NCERT BOOK I SOLUTIONS
FUNCTIONS ,RELATIONS AND BINARY OPERATIONS
Exercise 1.1
1.(i)
R   x, y  : 3x  y  0
R is not reflexive since 1,1  R as 31 1  0
R is not symmetric since 1,3  R
R is not transitive since 1,3  R
(ii)

3 1  3  0 but  3,1  R
3,9  R
but 1,9  R
R is not reflexive since 1,1  R as 1  1  5 is not true
R is not symmetric since 1,6  R But  6,1  R (Indeed 6 is not there in the domain)
R is transitive since if y  x  5

z  y  5 then on adding y  z  x  y  5
(iii)
z  x5 
R is certainly reflexive since  x, x   R for all x  A .
R is not symmetric since if  x, y   R then
 2,6  R
 y, x  may not belong to R . For instance
as 6 is divisible by 2 but  6, 2  R as 2 is not divisible by 6 .
It is transitive since
 x, y   R
 y   x ,  y, z   R  z   y  ,  are integers)
Then z  . x   x where  is not another integer 
(iv)
 x, z   R
 x, z   R
R is reflexive since x  x is an integer for all integer x ' s
(Indeed x  x is zero which is an integer for all x  R but this has no relevence here).
R is symmetric since if x  y is an integer then y  x is also an integer.
Finally R is transitive since if  x, y   R,  y, z   R then x  z  x  y  y  z =integer
(v)(a)
x and x work at the same place. If x and y work at same place then y and x also work at
the same place,
1
If x and y work at same place , y and z work at the same place then x and z work
at the same place. Thus R is reflexive, symmetric and transitive
 R is an equivalence relation.
(b)
Same as (a)
(c)
Cant be reflexive because no body can be taller himself by 7cm. R can not be symmetric
either (why?)
R is not transitive since if x is taller than y by 7cm. y is taller than z by 7cm. then x
is taller than z by 14cm. 
(d)
 x, z   R
No female is wife of herself !  R is not reflexive. Again if x is wife is y then y is
husband of x . 
R is not symmetric. Now comes a surprise!!. If  x, y   R then x
is wife of y .  y is a man and y can not be wife of anybody.
(e)
2.

No question of

R is transitive
 y, z  whether it belongs or does not belongs to R
arises.
Similar to (d)
a  a 2 is not true if a is a real number greater than 1 .
 R is not reflexive. If a  b 2 then b  a 2 is not true in general.
For example  2,5  R
2  25
But  5, 2   R as 5  22 is not true
 R is not symmetric.
Finally  5, 4  R,  4, 2  R .. But  5, 2   R  R is not transitive.
3.
a  a 1 is never true. R is not reflexive. If a  b  1 then b can not be equal to a  1
 R is not symmetric.
If a  b  1, b  c  1 then a  c  2
4.

 a, c   R 
R is not transitive.
a  a is always true .  R is reflexive. If a  b then b  a may not be true.
For example 3  5 but 5  3 is false.  R is not symmetric
Finally a  b, b  c  a  c 
R is transitive.
2
5.
a  a 3 is not true for all a (For example take a 
1
.
2
If a  b3 then b  a 3 may not be true. For example
 2,9  R
2  93
But  9, 2   R as 9 is not less than 23 .  R is not symmetric.
6.
R does not contain 1,1 and  2, 2   R is not reflexive.
Again 1, 2  R ,  2,1  R
and vice versa  R is symmetric.
Now 1, 2  R,  2,1  R but 1,1  R
7.
8.
 R is not transitive.
Follow V(a) above.
a  a  0 which is even . If a  b is even then b  a is also even since a  b  b  a
Now let a  b , b  c be even then a  c  a  b  b  c  even  even  even
 R is transitive.
Since difference of any two odd integers is an integer all the elements of 1,3,5 are related
to each other. Again difference of any two even integers is even, therefore all the
elements of 2, 4 are related to each other.
Finally difference of an odd and an even integer is not even.

9.(i)
No element of 1,3,5 is related to an element of 2, 4
a  a  0 which is a multiple of 9 . 
R is reflexive
Next, a  b  b  a
 if a  b is a multiple of 4 then b  a which is equal to a  b is a also a multiple of 4 .
Finally if a  b  4m, b  c  4l then
a c  a b bc
 4m  4l which is a multiple of 4 . 
R is transitive.
Thus R is an equivalence relation.
(ii)
R is reflexive since a  a for all a
3
R is symmetric since a  b  b  a
R is transitive since a  b, b  c  a  c  R is an equivalence relation.
10.(i)
(iii)
R  1, 2 ,  2,1 on 1, 2
(ii)
R  1, 2 ,  2,3 , 1,3 on 1, 2,3
R  1,1 ,  2, 2,  ,  3,3 , 1, 2 ,  2,1 , 1,3 , 3,1 on 1, 2,3
Note that R is reflexive and symmetric but R is not transitive as  2,1  R, 1,3  R but
 2,3  R
(iv)
R  1,1 ,  2, 2 , 1, 2  on 1, 2 is not transitive.
(v)
R  1,1 ,  2,1 , 1, 2 ,  2, 2  on 1, 2,3 is symmetric and transitive but not reflexive.
11.
R is reflexive since P and P are equidistant. R is symmetric since if P and Q are
equidistant then so are Q and P . Finally R is transitive since if P, Q are equidistant from
origin, Q and R are equidistant from origin then so are P and R . Thus R is an
equivalence relation. If P is a point which is not origin then all the points on the circle with
centre at origin and radius as OP will be related to the point P . For example if P is  3, 4
(as a co-ordinates) then equivalence class generated by  3, 4 is the circle x 2  y 2  25 .
12.
R is an equivalence relation easily follows. Now a triangle with side 3, 4,5 can not be similar
to a triangle having sides 5,12,13 but is definitely similar to the triangle having sides 6,8,10
since
13.
3 4 5
  .
6 8 10
R is an equivalence relation easily follows. The fact that 3, 4,5 are the sides of a right angled
triangle has nothing to do with relation. It will be indeed related to all polysons (triangles)
having three sides.
14.
The set of lines related to y  2 x  4 will be set of all lines which are parallel to y  2 x  4 .

15.
Such lines are y  2 x   where  is an arbitrary constant.
(A) is incorrect as R is not symmetric
3, 2  R,  2,3  R
(C) is incorrect as R is not symmetric
(D) is incorrect since R is not symmetric
(B) is correct as R is reflexive and transitive but not symmetric.
4
16.
(A) is false since 4 is not greater than 6
(B) is false since  3,8  R as 3  8  2 is not true.
(C) is correct since 6  8  2
(D) is false since 8  7  2
EXERCISE 1.2
1.
f  x  is clearly one to one since f  x1   f  x2  

x1  x2 . Also
1
1

x1 x2
1
attains all values in  ,0   0,    f  x  is onto.
x
If co-domain is replaced by R and domain is N then f will not attain all the values for
example 2 3 will never be attained as 2 3 is not reciprocal of any natural number.
2. (i)
f  x   x2 is clearly one-one
Since x12  x22

 x1  x2  x1  x2   0
x12  x22  0 
Now x1  x2 can not be zero as x1 and x2 are natural numbers

x1  x2  0

x1  x2
f  x   x2 is not onto since f will attain only the values 12 , 22 ,32 ,.....
(ii)
f : Z  Z is not injective since  5    5 
2
2
f is not onto either since negative integer in right Z will not be attained.
(iii)
f : R  R given by f  x   x2 will not be injective as  5    5  and will not be onto
2
either since only  0,   will be attained.
(iv)
f : N  N given by f  x   x3 is injective since n13  n23
 n13  n23  0

n1  n2

 n1  n1   n12  n22  n1n2   0

(*)
n1  0, n2  0  n12  n22  n1n2  0 
But f will attain only perfect cubes therefore f will not be onto.
5
2
(v)
Following the solution of (iv) part we note that
2
1  3

n  n  n1n2   n1  n2   n22 which can be zero if and on it n1  n2  0 . Showing
2  4

injectivity and if this is not the case then n1  n2 from (*)
2
1
2
2
Since f will not attain non perfect cubes integers, f will not be onto.
NOTE:- f  x   x3 is injective on  ,  
3.
Since 1.2  1.3  1,
 x is not one-one. Moreover  x will attain only integral values
in R

4.
f : R  R given by f  x    x is neither one-one nor onto.
Since 5  5  5 ,
x is not injective. Moreover x will attain only non negative
in R  f : R  R given by f  x   x is neither one-one nor onto.
5.
f  5  f  6  0  f is not one-one. f attains only 3 values in R .

6.
f is neither one-one nor onto.
f 1  4, f  2  5, f  3  6

If x1 , x2 are different f  x1  and f  x2  are also different.  f is one-one.
(But f is not onto since f does not attain the value 7 in B ).
7. (i)
f is one-one since 3  4 x1  3  4 x2 

f is onto since y  3  4 x
x1  x2
x
3 y
4
 x exists for every y  R  f is bijective
(ii)
f is not one-one since 1   5   1   5 
2
2
f is not onto since f can not attain values which are small than 1 in R
8.
The element of A  B are of the form  a, b  where a  A, b  B .
6
values
Also f satisfies f  a, b    b, a 
f is clearly one-one since if f  a, b   f  c, d  then  b, a    d , c 

b  d,a  c 
 a, b    c, d 
Again any  p, q   B  A is image of  q, p   A showing f is onto
Thus f is a bijective function.
9.
f is not one-one since f  5  
5 1
6
 3 , f  6   3
2
2
But f is clearly onto since any m  N is image of some n  N
(Indeed there are exactly two elements in domain of N whose image is m . These are
2m and
2m 1
Since f is one-one but not onto f is not a bijection.
10.
f is one-one since

x1  2 x2  2

x1  3 x2  3
x1 x2  3x1  2 x2  6  x1 x1  2 x1  3x2  6  let y 
yx  3 y  x  2 or x 
3y  2

y 1
x2
then
x3
x exists for all y   R  1
 f is onto
Thus f is one-one and onto
11.
Similar to Q.2 (iii) due to even power f is neither one-one nor onto
12.
3x1  3x2  x1  x2  f is one-one .Every y  R on right R has

f is onto 
y
 R on left
3
f is one-one and onto
Execise 1.3
1.
g f is a function from 1,3, 4 to 1,3 . Therefore we must calculate ( g f ) images of
elements in 1,3, 4
7




Now ( g f ) (1)  g f 1  g  2  3 ,  g f  3  g f  3  g  5  1
 g f  4  g  f  4  g 1  3 , Thus  g f   1,3 , 3,1 ,  4,3
2.
 f  g h   x    f  g   h  x 
 f  h  x   g  h  x 
(By definition of f  g )
  f  h  x    g h  x 
 fg  h  x    fg   h  x 
3.(i)
 g f  x   g  f  x 
 g f  x   8x3 
13
  f  h  x  g h  x 
 g x  5 x 2
 f  g  x   f  g  x   
(ii)
 f  h  x  g  h  x 
5 x  2  5 x  2 (Note)
 2x ,
 f  g  x   8. x1 3 
3
 8x
4x  3
3
4x  3
34 x
6
x

4
f  x 
 f  f  x  

x
4x  3
6x  4
34
6.
4
6x  4
4.
4.

f  x   f 1  x  
4x  3
6x  4
Or f 1 can be found by solving y 
5.(i)
4x  3
4y  3
or x 
6x  4
6x  4
f is not one-one. Therefore it will not have inverse.

g  5  4, g  7   4  g is not invertible.
(ii)
g is not one-one
(iii)
h is clearly one-one and onto since distinct element have distinct images and all elements
7,9,11,13 have pre images
 f is invertible
6.
One-one ness follows by noting
x1
x
 2
x1  2 x2  2
Now range of f will not contain 1 . Since y 
8

x1  x2
x
2y
 x
x2
1 y
Showing 1 can not be attained by x .
x
x2
2y
g  y  given by g  y  
1 y
1,1  R  1 then f  x  is invertible with inverse
Thus if we define f  x  
7.
8.
f is clearly one-one and onto since 4 x1  3  4 x2  3  x1  x2
Also any y is
attainable through x 
Thus f 1  y  
y 3
.
4
y 3
which is, in fact inverse also.
4
In general f  x   x2  4 is not one-one. But over 0,  it is one-one since
 x12  x22  0 
 x1  x2  x1  x2   0
x12  4  x22  4
(*)
Now x1  x2 can be zero if and only x1  x2  0 . If this is not the case then from (*) x1  x2
Again, in general f  x   x2  4 is not onto but if co-domain is changed from R to  4, 
it is onto. This can also be observed by y  x 2  4  x 
f 1  y   y  4
9.
y  4 or
4,   0, 
Let us show that f  x  is one-one over 0, 
Let f  x1   f  x2  where x1 , x2 0,   then 9 x12  6 x1  5  9 x22  6 x2  5


9  x12  x22   6  x1  x2   0
 x1  x2  9  x1  x2   6  0
Again y  9 x 2  6 x  5 

1  1   5  y 
3
Thus f 1  y  

 x1  x2 since 9  x1  x2   6  6 for all x1 , x2  0,  
9 x2  6 x   5  y   0  x 
6  36  36  5  y 
18
1  y  6
we can not take  sign since x  0
3
1  y  6
3
9
Note that f 1  y  exists for y  6 and we are given co-domain as  5,  
10.
Following the hint given in the question

f  g1  y    f  g2  y  
But f is invertible 
11.
12.
 f  g1  y    f  g2  y 
g1  y   g2  y 
f is one-one therefore (*) 
f  1, a  ,  2, b  , 3, c  , f 1   a,1 ,  b, 2 ,  c,3 ,  f 1   1, a  ,  2, b  ,  3, c 
1
 
Let f 1  g then inverse of f is g . Now to prove f 1
1
 f it is sufficient to show `
g 1  f which is true from f 1  g
since f 1  g 
13.

g  f  x    x  f  g  x    f 1  g
f  f  x   3   f  x 

3 13
 3   3  x3 
13

x
(Also follows from the fact that f 1  x   3  x 3
14.

13

Solve x in terms of y
EXECISE 1.4
1.(i)
No: Since a  b can be negative integer.
(ii)
Yes, since product of two positive integers is a positive integer.
(iii)
Yes, since if a, b  R then ab 2  R
(iv)
Yes, since a  b is always a non negative integer.
(v)
Yes, since if a, b  Z 2 then a  Z is obvious.
2. (i)
a * b  a  b, b * a  b  a  a * b  b * a
 * is not commutative
Again a *  b * c   a *  b  c   a  b  c   a  b  c and  a * b  * c   a  b  * c  a  b  c

(ii)
a * b * c    a *b  * c 
* is not associative.
a * b  b * a is obvious
10
Now a *  b * c   a *  bc  1
 a  bc  1  1  abc  1
and  a * b  * c   ab  1  c   ab  1 c  1

 ab  c 1
 a * b  * c  a * b * c 
(iii)
Do yourself.
(iv)
a * b  2ab , b * a  2ba  2ab  a * b

* is commutative
Now a *  b * c   a * 2bc  2a.2
bc
 a * b  * c  2ab * c  22
(v)
ab
.c
 * is not associative
a * b  ab , b * a  ba  * is not commutative
 
 
Now a *  b * c   a * b c  a b and  a * b  * c  ab * c  ab
c

(vi)
3.
c
 abc
* is not associative
a
b

verify a *  b * c    a * b  * c
b 1 a 1
Observe the table for all possible operations results between elements of the set.

1 2 3 4 5
1
1 1 1 1 1
2
1 2 2 2 2
3
1 2 3 3 3
1 2 3 4 4
5
1 2 3 4 5
4.
Do yourself.
11
5.
Yes since , HCF of 1 and 1 is 1 , HCF of 1 and 2 is 1
HCF of 1 and 3 is 1
6.(i)
, HCF of 2 and 1 is 1 , HCF of 2 and 2 is 2 and so on.
5*7  LCM of 5 and 7  35 , 20*16  LCM of 20 and 16  80
(ii)
Yes since LCM of a and b is same as LCM of b and a .
(iii)
Yes, since LCM does not depend upon the order of numbers taken for calculating
(iv)
Now LCM of 1 and any arbitrary integer n is n only.
LCM .
 1 is the identity element.
(v)
If n is any non unity integer then there does not exist m such that LCM of n and m is 1

identity element does not exists.
But for 1, the inverse exists. Indeed 1 itself is inverse of 1.
8.
Like LCM , HCF also does not depend upon order, therefore * is commutative and
associative. Now let a  N then for the existence of identity there must exist e  N such
that a * e  a which is possible only when e  a

e depends upon a .  Identity element does not exist.
9.
Similar to Q.2. Do yourself.
10.
For(v) part in Q.9. a * e 

ae
a
4
e  4 which is independent of 4 
identity element exists and equals 4.
For any other part of Q.9 e dependents upon a .
11.
 a, b *  c, d    a  c, b  d 
 c, d  *  a, b   c  a, d  b 
  a  c, b  d 

addition is commutative in N )
  a, b  *  c, d  ,Thus * is commutative on N * N

Again  a, b  *  c, d  *  e, f
and

   a, b *  c  e, d  f 
  a  c  e, b  d  f 
 a, b *  c, d  * e, f    a  c, b  d  *  e, f    a  c  e, b  d  f 
* is associative.
12
For identity element e   ,   we must have  a, b  *  ,     a, b 

 a   , b      a, b 
a    a, b    b

 a  0, b  0 which is not possible since a and b are natural numbers.
12.(ii)
a * b * c   b * c  * a
* is commutative
  c *b * a
13.
* is commutative ,Thus statement is true.
* is certainly commutative.



Also a *  b * c   a * b3  c3  a3  b3  c3
 , a *b *c   a
3
3
 b3  * c   a 3  b3   c 3
3
 * is not associative.
Miscellaneous Execise
1.
The function g must be inverse of f
From y  10 x  7 we have x 
2.
f is one.

y7

7
g  y 
y7
7
Proof:- Let n1 , n2 be two odd numbers then f  n1   f  n2 
n1  1  n2  1

n1  n2
Again let n1 , n2 be two even numbers in then f  n1   f  n2 

n1  1  n2  1 
n1  n2
Now if n1 is odd and n2 is even then n1  n2 ,And also n1  1 can not be equal to n2  1
 n1 1  n2  1  n1  n2  2  odd  even, absurd 
Thus if n1  n2 then f  n1   f  n2 
f is onto since if m is odd then it is the image of m 1 , and if m is even then it is the f
image of m  1
Thus every element in right W has pre image in left W
13

f is one-one and onto
m  1 if
m  1 if
Thus f 1  m   
3.

m is odd
m is even
, Note that f 1 is f itself
f  f  x    f  x   3 f  x   2
2
  x 2  3x  2   3  x 2  3n  2   2
2
4.
f is invertible
Let
 x 4  6 x3  10 x 2  3 x
Thus y  f  x  
Solving y 
1  x1  0,1  x2  0 
x
is one-one function
1 x
 x
1  x
Let us now find range of y . Indeed y  
 x
1  x


x1
x2
where x1 , x2   1,1 then x1  x2

1  x1 1  x2
x
y
we get x 
1 x
y 1

x  0 or 1  x  0
x  0 or
0  x 1
x exists for all y  1
Range of function is  , 1  1,  
5.
x13  x23 
7.
Following the example given  g f  x   g f  x   g  x  1
x1  x2
(Recall Q.2 (v), Ex.1.2)

 x  1 if

if
 0
x 1  1
x 1  1

But x  0 is not possible since x  N
Now range of  g f
 is
6. HINT is good enough

 x  1 if
if
 0
 g f  x   
x0
x0
 g f  x   x  1 if

x0
subset 2,3, 4,5,....... of N . Every element in this set has a
pre-image. But f itself is not onto since 1 in N will not have any pre-image.
8.
If A is a subset of B then B is not subset of A
 R is not an equivalence relation.
14
 R is not symmetric
9.
Since for any A in P  X  ,
 A   X    X  A  X
We note that X is the identity element of *.
Now A  P  X  will have inverse B in P  X  if A  B  X which is possible only when
A  B  X . Thus X is the only invertible element.
10.
Any onto function from 1, 2,3,....n to itself must be one-one. Now number of one-one
functions = number of ways of distributing n distinct objects in n different boxes such that
each box contains exactly one-element.

Number of onto functions  n  n  1 n  2 ....3.2.1  n!
11.(i)
F 1  1, c  ,  2, b  , 3, a 
12.
Since a * b  a  b  b  a  b * a * is commutative
(ii)
But 3*  2*5   3* 2  5  3*  3 
Does not exist since F is not one-one.
 3*3  3  3  0
Again  3*2 *5  1*5  4  * is not associative
Again  is certainly not commutative but associative since a  bc   ab  a and
 abc  ac  a ,
Now LHS  a *  bc 
and RHS   a * b    a * c   a  b  a  c

 a *b  a  b
 a b
LHS  RHS  * distributes on 
Now a  b * c   a
But  ab  *  ac 
13.
 a*a  0 
LHS  RHS ,Thus  does not distributes *.
A * B   A  B    B  A
Now a set E will be identity element of P  X  if for any
set A of P  X 
A* E  A 
 A  E    E  A  A
which is possible
only when E  
15
(See the hint also). Thus empty  is the identity element.
Again element A will have an inverse B if A * B   or
Now  A  A   A  A  A * A  
14.

 A  B    B  A  
A itself in inverse of A .
We will do the problem by making multiplication table keeping mind
if
 ab
a *b  
a  b  6 if
ab  6
* 0 1 2 3 4 5
ab  6
0 0 1 2 3 4 5
1 1 2 3 4 5 0
(For example 4*4=4+4-6=2 etc.)
2 2 3 4 5 0 1
Since first row and first column remain same for 0, 0 is the left and
3 3 4 5 0 1 2
right identity. Hence
it is the identity element. Now for inverse we 4
proceed as follows.Let a 0,1, 2,3, 4,5
4 5 0 1 2 3
5 5 0 1 2 3 4
We must solve a * b  0 which will yield a  b  0 or a  b  6  0  b  6  a
Thus inverse of a will be 6  a .
15.
Note that x 2  x and 2 x 
Note:-
x 2  x and 2 x 
1
 1 are equal for x  1, 0,1, 2
2
1
 1 can not be equal on all sets. They are equal on the given
2
set A and takes same values.
16.
If a relation on A  1, 2,3 is reflexive it has to contain at least three ordered pairs
1,1 ,  2, 2 , 3,3 . Now R already contains 1, 2 and 1,3 . To make R symmetric we
add  2,1 and  3,1 so that R  1,1 ,  2, 2 ,  3,3 ,  2,1 , 3,1 , 1, 2  , 1,3
This relation is not transitive since  2,1  R, 1,3  R but  2,3  R . Now we can not
add any more ordered pairs into R since if
 2,3 is added then 3, 2 should also be
added
after which it becomes universal relation and hence transitive also.
17.
Any equivalence relation on A  1, 2,3 has to contain 1,1 ,  2, 2 ,  3,3 as every
equivalence relation is reflexive. If 1, 2  R then  2,1 must also be there since R has


to be symmetric also. Let us test R  1,1 ,  2, 2 ,  3,3 , 1, 2  ,  2,1 which is clearly
reflexive symmetric and transitive. If we add 1,3 into R then  3,1 will also be added
16
in order to be symmetric but it will then loose transitivity for which  2,3 shall have to be
added (From previous question) whence it looses symmetric character. To maintain
symmetric properly  3, 2 will have to be added whence it becomes A  A and R
becomes a universal relation on A . But universal relation is essentially an equivalence
relation. Thus number of equivalence relations on 1, 2,3 containing 1, 2  is 2 .
18.
1 x0

f  x    0 x  0, g  x    x  , 0  x  1,  f  g  x   f  g  x    f
1 x  0

Now, if 0  x  1 then  x   0 
If x  1,  x  1  f
f
 x   f  0   0
 x   f 1  1 ,

 x 
Thus


0 if
1 if
 f  g  x   
0  x 1
x 1

Again  g f  x   g f  x  , If x  0, then g f  x   g  0  0


If x  0 then g f  x   g 1  1 Thus  g f
Or
19.
0 if
1 if
 x   
x0
0  x 1
 f  g  x    g f  x 
Binary operation on S  a, b will be a function from S  S to S
The number of functions from S  S to S must be same as number of ways of distributing
four distinct object (four ordered pairs in S  S ) in two distinct boxes
 a or b 

2 4 operations are possible.
PAST YEAR CBSE QUESTIONS UP TO 2013
1.
Let A  {1,2,3,}, B  {4,5,7} and let f  {(1,4),(2,5),(3,6)} be a function from A to B. State
whether f is one-one or not.
Ans: one-one
2.
If f : R  R is defined by f ( x)  3x  2, define f[f(x)].
3.
Write fog, if f : R  R and g : R  R are given by : f ( x) | x | and g ( x)  | 5x  2 | .
Ans: 9x+8
Ans: ||5x-2||
4.
Write fog , if f : R  R and g : R  R are given by : f ( x)  8x3 and g ( x)  x1/3 Ans: 8x
17
5.
State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be
transitive. Ans: (1, 1) does not belong to R
6.
What is the range of the function f ( x) 
7.
If f : R  R and g : R  R are given by f ( x) =sin x and g ( x)  5x 2 , find gof (x). Ans: 5 sin2 x
8.
If f ( x)  27 x3 and g ( x)  x 3 , find gof(x).
9.
If the function f : R  R , defined by f ( x)  3x  4 is invertible, find f 1 . Ans:
10.
If f : R  R defined by f ( x) 
11.
State whether the function f : N  N , f ( x)  5 x is injective, surjective or both.
| x  1|
, x  1?
x 1
1
Ans: {-1, 1}
Ans: 3x
x+4
3
3x  5
2x-5
is an invertible function, find f 1 . Ans:
3
2
Ans: Injective only
12.
13.
14.
If f : R  R defined by f ( x) 
2x  7
4x+7
is an invertible function, find f 1 . Ans:
4
2
4x  3
2
Show that the function f in A = R    defined as f ( x) 
is one-one and onto. Hence
6x  4
3
3+4x
find f 1 .
Ans:
6x-4
Let A  R  {3} and B  R  {1} . Consider the function f : A  B defined by f ( x) 
Show that f is one-one and onto and hence find f 1 .
15.
Ans:
x2
.
x 3
3x-2
x-1
Let f : R  R be defined as f ( x)  10 x  7 . Find the function g : R  R , such that
gof  fog  I R .
Ans:
x-7
10
16.
Show that a function f : R  R given by f ( x)  ax  b, a, b  R, a  0 is a bijection
17.
If f : R  R be the function defined by f ( x)  4x3  7 , show that f is a bijection.
18.
If the function f : R  R is given by f ( x)  x2  3x  1 and g : R  R is given by 2x  3, find (i)
fog (ii) gof.
Ans: 4x2 -6x+1 , 2x2 +6x-1
18
19.
If the function f : R  R is given by f ( x) 
find (i) fog (ii) gof. Is f 1  g ?
20.
Ans :
x3
and g : R  R is given by g ( x)  2 x  3,
3
2x 2x  3
,
, no
3
3
Let Z be the set of all integers and R is the relation on Z defined as R  {(a, b) : a, b  Z and a-b is
divisible by 5}. Prove that R is an equivalence relation.
21.
Show that the relation S defined on set N  N by (a, b) S (c, d)  a  d  b  c is an equivalence
relation.
22.
Let* be a binary operation on N given by a*b=LCM (a, b) for all a, b  N. Find 5*7.
23.
Let*:R  R  R is defined as a*b=2a+b. Find (2*3)*4.
24.
If the binary operation * on the set of integers Z, is defined by a*b=a+3 b2 , then find the value of
8*3. Ans: 35
25.
If * is a binary operation on set of integers l, defined by a*b=3a+4b-2. Find value of 4*5. Ans: 30
26.
Let * is the binary operation on N given by a*b=HCF (a, b) where, a, b  N. Write the value of
22*4. Ans: 2
27.
Let * be a binary operation on set Q of rational numbers defined as a*b=
Ans:
Ans: 35
18
ab
. Write the identity
5
for *, if any. Ans: e=5
28.
Consider the binary operations *:R  R  R and o :R  R  R defined as a*b=|a-b| and aob=a. For
all a, b  R. Show that * is commutative but not associative, ‘o’ is associative but not
commutative.
29.
Let * be a binary operation on Q, defined by a * b 
associative. Also, find its identity, if it exists.
19
3ab
Show that * is commutative as well as
5
Ans e 
5
3