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Chapter 1 Class 12 Relation and Functions
Ex 1.2 , 10 Important You are here
Example 17 Important
Question 8 Important Deleted for CBSE Board 2024 Exams
Ex 1.3, 3 (i) Important Deleted for CBSE Board 2024 Exams
Ex 1.3 , 6 Deleted for CBSE Board 2024 Exams
Ex 1.3 , 8 Important Deleted for CBSE Board 2024 Exams
Ex 1.3 , 9 Important Deleted for CBSE Board 2024 Exams
Ex 1.3, 13 (MCQ) Important Deleted for CBSE Board 2024 Exams
Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2024 Exams
Ex 1.4, 11 Important Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Misc 1 Important
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Chapter 1 Class 12 Relation and Functions
Last updated at May 29, 2023 by Teachoo
Ex 1.2, 10 Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f (x) = ((x − 2)/(x − 3)) Is f one-one and onto? Justify your answer. f (x) = ((x − 2)/(x − 3)) Check one-one f (x1) = ((x"1 " − 2)/(x"1" − 3)) f (x2) = ((x"2 " − 2)/(x"2" − 3)) Putting f (x1) = f (x2) ((x"1 " − 2)/(x"1" − 3)) = ((x"2 " − 2)/(x"2" − 3)) Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 (x1 – 2) (x2 – 3) = (x1 – 3) (x2 – 2) x1 (x2 – 3) – 2 (x2 – 3) = x1 (x2 – 2) – 3 (x2 – 2) x1 x2 – 3x1 – 2x2 + 6 = x1 x2 – 2x1 – 3x2 + 6 – 3x1 – 2x2 = – 2x1 – 3x2 3x2 – 2x2 = – 2x1 + 3x1 x1 = x2 Hence, if f (x1) = f (x2), then x1 = x2 ∴ f is one-one Check onto f (x) = ((x − 2)/(x − 3)) Let f(x) = y such that y ∈ B i.e. y ∈ R – {1} So, y = ((x − 2)/(x − 3)) y(x – 3) = x – 2 xy – 3y = x – 2 xy – x = 3y – 2 x (y – 1) = 3y – 2 x = (3y − 2)/(y − 1) For y = 1 , x is not defined But it is given that y ∈ R – {1} Hence , x = (3y − 2)/(y − 1) ∈ R – {3} Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((3y − 2)/(y − 1)) = (((3y − 2)/(y − 1)) − 2)/(((3y − 2)/(y − 1)) −3) = ((((3y − 2) − 2(𝑦 −1))/(y − 1)))/((((3y − 2) − 3(𝑦 −1))/(y − 1)) ) = (3𝑦 − 2 − 2𝑦 + 2)/(3𝑦 − 2 − 3𝑦 + 3) = 𝑦/1 = y Thus, for every y ∈ B, there exists x ∈ A such that f(x) = y Hence, f is onto