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

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Ex1.3 , 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] Thus, g is inverse of f Inverse of f = g(y) = ﷐𝟐𝒚 ﷮﷐𝟏 − 𝒚﷯﷯