If the function f :
R
→
R
be defined by f(x) = 2x − 3 and g :
R
→
R
by g(x) = x
^{
3
}
+ 5, then find
fog
and show that
fog
is invertible. Also, find (
fog
)
^{
−1
}
, hence find (
fog
)
^{
−1
}
(9).

Question 24Question 24
A binary operation * is defined on the set β of real numbers by
a * b = {β(π, ππ π=0@|π|+π, ππ π β 0)β€ if at least one of a and b is 0, then Prove that π * b = b * π. Check whether * is commutative. Find the identity element for * , if it exists
If the function π : β β β be defined by π(x) = 2x β 3 and g : β β β by g(x) = x3 + 5, then find πβπ and show that π β g is invertible. Also, find (πβπ)β1, hence find (πβπ)β1 (9).
Given π(x) = 2x β 3 and g(x) = x3 + 5
πβπ = f(g(x))
= f(x3 + 5)
= 2(x3 + 5) β 3
= 2x3 + 10 β 3
= 2x3 + 7
β΄ πβπ = 2x3 + 7
Now, we need to check if πβπ is invertible and find its inverse
We check if it is invertible by checking one-one and onto
Let p(x) = πβπ = 2x3 + 7
Checking one-one
p(x1) = 2(x1)3 + 7
p(x2) = 2(x2)3 + 7
Putting p(x1) = p(x2)
2(x1)3 + 7 = 2(x2)3 + 7
2(x1)3 = 2(x2)3
x13 = x23
this is possible only if x1 = x2
Hence, if p(x1) = p(x2) , then x1 = x2
β΄ p is one-one
Check onto
p(x) = 2x3 + 7
Let p(x) = y such that y β R
Putting in equation
y = 2x3 + 7
y β 7 = 2x3
2x3 = y β 7
x3 = (π¦ β 7)/2
x = ((π¦ β 7)/2)^(1/3)
Thus,
For every y in range of p, there is a pre-image x in R
Hence, f is onto
Since the function is one-one and onto
β΄ It is invertible
Calculating inverse
For finding inverse, we put f(x) = y and find x in terms of y
We have done that while proving onto
x = ((π¦ β 7)/2)^(1/3)
Let g(y) = ((π¦ β 7)/2)^(1/3)
So, inverse of p = pβ1 =((π¦ β 7)/2)^(1/3)
i.e. Inverse of πβπ = (πβπ)β1 =((π¦ β 7)/2)^(1/3)
Also, we need to find (πβπ)β1 (9)
(πβπ)β1 = ((π¦ β 7)/2)^(1/3)
Putting y = 9
(πβπ)β1 (9) = ((9 β 7)/2)^(1/3)
(πβπ)β1 (9) = (2/2)^(1/3) = 1^(1/3) = 1
β΄ (πβπ)β1 (9) = 1
First, let us prove
if at least one of a and b is 0, then a * b = b * a
There will be 3 cases
a = 0, b β 0
a β 0, b = 0
a = 0, b = 0
Let a = 0, b β 0
Then,
a * b = 0 * a = |a| + b = |0| + b = b
b * a = b * 0 = b
Thus, a * b = b * a
Let a β 0, b = 0
Then,
a * b = a * 0 = a
b * a = 0 * a = |0| + a = a
Thus, a * b = b * a
Let both a = 0, b = 0
Then,
a * b = 0 * 0 = 0
b * a = 0 * 0 = 0
Thus, a * b = b * a
β΄ If at least one of a and b is 0, then a * b = b * a
Hence proved
Now,
Letβs check commutative
a * b is commutative if
a * b = b * a
for all values of a, b
a * b = 0 * 0 = 0
b * a = 0 * 0 = 0
Thus, a * b = b * a
β΄ If at least one of a and b is 0, then a * b = b * a
Hence proved
Now,
Letβs check commutative
a * b is commutative if
a * b = b * a
for all values of a, b
There will be total 4 cases
a = 0, b β 0
a β 0, b = 0
a = 0, b = 0
a β 0, b β 0
We proved that a * b = b * a in first 3 cases,
Letβs check the fourth case
Let a β 0, b β 0
Then,
a * b = |a| + b
b * a = |b| + a
Thus, a * b β b * a
There will be total 4 cases
a = 0, b β 0
a β 0, b = 0
a = 0, b = 0
a β 0, b β 0
We proved that a * b = b * a in first 3 cases,
Letβs check the fourth case
Let a β 0, b β 0
Then,
a * b = |a| + b
b * a = |b| + a
Thus, a * b β b * a
Lets take an example,
Let a = β1, b = 2
a * b = |a| + b = |β1| + 2 = 1 + 2 = 3
b * a = |b| + a = 2 + (β1) = 2 β 1 = 1
β΄ a * b β b * a
Thus, * is not commutative
Lets find identity element
Identity Element
e is the identity of * if
a * e = e * a = a
Given
a * b = {β(π, ππ π=0@|π|+π, ππ π β 0)β€
Let e = 0
Then
a * e = a * 0 = a
e * a = 0 * a = |0| + a = a
β΄ a * e = e * a = e = 0
Thus, 0 is the identity element of *

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.