH×G will be,

H×G={(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}H×G={(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

(i) If P={m,n}P={m,n} and Q={n,m}Q={n,m} , then P×Q={(m,n),(n,m)}P×Q={(m,n),(n,m)} .

We have the value as, P={m,n}P={m,n} and Q={n,m}Q={n,m}.

Thus, P×Q={(m,m),(m,n),(n,m),(n,n)}P×Q={(m,m),(m,n),(n,m),(n,n)}

We know, B∩∅=∅B∩∅=∅

Thus, we have, A×{B∩∅}=A×∅=∅A×{B∩∅}=A×∅=∅.

A×A×A={(p,q,r):p,q,r∈A}A×A×A={(p,q,r):p,q,r∈A}

Now, we are provided with the fact that, A={−1,1}A={−1,1}

Thus,

A×A×A={(−1,−1,−1),(−1,−1,1),(−1,1,−1),(−1,,1,1),(1,−1,−1),(1,−1,1),(1,1,−1),(1,1,1)}A×A×A={(−1,−1,−1),(−1,−1,1),(−1,1,−1),(−1,,1,1),(1,−1,−1),(1,−1,1),(1,1,−1),(1,1,1)}

On the other hand, the Cartesian product of two non-empty sets AA and BB is defined as A×B={(a,b):a∈Aandb∈B}A×B={(a,b):a∈Aandb∈B}

As we can see, AA is the set of all the first elements and BB is the set of all the second elements.

So, we will have, A={a,b}A={a,b} and B={x,y}B={x,y} .

(i) A×(B∩C)=(A×B)∩(A×C)A×(B∩C)=(A×B)∩(A×C)

To start with, we will have, B∩C=∅B∩C=∅ , as there are no elements in common between these sets.

Thus, we have, A×(B∩C)=A×∅=∅A×(B∩C)=A×∅=∅

For the right hand side, we have,

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

And similarly,

A×C={(1,5),(1,6),(2,5),(2,6)}A×C={(1,5),(1,6),(2,5),(2,6)}

Again, we can see there are no elements in common between these sets. So, we have, (A×B)∩(A×C)=∅(A×B)∩(A×C)=∅

So, we get, L.H.S = R.H.S.

So, we have, A×C={(1,5),(1,6),(2,5),(2,6)}A×C={(1,5),(1,6),(2,5),(2,6)}

And similarly,

B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}

We can easily see, every element of A×CA×C is an element of B×DB×D . So, A×CA×C is a subset of B×DB×D .

Thus, we have, A×B={(1,3),(1,4),(2,3),(2,4)}A×B={(1,3),(1,4),(2,3),(2,4)}

So, the set A×BA×B has 4 elements.

Now, this is also known to us that, if a set AA has nn elements, then the number of subsets of AAis 2n2n .

We can thus conclude that, A×BA×Bwill have 24=1624=16 subsets.

Now, noting down the subsets of A×BA×B we get,

∅,{(1,3)},{(1,4)},{(2,3)},{(2,4)},{(1,3),(1,4)},∅,{(1,3)},{(1,4)},{(2,3)},{(2,4)},{(1,3),(1,4)},

{(1,3),(2,3)},{(1,3),(2,4)},{(1,3),(2,3)},{(1,3),(2,4)},

{(1,4),(2,3)},{(1,4),(2,4)},{(2,3),(2,4)},{(1,4),(2,3)},{(1,4),(2,4)},{(2,3),(2,4)},

{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},

{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)},{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)},

{(1,3),(1,4),(2,3),(2,4)}{(1,3),(1,4),(2,3),(2,4)}

(x,1),(y,2),(z,1)(x,1),(y,2),(z,1)

are in A×BA×B .

We also know that, AA is the set of all the first elements and BB is the set of all the second elements.

So, we can conclude, AA having elements x,y,zx,y,z and BB having elements 1,21,2 .

Thus, we get, n(A)=3,n(B)=2n(A)=3,n(B)=2.

So, A={x,y,z},B={1,2}A={x,y,z},B={1,2} .

We also know that, if n(A)=a,n(B)=bn(A)=a,n(B)=b , then n(A×B)=abn(A×B)=ab

As it is given that, n(A×A)=9n(A×A)=9

It can be written as,

n(A)×n(A)=9n(A)×n(A)=9

⇒n(A)=3⇒n(A)=3

And it is also given that (−1,0),(0,1)(−1,0),(0,1) are the two elements of A×AA×A .

Again, the fact is also known that, A×A={(a,a):a∈A}A×A={(a,a):a∈A} . And also −1,0,1−1,0,1 are the elements of AA .

Also, n(A)=3n(A)=3 , implies A={−1,0,1}A={−1,0,1} .

So, (−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)(−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1) are the remaining elements of A×AA×A .

So, we can write RR as, R={(1,3),(2,6),(3,9),(4,12)}R={(1,3),(2,6),(3,9),(4,12)} .

Thus, the domain of RR is, {1,2,3,4}{1,2,3,4} .

And similarly, the range of RR is, {3,6,9,12}{3,6,9,12} .

And also, the codomain of RR is, A={1,2,3,……,14}A={1,2,3,……,14} .

We have the value of xx as 1,2,31,2,3 are it must be less than 4.

So, the relation RR will look like, R={(1,6),(2,7),(3,8)}R={(1,6),(2,7),(3,8)}

The domain of RR will be, ={1,2,3}={1,2,3}

And similarly, the range of RR will be, ={6,7,8}={6,7,8} .

We are also given that, R=(x,y):the difference between x and y is odd; x in A,y in B

Simply, writing down according to the given condition,

R={(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}R={(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}

(i) In set-builder form

Now, writing the relation in the set-builder form,

(i)R={(x,y):y=x−2;x∈P}(i)R={(x,y):y=x−2;x∈P}

And in another form,

R={(x,y):y=x−2;x∈5,6,7}R={(x,y):y=x−2;x∈5,6,7}

What is its domain and range?

And again, in roster form,

(ii)R={(5,3),(6,4),(7,5)}(ii)R={(5,3),(6,4),(7,5)}

Where the domain of RR is{5,6,7}{5,6,7} and range of RR is{3,4,5}{3,4,5} .

(i) Write RR in roster form.

Using the conditions given in the problem, we get,

R={(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),R={(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),

(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}

And this is the roster form of the relation.

Using the condition given,

We can clearly write that, R={(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)}R={(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)}

And this is our needed relation.

Now, it can be clearly observed, that the domain of RR is, {(x:x∈(0,1,2,3,4,5)}{(x:x∈(0,1,2,3,4,5)}. And similarly, the range of RR is, {(y:y∈(5,6,7,8,9,10)}{(y:y∈(5,6,7,8,9,10)}.

R={(x,x3):xisaprimenumberlessthan10}R={(x,x3):xisaprimenumberlessthan10} .

We know, the prime numbers less than 1010 are 2,3,5,72,3,5,7 .

Thus, the relation can be written as,

R={(2,8),(3,27),(5,125),(7,343)}R={(2,8),(3,27),(5,125),(7,343)}

Now, we will try to find out the Cartesian product of these to sets, A×B={(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)}A×B={(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)}

Thus, we see, the number of elements in A×BA×B is 66 .

So, the number of subsets, 2626 .

Then, the number of relations from AA to BB is 2626 .

And, we know the fact that, the difference of two given integers in always an integer.

Thus, it can be concluded that, Domain of RR is ZZ and similarly, the range of RR is also ZZ .

(i) {(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)}{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)}

Thus, we can see, the domain of the relation consists of {2,5,8,11,14,17}{2,5,8,11,14,17} and range is {1}{1} .

And we also have, every element of the domain is having their unique images, then it is a function.

Ans: We have our given relation, {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)}{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)}.

Thus, we have our domain as, {2,4,6,8,10,12,14}{2,4,6,8,10,12,14} and range as, {1,2,3,4,5,6,7}{1,2,3,4,5,6,7} .

Every element of the domain is having their unique images, so this is a function.

From the domain of the relation the element 11 is having two different images 3,53,5 .

So,every element of the domain is not having their unique images. So, this is not a function.

(i)f(x)=−|x|(i)f(x)=−|x|

It is also know that, $\left| x \right|=\left\{ \begin{matrix} x,if\,x\ge 0 \\

-x,if\,x<0 \\ \end{matrix} \right.$

Thus, $f(x)=-\left| x \right|=\left\{ \begin{matrix} -x,if\,x\ge 0 \\ x,if\,x<0 \\ \end{matrix} \right.$ As the function is a real function, the domain of the function is RR . And again, we can see that the function is giving values of all real numbers except positive ones. So, the range of the function is, (−∞,0](−∞,0] .

Let us try to write the value of the given function in a tabular form as,

xx

0.010.01

0.10.1

0.90.9

11

22

2.52.5

44

55

…

f(x)f(x)

1.971.97

1.71.7

−0.7−0.7

−1−1

−4−4

−5.5−5.5

−10−10

−13−13

…

We can now see, it can be seen that the elements of the range is less than 2.

So, the range will be, f=(−∞,2)f=(−∞,2)

Alternative way to solve

Let us take, x>0x>0

We can again go forward by writing,

3x>03x>0

⇒2−3x<2⇒2−3x<2

⇒f(x)<2⇒f(x)<2 So, the range of ff is (−∞,2)(−∞,2)

After further simplification,

x2+2≥2×2+2≥2

⇒f(x)≥2⇒f(x)≥2

Thus, the range of the function is =[2,∞)=[2,∞)

Now, we can clearly have, that the range of the function is the set of all the numbers.

So, the range of the function will be, RR .

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