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Byju's Answer
Standard XII
Mathematics
Derivative of One Function w.r.t Another
Let A=1,2,3,4...
Question
Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f : A → B, g : B → C be defined as f(x) = 2x + 1 and g(x) = x
2
− 2. Express (gof)
−1
and f
−1
og
−1
as the sets of ordered pairs and verify that (gof)
−1
= f
−1
og
−1
.
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Solution
f
x
=
2
x
+
1
⇒
f
=
1
,
2
1
+
1
,
2
,
2
2
+
1
,
3
,
2
3
+
1
,
4
,
2
4
+
1
=
1
,
3
,
2
,
5
,
3
,
7
,
4
,
9
g
x
=
x
2
-
2
⇒
g
=
3
,
3
2
-
2
,
5
,
5
2
-
2
,
7
,
7
2
-
2
,
9
,
9
2
-
2
=
3
,
7
,
5
,
23
,
7
,
47
,
9
,
79
Clearly
f
and
g
are bijections and, hence,
f
-
1
:
B
→
A
and
g
-
1
:
C
→
B
exist.
So
,
f
-
1
=
3
,
1
,
5
,
2
,
7
,
3
,
9
,
4
and
g
-
1
=
7
,
3
,
23
,
5
,
47
,
7
,
79
,
9
Now,
f
-
1
o
g
-
1
:
C
→
A
f
-
1
o
g
-
1
=
7
,
1
,
23
,
2
,
47
,
3
,
79
,
4
.
.
.
1
Also,
f
:
A
→
B
and
g
:
B
→
C
,
⇒
g
o
f
:
A
→
C
,
g
o
f
-
1
:
C
→
A
So,
f
-
1
o
g
-
1
and
g
o
f
-
1
have same domains.
g
o
f
x
=
g
f
x
=
g
2
x
+
1
=
2
x
+
1
2
-
2
⇒
g
o
f
x
=
4
x
2
+
4
x
+
1
-
2
⇒
g
o
f
x
=
4
x
2
+
4
x
-
1
Then
,
g
o
f
1
=
g
f
1
=
4
+
4
-
1
=
7
,
g
o
f
2
=
g
f
2
=
4
+
4
-
1
=
23
,
g
o
f
3
=
g
f
3
=
4
+
4
-
1
=
47
and
g
o
f
4
=
g
f
4
=
4
+
4
-
1
=
79
So
,
g
o
f
=
1
,
7
,
2
,
23
,
3
,
47
,
4
,
79
⇒
g
o
f
-
1
=
7
,
1
,
23
,
2
,
47
,
3
,
79
,
4
.
.
.
2
From
1
and
2
, we get:
g
o
f
-
1
=
f
-
1
o
g
-
1
Suggest Corrections
1
Similar questions
Q.
If
A
=
{
1
,
2
,
3
,
4
}
,
B
=
{
3
,
5
,
7
,
9
}
,
C
=
{
7
,
23
,
47
,
79
}
and
f
:
A
→
B
,
f
(
x
)
=
2
x
+
1
,
g
:
B
→
C
,
g
(
x
)
=
x
2
−
2
then write
(
g
o
f
)
−
1
and
f
−
1
o
g
−
1
in the form of ordered pair.
Q.
Let f : A → B and g : B → C be the bijective functions. Then, (gof)
–1
=
(a) f
–1
o g
–1
(b) fog
(c) g
–1
of
–1
(d) gof
Q.
Let
f
:
A
→
B
a
n
d
g
:
B
→
A
be two functions such that
g
o
f
=
I
A
a
n
f
f
o
g
=
I
B
.
Then, which of the following statement/s is/are true?
