{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 
Answer:
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.
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