Get live Maths 1-on-1 Classs - Class 6 to 12

Ex 1.3

Ex 1.3, 1
Deleted for CBSE Board 2023 Exams

Ex 1.3, 2 Deleted for CBSE Board 2023 Exams

Ex 1.3, 3 (i) Important Deleted for CBSE Board 2023 Exams

Ex 1.3, 3 (ii) Deleted for CBSE Board 2023 Exams

Ex 1.3 , 4 Deleted for CBSE Board 2023 Exams

Ex 1.3, 5 (i) Deleted for CBSE Board 2023 Exams

Ex 1.3, 5 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 1.3, 5 (iii) Important Deleted for CBSE Board 2023 Exams

Ex 1.3 , 6 Deleted for CBSE Board 2023 Exams You are here

Ex 1.3 , 7 Deleted for CBSE Board 2023 Exams

Ex 1.3 , 8 Important Deleted for CBSE Board 2023 Exams

Ex 1.3 , 9 Important Deleted for CBSE Board 2023 Exams

Ex 1.3, 10 Important Deleted for CBSE Board 2023 Exams

Ex 1.3, 11 Deleted for CBSE Board 2023 Exams

Ex 1.3, 12 Deleted for CBSE Board 2023 Exams

Ex 1.3, 13 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 1 Class 12 Relation and Functions

Serial order wise

Last updated at March 16, 2023 by Teachoo

Ex 1.3, 6 Show that f: [−1, 1] → R, given by f(x) = 𝑥/(𝑥 + 2) is one-one. Find the inverse of the function f: [−1, 1] → Range f. (Hint: For y ∈ Range f, y = f(x) = 𝑥/(𝑥 + 2) , for some x in [−1, 1], i.e., x = 2𝑦/(1 − 𝑦) ) f(x) = x/(x+2) Check one-one f(x1) = 𝑥1/(𝑥1 + 2) f(x2) = 𝑥2/(𝑥2 + 2) Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Putting f(x1) = f(x2) 𝑥1/(𝑥1 + 2) = 𝑥2/(𝑥2 + 2) x1(x2 + 2) = x2(x1 + 2) x1x2 + 2x1 = x2x1 + 2x2 x1x2 – x2x1 + 2x1 = 2x2 0 + 2x1 = 2x2 2x1 = 2x2 x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking onto f(x) = 𝑥/(𝑥 + 2) Putting f(x) = y y = 𝑥/(𝑥 + 2) y(x + 2) = x yx + 2y = x yx – x = –2y x(y – 1) = –2y x = (−2𝑦 )/(𝑦 −1) x = (−2𝑦 )/(−1(−𝑦 + 1) ) x = (2𝑦 )/((1 − 𝑦) ) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((2𝑦 )/((1 − 𝑦) )) = ((2𝑦 )/((1 − 𝑦) ))/((2𝑦 )/((1 − 𝑦) ) + 2) = ((2𝑦 )/((1 − 𝑦) ))/((2𝑦 + 2(1 − 𝑦) )/((1 − 𝑦) )) = 2𝑦/(2𝑦 + 2 − 2𝑦) = y Thus, for every y ∈ Range f, there exists x ∈ [−1, 1] such that f(x) = y Hence, f is onto Since f(x) is one-one and onto, So, f(x) is invertible And Inverse of x = 𝑓^(−1) (𝑦) = (2𝑦 )/((1 − 𝑦) ) , y ≠ 1 Note: Here, y ∈ Range f is important Inverse is not defined for y ∈ R Because denominator in (2𝑦 )/((1 − 𝑦) ) will be 0 if y = 1