Misc 1 - Chapter 1 Class 12 Relation and Functions (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 1 Class 12 Relation and Functions
Ex 1.2 , 10 Important
Example 17 Important
Question 8 Important Deleted for CBSE Board 2025 Exams
Ex 1.3, 3 (i) Important Deleted for CBSE Board 2025 Exams
Ex 1.3 , 6 Deleted for CBSE Board 2025 Exams
Ex 1.3 , 8 Important Deleted for CBSE Board 2025 Exams
Ex 1.3 , 9 Important Deleted for CBSE Board 2025 Exams
Ex 1.3, 13 (MCQ) Important Deleted for CBSE Board 2025 Exams
Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2025 Exams
Ex 1.4, 11 Important Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Misc 1 Important You are here
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 11 Deleted for CBSE Board 2025 Exams
Chapter 1 Class 12 Relation and Functions
Last updated at April 16, 2024 by Teachoo
Misc 1 Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = x/(1 + |𝑥| ) , x ∈ R is one-one and onto function. f: R → {x ∈ R: −1 < x < 1} f(x) = x/(1 + |𝑥| ) We know that |𝑥| = {█( 𝑥 , 𝑥≥0 @−𝑥 , 𝑥<0)┤ So, 𝑓(𝑥)={█(𝑥/(1 + 𝑥), 𝑥≥0@&𝑥/(1 − 𝑥), 𝑥<0)┤ For x ≥ 0 f(x1) = 𝑥_1/(1 + 𝑥_1 ) f(x2) = 𝑥_2/(1 + 𝑥_2 ) Putting f(x1) = f(x2) 𝑥_1/(1 + 𝑥_1 ) = 𝑥_2/(1 + 𝑥_2 ) 𝑥_1 (1 + 𝑥_2)= 𝑥_2 (1 + 𝑥_1) 𝑥_1+𝑥_1 𝑥_2= 𝑥_2 +𝑥_2 𝑥_1 𝑥_1= 𝑥_2 For x < 0 f(x1) = 𝑥_1/(1 − 𝑥_1 ) f(x2) = 𝑥_2/(1 − 𝑥_2 ) Putting f(x1) = f(x2) 𝑥_1/(1 − 𝑥_1 ) = 𝑥_2/(1 − 𝑥_2 ) 𝑥_1 (1 − 𝑥_2)= 𝑥_2 (1 − 𝑥_1) 𝑥_1−𝑥_1 𝑥_2= 𝑥_2 −𝑥_2 𝑥_1 𝑥_1= 𝑥_2 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking onto 𝑓(𝑥)={█(𝑥/(1 + 𝑥), 𝑥≥0@&𝑥/(1 − 𝑥), 𝑥<0)┤ For x ≥ 0 f(x) = 𝑥/(1 + 𝑥) Let f(x) = y, "y = " 𝑥/(1 + 𝑥) y(1 + x) = x y + xy = x y = x – xy x – xy = y x(1 – y) = y x = 𝑦/(1 − 𝑦) For x < 0 f(x) = 𝑥/(1 − 𝑥) Let f(x) = y "y = " 𝑥/(1 − 𝑥) y(1 – x) = x y – xy = x y = x + xy x + xy = y x(1 + y) = y x = 𝑦/(1 + 𝑦) Thus, x = 𝑦/(1 − 𝑦) , for x ≥ 0 & x = 𝑦/(1 + 𝑦) , for x < 0 Here, y ∈ {x ∈ R: −1 < x < 1} i.e. Value of y is from –1 to 1 , – 1 < y < 1 If y = 1, x = 𝑦/(1 − 𝑦) will be not defined, If y = –1, x = 𝑦/(1 + 𝑦) will be not defined, But here – 1 < y < 1 So, x is defined for all values of y. & x ∈ R ∴ f is onto Hence, f is one-one and onto.