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

Exercise 2.1

1. If (x3+1,y−23)=(53,13)(x3+1,y−23)=(53,13) , find the values of xx and yy .

Ans: We are provided with the fact that (x3+1,y−23)=(53,13)(x3+1,y−23)=(53,13)
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.

2. If the set A has 3 elements and the set B={3,4,5} , then find the number of elements in (A×B)(A×B) ?

Ans: We are provided with the fact that the set AA has 33 elements and the set BB is given as {3,4,5}{3,4,5} .
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 .

3. If G={7,8}G={7,8} and H={5,4,2}H={5,4,2} , find G×HG×H and H×GH×G .

Ans: We have the sets G={7,8}G={7,8} and H={5,4,2}H={5,4,2} .
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)}

4. State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.
(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)} .

Ans: The statement is False.
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)}

(ii) If AA and BB are non-empty sets, then A×BA×B is a non-empty set of ordered pairs (x,y)(x,y) such that x∈Ax∈A and y∈By∈B.

Ans: The statement is True.

(iii) If A={1,2},B={3,4}A={1,2},B={3,4} , then A×{B∩∅}=∅A×{B∩∅}=∅ .

Ans: The statement is True.
We know, B∩∅=∅B∩∅=∅
Thus, we have, A×{B∩∅}=A×∅=∅A×{B∩∅}=A×∅=∅.

5. If A={−1,1}A={−1,1} , find A×A×AA×A×A .

Ans: For any non-empty set AA , the set A×A×AA×A×A is defined by,
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)}

6. If A×B={(a,x),(a,y),(b,,x),(b,y)}A×B={(a,x),(a,y),(b,,x),(b,y)} . Find AA and BB .

Ans: We are provided with the fact that A×B={(a,x),(a,y),(b,x),(b,y)}A×B={(a,x),(a,y),(b,x),(b,y)}
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} .

7. Let A={1,2},B={1,2,3,4},C={5,6}A={1,2},B={1,2,3,4},C={5,6} and D={5,6,7,8}D={5,6,7,8} . Verify that
(i) A×(B∩C)=(A×B)∩(A×C)A×(B∩C)=(A×B)∩(A×C)

Ans: We are provided with 3 sets and we have to prove 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.

(ii) A×CA×C is a subset of B×DB×D

Ans: Again, we are to verify, A×CA×C is a subset of B×DB×D
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 .

8. Let A={1,2}A={1,2} and B={3,4}B={3,4} . Write A×BA×B . How many subsets will A×BA×B have? List them.

Ans: We are provided with the fact that A={1,2}A={1,2} and B={3,4}B={3,4}
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)}

9. Let AA and BB be two sets such that n(A)=3n(A)=3 and n(B)=2n(B)=2 . If (x,1),(y,2),(z,1)(x,1),(y,2),(z,1) are in A×BA×B , find AA and BB , where x,yx,y and zz are distinct elements.

Ans: We are provided with the fact that n(A)=3n(A)=3 and n(B)=2n(B)=2 ; and
(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} .

10. The Cartesian product A×AA×A has 9 elements among which are found (−1,0)(−1,0) and (0,1)(0,1) . Find the set AA and the remaining elements of A×AA×A .

Ans: We are provided with, n(A×A)=9n(A×A)=9 .
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 .

Exercise 2.2

1. Let A={1,2,3,……,14}A={1,2,3,……,14} . Define a relation RR from AA to AA by R={(x,y):3x−y=0}R={(x,y):3x−y=0} , where x,y∈Ax,y∈A . Write down its domain, codomain and range.

Ans: We are given with the relation RR from AA to AA as, R={(x,y):3x−y=0}R={(x,y):3x−y=0} where x,y∈Ax,y∈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} .

2. Define a relation R on the set N of natural numbers by R=(x,y) : y=x+5x is a natural number less than 4; x,y in N . Depict this relationship using roster form. Write down the domain and the range.

Ans: We are given with the fact that, R=(x,y): y=x+5,x is a natural number less than 4; x,y in N.
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} .

3. A={1,2,3,5} and B={4,6,9} . Define a relation R from A to B by R=(x,y):the difference between x and y is odd; x in A,y in B . Write R in roster form.

Ans: We are provided with the fact that A={1,2,3,5}A={1,2,3,5} and B={4,6,9}B={4,6,9} .
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)}

4. The given figure shows a relationship between the sets PP and QQ . Write this relation
(i) In set-builder form

Ans: From the given figure in the problem, we have, P={5,6,7},Q={3,4,5}P={5,6,7},Q={3,4,5}
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}

(ii) In roster form
What is its domain and range?

Ans: From the given figure in the problem, we have, P={5,6,7},Q={3,4,5}P={5,6,7},Q={3,4,5}
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} .

5. Let A={ 1,2,3,4,6 } . Let R be the relation on A defined by (a,b):a,b in A,b is exactly divisible by a .
(i) Write RR in roster form.

Ans: We are provided with the fact that,$A=\{1,2,3,4,6\},R=(a,b):a,b in A,b is exactly divisible by a
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.

(ii) Find the domain of R

Ans: We can clearly see, the domain of RR is, {1,2,3,4,6}{1,2,3,4,6}

(iii) Find the range of R.

Ans: And similarly, the range of RR is, {1,2,3,4,6}{1,2,3,4,6}

6. Determine the domain and range of the relation RR defined by R={(x,x+5):x∈{0,1,2,3,4,5}}R={(x,x+5):x∈{0,1,2,3,4,5}}

Ans: We are provided with the fact, R={(x,x+5):x∈{0,1,2,3,4,5}}R={(x,x+5):x∈{0,1,2,3,4,5}}
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)}.

7. Write the relationR=(x,x3)R=(x,x3): x is a prime number less than 10 in roster form.

Ans: We are provided with the fact that,
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)}

8. Let A={x,y,z}A={x,y,z} and B={1,2}B={1,2} . Find the number of relations from AA to BB .

Ans: The facts provided to us are, A={x,y,z}A={x,y,z} and B={1,2}B={1,2} .
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 .

9. Let RR be the relation on ZZ defined by R={(a,b):a,b∈Z,a−bisaninteger}R={(a,b):a,b∈Z,a−bisaninteger} . Find the domain and range of RR .

Ans: The relation is given as, R={(a,b):a,b∈Z,a−bisaninteger}R={(a,b):a,b∈Z,a−bisaninteger} .
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 .

Exercise 2.3

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(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)}

Ans: We have the given relation as, {(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.

(ii) {(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)}
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.

(iii) {(1,3),(1,5),(2,5)}{(1,3),(1,5),(2,5)}

Ans: Our given relation is, {(1,3),(1,5),(2,5)}{(1,3),(1,5),(2,5)}.
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.

2. Find the domain and range of the following real function:
(i)f(x)=−|x|(i)f(x)=−|x|

Ans: We have the given function as, f(x)=−|x|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] .

(ii)f(x)=9−x2−−−−−√(ii)f(x)=9−x2

Ans: The function is given as, f(x)=9−x2−−−−−√f(x)=9−x2 We can clearly see that the function is well defined for all the real numbers which are greater than or equal to −3−3 and less than or equal to 33 , thus, the domain of the function is, {x:−3≤x≤3}{x:−3≤x≤3} or [−3,3][−3,3] . And for such value of xx , the value of the function will always be between 00 and 33. Thus, the range is, {x:0≤x≤3}{x:0≤x≤3} or [0,3][0,3] .

3. A function is defined by f(x)=2x−5f(x)=2x−5 (i) f(0)f(0)

Ans: We have the given function as, f(x)=2x−5f(x)=2x−5 So, the value of, f(0)=2×0−5=−5f(0)=2×0−5=−5

(ii) f(7)f(7)

Ans: We have the given function as, f(x)=2x−5f(x)=2x−5 So, the value of, f(7)=2×7−5=14−5=9f(7)=2×7−5=14−5=9

(iii) f(−3)f(−3)

Ans: We have the given function as, f(x)=2x−5f(x)=2x−5 So, the value of, f(−3)=2×(−3)−5=−6−5=−11f(−3)=2×(−3)−5=−6−5=−11

4. The function ′t′′t′ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C)=9C5+32t(C)=9C5+32 . Find

(i) t(0)t(0)

Ans: We have our given function as, t(C)=9C5+32t(C)=9C5+32 . Thus, to find our needed values of the function we just have to put the values in the given function and simplify it. So, we get now, t(0)=9×05+32=32t(0)=9×05+32=32

(ii) t(28)t(28)

Ans: We have our given function as, t(C)=9C5+32t(C)=9C5+32 . Thus, to find our needed values of the function we just have to put the values in the given function and simplify it. So, we get now, t(28)=9×285+32t(28)=9×285+32 =252+1605=252+1605 =4125=4125 =82.4=82.4

(iii) t(−10)t(−10)

Ans: We have our given function as, t(C)=9C5+32t(C)=9C5+32 . Thus, to find our needed values of the function we just have to put the values in the given function and simplify it. So, we get now, t(−10)=9×(−10)5+32t(−10)=9×(−10)5+32 =−18+32=−18+32 =14=14

(iv) The value of CC, when t(C)=212t(C)=212 .

Ans: We have our given function as, t(C)=9C5+32t(C)=9C5+32 . For this problem, we are given that, t(C)=212t(C)=212 . So, it can be written as, 9C5+32=2129C5+32=212 Simplifying further, 9C5=212−329C5=212−32 ⇒9C5=180⇒9C5=180 ⇒C=9009=100⇒C=9009=100 Thus, it can be said that, for t(C)=212t(C)=212, the value of tt is 100100 .

5. Find the range of each of the following functions:

(i) f(x)=2−3x,x∈R,x>0f(x)=2−3x,x∈R,x>0 .

Ans: We have the given function as, f(x)=2−3x,x∈R,x>0f(x)=2−3x,x∈R,x>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)

(ii) f(x)=x2+2,xisarealnumber.f(x)=x2+2,xisarealnumber.

Ans: We have our given function that,f(x)=x2+2f(x)=x2+2 Let us try to write the value of the given function in a tabular form as, xx 00 ±0.3±0.3 ±0.8±0.8 ±1±1 ±2±2 ±3±3 … f(x)f(x) 22 2.092.09 2.642.64 33 66 1111 … So, we see that the range of the function ff is the set of all numbers which are greater than or equal to 2. Thus, we can conclude that the range of the function is, [2,∞)[2,∞) . Alternative Method: Let xx be any real number. So, x2>0x2>0
After further simplification,
x2+2≥2×2+2≥2
⇒f(x)≥2⇒f(x)≥2
Thus, the range of the function is =[2,∞)=[2,∞)

(iii) f(x)=x,xf(x)=x,x is a real number .

Ans: We have our given function as, f(x)=x,xisarealnumberf(x)=x,xisarealnumber .
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 .