Miscellaneous

Chapter 1 Class 12 Relation and Functions
Serial order wise

### Transcript

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.

#### Davneet Singh

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.