Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions

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Ex 1.3 , 6 Show that f: [−1, 1] → R, given by f(x) = 𝑥/(𝑥 + 2) is one-one. Find the inverse of the function f: [−1, 1] → Range f. (Hint: For y ∈ Range f, y = f(x) = 𝑥/(𝑥 + 2) , for some x in [−1, 1], i.e., x = 2𝑦/(1 − 𝑦) ) f(x) = x/(x+2) Check one-one f(x1) = 𝑥1/(𝑥1 + 2) f(x2) = 𝑥2/(𝑥2 + 2) Putting f(x1) = f(x2) 𝑥1/(𝑥1 + 2) = 𝑥2/(𝑥2 + 2) x1(x2 + 2) = x2(x1 + 2) x1x2 + 2x1 = x2x1 + 2x2 x1x2 – x2x1 + 2x1 = 2x2 0 + 2x1 = 2x2 2x1 = 2x2 ⇒ x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Finding inverse f(x) = 𝑥/(𝑥 + 2) Putting f(x) = y y = 𝑥/(𝑥 + 2) y(x + 2) = x yx + 2y = x yx – x = –2y x(y – 1) = –2y x = (−2𝑦 )/(𝑦 −1) x = (−2𝑦 )/(−1(−𝑦 + 1) ) x = (2𝑦 )/((1 − 𝑦) ) Let g(y) = (2𝑦 )/((1 − 𝑦) ) where g: R → [−1, 1], y ≠ 1 Thus, g is inverse of f Inverse of f = g(y) = (𝟐𝒚 )/((𝟏 − 𝒚) ) , y ≠ 1