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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.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


If f(x) = 2x - 3, g(x) = x^3 + 5, then find fog and show that fog is

Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 9 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 10 Question 24 - CBSE Class 12 Sample Paper for 2018 Boards - Part 11

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 24Question 24 A binary operation * is defined on the set ℝ of real numbers by a * b = {β–ˆ(π‘Ž, 𝑖𝑓 𝑏=0@|π‘Ž|+𝑏, 𝑖𝑓 𝑏 β‰ 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 β†’ ℝ be defined by 𝑓(x) = 2x βˆ’ 3 and g : ℝ β†’ ℝ by g(x) = x3 + 5, then find π‘“βˆ˜π‘” and show that 𝑓 ∘ g is invertible. Also, find (π‘“βˆ˜π‘”)βˆ’1, hence find (π‘“βˆ˜π‘”)βˆ’1 (9). Given 𝑓(x) = 2x βˆ’ 3 and g(x) = x3 + 5 π‘“βˆ˜π‘” = f(g(x)) = f(x3 + 5) = 2(x3 + 5) – 3 = 2x3 + 10 – 3 = 2x3 + 7 ∴ π’‡βˆ˜π’ˆ = 2x3 + 7 Now, we need to check if π‘“βˆ˜π‘” is invertible and find its inverse We check if it is invertible by checking one-one and onto Let p(x) = π’‡βˆ˜π’ˆ = 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 βˆ’ 7)/2 x = ((𝑦 βˆ’ 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 βˆ’ 7)/2)^(1/3) Let g(y) = ((𝑦 βˆ’ 7)/2)^(1/3) So, inverse of p = p–1 =((𝑦 βˆ’ 7)/2)^(1/3) i.e. Inverse of π‘“βˆ˜π‘” = (π‘“βˆ˜π‘”)βˆ’1 =((𝑦 βˆ’ 7)/2)^(1/3) Also, we need to find (π‘“βˆ˜π‘”)βˆ’1 (9) (π‘“βˆ˜π‘”)βˆ’1 = ((𝑦 βˆ’ 7)/2)^(1/3) Putting y = 9 (π‘“βˆ˜π‘”)βˆ’1 (9) = ((9 βˆ’ 7)/2)^(1/3) (π‘“βˆ˜π‘”)βˆ’1 (9) = (2/2)^(1/3) = 1^(1/3) = 1 ∴ (π‘“βˆ˜π‘”)βˆ’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 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 ∴ 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 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, Let’s 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, Let’s 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 = –1, b = 2 a * b = |a| + b = |–1| + 2 = 1 + 2 = 3 b * a = |b| + a = 2 + (–1) = 2 – 1 = 1 ∴ 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–ˆ(π‘Ž, 𝑖𝑓 𝑏=0@|π‘Ž|+𝑏, 𝑖𝑓 𝑏 β‰ 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 *

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.