Misc 4 - Show f(x) = x / 1 + |x| is one-one onto  - Miscellaneous

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Misc. 4 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 𝑥﷯ = 𝑥 , 𝑥≥0 ﷮−𝑥 , 𝑥<0﷯﷯ So, 𝑓 𝑥﷯= 𝑥﷮1 + 𝑥﷯, 𝑥≥0﷮& 𝑥﷮1 − 𝑥﷯, 𝑥<0﷯﷯ Checking one-one Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one 𝑓 𝑥﷯= 𝑥﷮1 + 𝑥﷯, 𝑥≥0﷮& 𝑥﷮1 − 𝑥﷯, 𝑥<0﷯﷯ Checking onto 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, 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.

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