# Misc. 4

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc. 4 Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = x1 + 𝑥 , x ∈ R is one-one and onto function. f: R → {x ∈ R: −1 < x < 1} f(x) = x1 + 𝑥 We know 𝑥 = 𝑥 , 𝑥≥0 −𝑥 , 𝑥<0 So, 𝑓 𝑥= 𝑥1 + 𝑥, 𝑥≥0& 𝑥1 − 𝑥, 𝑥<0 Checking one-one Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one 𝑓 𝑥= 𝑥1 + 𝑥, 𝑥≥0& 𝑥1 − 𝑥, 𝑥<0 Checking onto Thus, x = 𝑦1 − 𝑦 , for x ≥ 0 & x = 𝑦1 + 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 = 𝑦1 − 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.

Chapter 1 Class 12 Relation and Functions

Serial order wise

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