} is neither
reflexive nor symmetric nor transitive.
b} is
reflexive and transitive
but not symmetric.
} is reflexive,
symmetric or transitive.
given by
R
= {(a, b) : |a ? b| is a multiple of 4}
(ii)
R = {(a, b) : a = b}
is an equivalence relation. Find the set of all
elements related to 1 in each case.
Is
the circle passing through P with origin as centre.
is similar to
T2}, is equivalence relation. Consider three right angle triangles T1
with sides 3, 4, 5. T2 with sides 5, 12, 13 and T3 with
sides 6, 8, 10. Which triangles among T1, T2 and T3
are related ?
R (B)
(3, 8) R
(C) (6, 8) R (D)
(8, 7) R
defined
by 
is one-one
and onto, where R* is the set of all non zero real numbers. Is the result true,
if the domain R* is replaced by N with co-domain being same as R* ?
N given by
f(x) = x2
(ii)
f : Z Z given by
f(x) = x2
(iii)
f : R R given by
f(x) = x2
(iv) f : N N given by
f(x) = x3
(v) f : Z Z given by
f(x) = x3
R given by
f(x) = [x], is neither one one nor onto, where [x] denotes the greatest integer
less than or equal to x.
given by f(x)
= |x|, is neither one one nor onto, where |x| is, if x position or 0 and |x| is
? x, if x is negative.
given by 
is neither
one-one nor onto.
defined by
f(x) = 3 ? 4x
(ii) defined by
f(x) = 1 + x2
such
that f(a, b) = (b, a) is bijective functrion.
such
that f(a, b) = (b, a) is bijective functrion.
N be defined
by
for
all 
, state
whether the function f is bijective. Justify your answer.
defined by 
Is f one-one
and onto? Justify your answer.
be defined
as f(x) = 3x, choose the correct answer
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not
onto
(d) f is neither one-one
onto.
{1,
2, 5} and g : {1, 2, 5} {1,
3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.
Write down gof.
show that
(fof) (x) = x for all 
What
is the inverse of f?
{10} with
f = {(1, 10),
(2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7,
8} {1, 2, 3, 4}
with
g = {(5, 4), (6,
3), (7, 4) (8, 2)}
(iii) h : {2, 3, 4,
5} {7, 9, 11,
13} with
h : {(2, 7), (3,
9), (4, 11), (5, 13)}.
, given by 
is one-one.
Find the inverse of the function f : [?1, 1] 
Rnage f.
R given by
f(x) = 4x +| 3. Show that f is invertible. Find the inverse of f.
is given by
f(x) = x2 + 4 show that f is invertible with the inverse f-1
of given by 
where R+
is set of all non negative real numbers.
given
by f(x) = 9x2 + 6x ?5. Show that f is invertible with 
{1,
2, 5} and g : {1, 2, 5} {1,
3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.
Write down gof.
Sol.
Given f = {(1, 2), (3, 5), (4, 1)}
Here
Rf = {1, 2, 5} = Dg and Df = {1, 3, 4}
Now,
(gof) (1) = g(f(1)) = g(2) = 3
(gof)(3)
= g(f(3)) = g(5) = 1
(fog)
(4) = g(f(4)) = g(1) = 3
So,
gof = {(1, 3), (3,1), (4,3)}
E2. Let
f, g and h be function from R to R.
show
that (f + g) oh = foh + goh
(f.g)oh
= (foh).(goh)
Sol.
Let 
be any
element
=
f(h(x)+ g(h(x))
=
(foh) (x) + (goh) (x)
=
(foh + goh) (x)
=
(foh + goh) (x)
Also,
((f.g) oh) (x) = (f.g) (h(x))
=
f(h(x)).g(h(x))
=
(foh)(x). (goh)(x)
=
((foh).(goh) (x)
which
is proved.
E3. Find gof and fog. If
(i)
f(x) = |x| and g(x) = |5x ? 2|
(ii)
f(x) = 8x3 and g(x) = x1/3
Sol. (i) Given
f(x)
= |x|, g(x) = |5x ? 2|
(gof)
(x) = g(f(x)) = g(|x|) = |5|x|?2|
(fog)
(x) = g(g(x)) = f(|5x ? 2|)
=
||5x ? 2|| = |5x ? 2|
(ii)
Given f(x) = 8x3 and g(x) = x1/3
(gof) (x) = g(f(x)) = g(8x3) = 
E4.
If 
show that
(fof) (x) = x for all 
What
is the inverse of f?
Sol.
Given f(x) = 
(fof) = f(f(x)) = 
E5. State
with reason whether following functions have inverse.
(i) f : {1, 2,
3, 4} {10} with
f = {(1, 10),
(2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7,
8} {1, 2, 3, 4}
with
g = {(5, 4), (6,
3), (7, 4) (8, 2)}
(iii) h : {2, 3, 4,
5} {7, 9, 11,
13} with
h : {(2, 7), (3,
9), (4, 11), (5, 13)}.
Sol. (i) f : {1,
2, 3, 4} {10}
f(1) = 10, f(2)
= 10, f(3) = 10, f(4) = 10
is not
one-one.
(ii) g : {5, 6, 7, 8}
{1, 2, 3, 4}
Here g(5) = 4 and g(7) = 4
for 
is not
one-one.
(iii) h : {2, 3, 4, 5} {7, 9, 11,
13}
= 7, h(3) =
9, h(4) = 11, h (5) = 13
is one-one.
Also for every 
there exists
distinct value of x in {2, 3, 4, 5}
function ?h?
is onto.
Hence function
?h? is invertible.
is given by
E6. Show that f
: [?1, 1] 
, given by 
is one-one.
Find the inverse of the function f : [?1, 1] Rnage f.
Sol.
To find range of
f : Let y = f(x)
Injectivity :
Let x1, x2 
To find inverse
of f : Let y = f(x)
i.e., 
E7. Consider f
: R R given by
f(x) = 4x +| 3. Show that f is invertible. Find the inverse of f.
Sol. Given f(x)
= 4x + 3, Df =
Injectivity :
Let x1, x2 be any two elements of R
is one-one.
Surjectivity:
Let 
then y = f(x)
Corresponding
to each element y of R, there exists a unique element 
of R such
that 
is onto.
is one-one
and onto function.
exists.
To find inverse of f :
E8. Consider
f : R+
is
given by f(x) = x2 + 4 show that f is invertible with the inverse f-1
of given by 
where R+
is set of all non negative real numbers.
Sol.
Given f(x) = x2 + 4, Df = R+ = [0, 
)
Injectivity
: Let x1, x2 be any two elements of R+
such that f(x1) = f(x2)
But
both x1 and x2 are positive real numbers
Therefore
x1 = x2
So
f : R+ is
one-one.
Surjectivity
: Let 
then y = f(x)
Corresponding
to every element y of 
there
exists a unique element 
such
that 
is onto.
Hence
f is both one-one and onto
To
find inverse of f : 
E9.
Consider f : R+ 
given
by f(x) = 9x2 + 6x ?5. Show that f is invertible with 
Sol.
Given f(x) = 9x2 + 6x ? 5, Df = R+
Injectivity
: Let x1, x2 be any two elements of R+
such that f(x1) = f(x2)
Surjective
: Let y be any element of [?5, )
then
y
= f(x) 
Since
, therefore 
is not
possible.
co-domain
is onto.
is both
one-one and onto.
exists.
To
find inverse of f :
E10.
Let f : 
be an
invertible function. Show that f has unique inverse.
Sol.
Let g1 and g2 are two inverses of f
Therefore
for al 
(fog1)
(y) = f(g1(y)) = y
is
invertible function is
one-one]
inverse of f
is unique.
E11. Consider f
: {1, 2, 3} {a, b, c}
given by f(1) = a, f(2) = b and f(3) = c. Find f?1 and show that (f?1)?1
= f.
Sol.
Given f : {1, 2, 3} {a,
b, c}
for
distinct values of x in {1, 2, 3}, there are distinct values of y in {a,
b, c}.
is onto.
Hence
f is both one-one and onto.
is
invertible.
and
Now
for distinct values of x in {a, b, c}, there are distinct values of y in {1, 2,
3}
is one-one.
Also
for every 
there exists
a unique element 
{a, b, c}
given by f(1) = a, f(2) = b and f(3) = c. Find f?1 and show that (f?1)?1
= f.
be
an invertible function. Show that the inverse of f?1 is f i.e. (f?1)?1
= f
be given by
f(x)
= (3 ? x3)1/3, (fof) (x) is
(a)
x1/3 (b) x3
(c)
x (d) (3 ? x3)
R = {(1, 3),
(2, 6), (3, 9), (4, 12)}
Reflexive : As 1 
A but (1, 1) 
R.
R is not
reflexive.
Symmetric : As (1, 3) 
but
(3, 1) 
R is not
symmetric.
Transitive : As (1, 3) 
and
(3, 9) but (1, 9) ,
R is not
transitive.
Hence relation R is neither reflexive, nor
symmetric, nor transitive.
(ii) N = {1, 2, 3, 4, 5 6 ??}
R = {(x, y) : y = x + 5 and x < 4}
= {(1, 6),
(2, 7), (3, 8)}
Reflexive : As 
but (1, 1), 
is
not reflexive.
Symmetric : As (1, 6) but
(6, 1).
is not
symmetric.
Transitive : Clearly R is transitive
since it is not contradicted here.
Hence relation R is transitive but neither
reflexive nor symmetric.
(iii)
A = {1, 2, 3, 4, 5, 6}
R = {(x, y) : y is divisible by x}
R = {(1, 1),
(2, 2), (3, 3), (4, 4), (5, 5), (6, 6),
(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6), (3, 6)}
Reflexive : As (a, a) 
is reflexive
Symmetric : A (1, 2) 
but (2, 1)
is not
symmetric.
Transitive : As (a, b) and (b, c) 
is divisible
by a and c is divisible by b
is divisible
by 
(a, c) 
is
transitive.
Hence relation R is reflexive, transitive but not symmetric.
(iv) Z = {?.., ?3, ?2, ?1, 0, 1, 2, 3, ?}
R = {(x, y) : x ? y is an integer.
Reflexive : As a ? a = 0 is an integer 
is symmetric.
Symmetric : As a ? b and b ? a are integers 
is symmetric.
Transitive : As a ? b and b ? c are integers and
(a ? b) + (b ? c) = a ? c is also an integer.
and (b, c) 
is transitive.
Hence R is reflexive, symmetric and transitive.
(v) (a) Clearly R is reflexive, symmetry and transitive. (b)
Clearly R is reflexive, symmetric and transitive.
(c) A = {x : x is human being in a town}
R = {(x, y) : x is exactly 7 cm taller than y}
Reflexive : As a is not 7 cm taller than a.
is not
reflexive.
Symmetric : If a is exactly 7 cm taller than
b, then b cannot be 7 cm taller, than a
R is not
symmetric.
Transitive : If a exactly 7 cm taller
than b and b is exactly 7 cm taller than c then a is exactly 14 cm taller than
c.
is not
transitive.
Hence R is neither reflexive, nor symmetric
nor transitive.
(d) A = {x : x is human being}
R = {(x, y) : x is a wife of y}
Reflexive : As a is not wife of 
is not
reflexive.
Symmetric : If a is a wife of b then b cannot
be wife of a.
R is not
symmetric.
Transitive : If a is a wife of b then b
is a male ad a male cannot be a wife.
(a, b) 
R is
transitive as it is not contradicted here.
R is
transitive but neither reflexive nor symmetric.
(e) Clearly R is neither reflexive nor
symmetric nor transitive.
E2. Show that the relation R in the set R of
real numbers, defined as R = {(a, b) : a 
} is neither
reflexive nor symmetric nor transitive.
Sol. R = {(a, b) : a 
a,
b, 
}
Reflexive : 
i.e. 
(a, a) 
Hence R is not reflexive.
Symmetric : As 2, 5 and
2 < 25 i.e. 2 < 52.
(2, 5)
But 
4 i.e. 
but (5, 2)
Hence R is not symmetric.
Transitive : As 3, ?2 and ?1 and
3 < (2)2 and ?2 < (?1)2.
and (?2,
?1)
But 
is not
symmetric.
Transitive : (a, b) and
(b, c) 
and 
is
transitive.
Hence R is reflexive and transitive but not
symmetric.
Example E 3. Check whether the relation R defined in the set {1,2, 3,4,
5, 6} as
R = {(a, b) : b = a + 1} is reflexive,
symmetric or transitive.
Sol. t A = {1, 2, 3,
4, 5. 6}
R = {(a,
b) : b = a + 1, a, b A}.
R =
{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.
Reflexive
: As 1 e A and (1, 1) R
R
is not reflexive.
Symmetric
: As 2, 3 A and (2, 3) R.
But
R
is not symmetric,
Transitive
: As 2. 3 and 4 A and (2, 3) R,
(3, t (2,4) R '
R
is not transitive.
Hence
R is neither reflexive, nor symmetric, nor transitive.
E 4. Show that the relation R defined
as R = {(a, b) : a 
b} is
reflexive and transitive
but not symmetric.
Sol. R
= {(a, b) : a b, a, b R
}
Reflexive
: As a a 
a
(a,
a) a
R
is reflexive.
Symmetric
: As 1, 2 and 1 2
(1,
2)
But
is
not symmetric.
Transitive
: (a, b) and (b, c) 
a,
b,c
and
R
is transitive.
Hence
R is reflexive and transitive but not
symmetric.
E5. Check whether the relation R in R defined
by R = {(a, b) : a 
} is
reflexive, symmetric or transitive.
Sol. R = {(a, b) : 
}
Reflexive : as 
i.e. 
is not
reflexive.
Symmetric : As 1 < 27 for 1, 3
i.e. (1, 3) for
1, 3
But 
is not
symmetric.
Transitive : As 100 < 125 and 5 <
8
i.e. 100 < 53 and 5 < 23
(100, 5) and
(5, 2)
But 100 
8 i.e. 100
< 23
(100, 2) 
(100, 5) ,
(5, 2) 
R is not
transitive.
Hence R is neither reflexive nor symmetric nor
transatie.
E6. Show that the
relation R is the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but
neither reflexive nor transitive.
Sol. Let A = {1, 2, 3} : R = {(1, 2), (2, 1)}
Reflexive : As 
and
(1, 1)
is not
reflexive.
Symmetric : As 1, 2 and
(1, 2)
(iv) On
Z+, define a * b = 2ab
(v) On
Z+, define a * b = ab
(vi) On
R ? {?1} define a * b 
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
(vi)
a * b = ab2
Find
which of i.e. binary operations are commutative and which are associative.
(ii)
If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a.
be
defined as f(x) = 10x + 7. Find the function g : 
such that gof
= fog = IR.
be defined as
f(n) = n ? 1, if n is odd and f(n) = n + 1, if n is even. Show that f is
invertible. Find the inverse of f. Here, W is the set of all whole numbers.
is defined by
f(x) = x2 ? 3x + 2 find f(f(x)).
defined by
is
one-one and onto function.
R
given by f(x) = x3 is injective.
Z
and g : Z Z such that
gof is injective but g is not injective.
N
and
g
: N N such that
gof is onto but f is not onto.
. Is R an
equivalence relation on P(X). Justify your answer.
given by A *
B =
where P(X) is
the power set of X. Show that X is the identity element for this operation and
X is the only invertible element in P(X) with respect to the operation.
defined
as a * b = |a ? b| and a o b = a, 
Show
that * is commutative but not associative, o is associative but not
commutative. Further, showthat 
[If
it is so, we say that the operation * distributes over the operation o.] Does o
distribute over * ? Justify your answer.
be defined
as A * B = (A ? B) 
Show
that the empty set 
is
the identity for the operation * and all the elements A of P(X) are invertible
with A?1 = A. (Hint : 
and (A ? A) 
Show
that 
Zero
is the identity for this operation and each element a fo the set is invertible
with 6-a being the inverse of a.
B be
functions defined by f(x) = x2 ? x, 
and g(x) = 
1, 
Are f and g
equal ? Justify your answer.
(Hint
: One may note that two functions f : A B and
g : A B such that
f(a) = g (a) 
are called
equal functions.
R be the
signum functions defined as 
and
g : R R be the
Greatest Integer Function given by g(x) = [x], where [x] is greatest integer
less than or equal to x. Then, does fog and gof coincide in (0, 1] ?
(B)
(C) 0< y <
(D)
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