Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

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).

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

Question 24Question 24
A binary operation * is defined on the set ℝ of real numbers by
a * b = {█(𝑎, 𝑖𝑓 𝑏=0@|𝑎|+𝑏, 𝑖𝑓 𝑏 ≠0)┤ if at least one of a and b is 0, then Prove that 𝒂 * b = b * 𝒂. Check whether * is commutative. Find the identity element for * , if it exists
If the function 𝑓 : ℝ → ℝ be defined by 𝑓(x) = 2x − 3 and g : ℝ → ℝ by g(x) = x3 + 5, then find 𝑓∘𝑔 and show that 𝑓 ∘ g is invertible. Also, find (𝑓∘𝑔)−1, hence find (𝑓∘𝑔)−1 (9).
Given 𝑓(x) = 2x − 3 and g(x) = x3 + 5
𝑓∘𝑔 = f(g(x))
= f(x3 + 5)
= 2(x3 + 5) – 3
= 2x3 + 10 – 3
= 2x3 + 7
∴ 𝒇∘𝒈 = 2x3 + 7
Now, we need to check if 𝑓∘𝑔 is invertible and find its inverse
We check if it is invertible by checking one-one and onto
Let p(x) = 𝒇∘𝒈 = 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
∴ p is one-one
Check onto
p(x) = 2x3 + 7
Let p(x) = y such that y ∈ R
Putting in equation
y = 2x3 + 7
y – 7 = 2x3
2x3 = y – 7
x3 = (𝑦 − 7)/2
x = ((𝑦 − 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
∴ 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 = ((𝑦 − 7)/2)^(1/3)
Let g(y) = ((𝑦 − 7)/2)^(1/3)
So, inverse of p = p–1 =((𝑦 − 7)/2)^(1/3)
i.e. Inverse of 𝑓∘𝑔 = (𝑓∘𝑔)−1 =((𝑦 − 7)/2)^(1/3)
Also, we need to find (𝑓∘𝑔)−1 (9)
(𝑓∘𝑔)−1 = ((𝑦 − 7)/2)^(1/3)
Putting y = 9
(𝑓∘𝑔)−1 (9) = ((9 − 7)/2)^(1/3)
(𝑓∘𝑔)−1 (9) = (2/2)^(1/3) = 1^(1/3) = 1
∴ (𝑓∘𝑔)−1 (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 ≠ 0
a ≠ 0, b = 0
a = 0, b = 0
Let a = 0, b ≠ 0
Then,
a * b = 0 * a = |a| + b = |0| + b = b
b * a = b * 0 = b
Thus, a * b = b * a
Let a ≠ 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
∴ If at least one of a and b is 0, then a * b = b * a
Hence proved
Now,
Let’s 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
∴ If at least one of a and b is 0, then a * b = b * a
Hence proved
Now,
Let’s 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 ≠ 0
a ≠ 0, b = 0
a = 0, b = 0
a ≠ 0, b ≠ 0
We proved that a * b = b * a in first 3 cases,
Let’s check the fourth case
Let a ≠ 0, b ≠ 0
Then,
a * b = |a| + b
b * a = |b| + a
Thus, a * b ≠ b * a
There will be total 4 cases
a = 0, b ≠ 0
a ≠ 0, b = 0
a = 0, b = 0
a ≠ 0, b ≠ 0
We proved that a * b = b * a in first 3 cases,
Let’s check the fourth case
Let a ≠ 0, b ≠ 0
Then,
a * b = |a| + b
b * a = |b| + a
Thus, a * b ≠ b * a
Lets take an example,
Let a = –1, b = 2
a * b = |a| + b = |–1| + 2 = 1 + 2 = 3
b * a = |b| + a = 2 + (–1) = 2 – 1 = 1
∴ a * b ≠ 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 = {█(𝑎, 𝑖𝑓 𝑏=0@|𝑎|+𝑏, 𝑖𝑓 𝑏 ≠0)┤
Let e = 0
Then
a * e = a * 0 = a
e * a = 0 * a = |0| + a = a
∴ a * e = e * a = e = 0
Thus, 0 is the identity element of *

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!