1. Chapter 1 Class 12 Relation and Functions
2. Serial order wise

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Ex1.3 , 8 (Method 1) Consider f: R+ → [4, 𕔴 given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = ﷐﷮y−4﷯ , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 Step 1 Put f(x) = y y = x2 + 4 y – 4 = x2 x2 = y – 4 x = ±﷐﷮𝑦−4﷯ Since f: R+ → [4,𕔴) x ∈ R+ , so x is positive Hence we cannot take x = −﷐﷮𝑦−4﷯ ⇒ x = ﷐﷮𝑦−4﷯ Let g(y) = ﷐﷮𝑦−4﷯ where g: [4,𕔴 → R+ Step 2: gof = g(f(x)) = g(x2 + 4) = ﷐﷮﷐x2 + 4﷯ −4﷯ = ﷐﷮﷐𝑥﷮2﷯﷯ = ﷐𝑥﷮﷐1﷮2﷯ × 2﷯ = ﷐𝑥﷮1﷯ = x Hence, gof(x) = IX = x Step 3: fog = f(g(y)) = f(﷐﷮𝑦−4﷯ ) = (﷐﷮𝑦−4﷯ )2 + 4 = ﷐(𝑦 −4)﷮﷐1﷮2﷯ × 2﷯+ 4 = ﷐(𝑦 −4)﷮1﷯ + 4 = y – 4 + 4 = y Hence, fog(y) = IY = y Since gof = IX and fog = IY, f is invertible & Inverse of f = g(y) = ﷐﷮𝑦−4﷯ So, f-1 = ﷐﷮𝒚−𝟒﷯ Ex1.3 , 8 (Method 2) Consider f: R+ → [4, 𕔴 given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = ﷐﷮y−4﷯ , where R+ is the set of all non-negative real numbers. f(x) = x2 + 4 f is invertible if f is one-one and onto Checking one-one f (x1) = (x1)2 + 4 f (x2) = (x2)2 + 4 Putting f (x1) = f (x2) ⇒ (x1)2 + 4 = (x2)2 + 4 ⇒ (x1)2 = (x2)2 ⇒ x1 = x2 or x1 = –x2 Since f: R+ → [4,𕔴 So x ∈ R+ i.e. x is always positive, Hence x1 = –x2 is not true So, x1 = x2 ∴ f is one-one Check onto f(x) = x2 + 4 Let f(x) = y such that y ∈ [4,𕔴 y = x2 + 4 y – 4 = x2 x2 = y – 4 x = ±﷐﷮𝑦−4﷯ Since f: R+ → [4,𕔴 x ∈ R+ , so x is positive Hence we cannot take x = −﷐﷮𝑦−4﷯ ⇒ x = ﷐﷮𝑦−4﷯ Since, y ∈ [4,𕔴 i.e. y is greater than or equal to 4 i.e. y ≥ 4 y – 4 ≥ 0 ⇒Hence the value inside root is positive Hence ﷐﷮𝑦−4﷯ ≥ 0 ⇒ x ≥ 0 Hence x is a real number which is greater than or equal to 0. ⇒ x ∈ R+ Hence, the function f is onto Since the function is one-one and onto ∴ It is invertible Calculating inverse For finding inverse, we put f(x) = y and find x in terms of y We have done that while proving onto x = ﷐﷮𝑦−4﷯ Let g(y) = ﷐﷮𝑦−4﷯ where g: [4,𕔴 → R+ ∴ Inverse of f = g(y) = ﷐﷮𝑦−4﷯ So, f-1 = ﷐﷮𝒚−𝟒﷯