Example 25 - Chapter 1 Class 12 Relation and Functions

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Example 25 (Method 1) Let f : N → R be a function defined as f (x) = 4x2 + 12x + 15. Show that f : N→ S, where, S is the range of f, is invertible. Find the inverse of f. f(x) = 4x2 + 12x + 15 Step 1: Let f(x) = y y = 4x2 + 12x + 15 0 = 4x2 + 12x + 15 – y 4x2 + 12x + 15 – y = 0 4x2 + 12x + (15 – y) = 0 x = −𝑏 ± ﷮ 𝑏﷮2﷯ − 4𝑎𝑐﷯﷮2𝑎﷯ Putting values x = − 12 ± ﷮ 12﷮2﷯ − 4 4﷯ 15 − 𝑦﷯﷯﷮2 4﷯﷯ x = − 12 ± ﷮144 − 16 15 − 𝑦﷯﷯﷮8﷯ = − 12 ± ﷮16(9 −(15 − 𝑦)﷯﷮8﷯ = − 12 ± ﷮16(9 −15 + 𝑦)﷯﷮8﷯ = − 12 ± ﷮16(𝑦 − 6)﷯﷮8﷯ = − 12 ± ﷮16﷯ ﷮𝑦 − 6﷯﷮8﷯ = − 12 ± ﷮ 4﷮2﷯﷯ ﷮𝑦 − 6﷯﷮8﷯ = − 12 ± 4 ﷮𝑦 − 6﷯﷮8﷯ = 4 − 3 ± ﷮𝑦 − 6﷯﷯﷮8﷯ = − 3 ± ﷮𝑦 − 6﷯﷮2﷯ So, x = − 3 + ﷮𝑦 − 6﷯﷮2﷯ or − 3 − ﷮𝑦 − 6﷯﷮2﷯ As x ∈ N , So, x is a positive real number x cannot be equal to − 3 − ﷮𝑦 − 6﷯﷮2﷯ Hence, x = − 𝟑 + ﷮𝒚 − 𝟔﷯﷮𝟐﷯ Let g(y) = − 3 + ﷮𝑦 − 6﷯﷮2﷯ where g: S → N Step 2: gof = g(f(x)) = g(4x2 + 12x + 15) = −3 + ﷮4 𝑥﷮2﷯ + 12𝑥 + 15 − 6﷯﷮2﷯ = −3 + ﷮4 𝑥﷮2﷯ +12𝑥 + 9﷯﷮2﷯ = −3 + ﷮ (2𝑥)﷮2﷯+ 3﷮2﷯ +2 2𝑥﷯ ×3﷯﷮2﷯ = −3 + ﷮ 2𝑥 + 3﷯﷮2﷯﷯﷮2﷯ = −3 + 2𝑥 +3﷮2﷯ = 2𝑥﷮2﷯ = x Hence, gof = x = IN Step 3: fog = f(g(x)) = f( − 3 + ﷮𝑦 − 6﷯﷮2﷯) = 4 − 3 + ﷮𝑦 − 6﷯﷮2﷯﷯﷮2﷯ + 12 − 3 + ﷮𝑦 − 6﷯﷮2﷯﷯ + 15 = 4 −3 + ﷮𝑦 − 6﷯﷯﷮2﷯﷮4﷯ + 6 −3 + ﷮𝑦 −6﷯﷯ + 15 = −3 + ﷮𝑦 −6﷯﷯﷮2﷯– 18 + 6 ﷮𝑦 −6﷯ + 15 = (–3)2 + ﷮𝑦 −6﷯﷯﷮2﷯– 6 ﷮𝑦 −6﷯ – 18 + 6 ﷮ 6 + 𝑦﷯﷯ + 15 = 9 + y – 6 – 18 + 15 = y Hence, fog = y = IS Since, gof = IN & fog = IS f is invertible, and inverse of f = g(y) = −𝟏 + ﷮ 𝟔 + 𝒚﷯﷯﷮𝟑﷯ Example 25 (Method 2) Let f : N → R be a function defined as f (x) = 4x2 + 12x + 15. Show that f : N→ S, where, S is the range of f, is invertible. Find the inverse of f. f(x) = 4x2 + 12x + 15 f is invertible if it is one-one and onto Checking one-one f (x1) = 4(x1)2 + 12x1 + 15 f (x2) = 4(x2)2 + 12x2 + 15 Putting f (x1) = f (x2) 4(x1)2 + 12x1 + 15 = 4(x2)2 + 12x2 + 15 4(x1)2 – 4(x2)2 + 12x1 – 12x2 = 15 – 15 4(x1)2 – 4(x2)2 + 12x1 – 12x2 = 0 4[(x1)2 – (x2)2 ]+ 12[x1 – x2] = 0 4[(x1 – x2) (x1 + x2) ]+ 12[x1 – x2] = 0 4(x1 – x2) [(x1 + x2) + 3] = 0 (x1 – x2) [x1 + x2 + 3] = 0﷮4﷯ (x1 – x2) [x1 + x2 + 3] = 0 Hence, if f (x1) = f (x2) , then x1 = x2 ∴ f is one-one Check onto f(x) = 4x2 + 12x + 15 Let f(x) = y such that y ∈ S Putting in equation y = 4x2 + 12x + 15 0 = 4x2 + 12x + 15 – y 4x2 + 12x + 15 – y = 0 4x2 + 12x + (15 – y) = 0 x = −𝑏 ± ﷮ 𝑏﷮2﷯ − 4𝑎𝑐﷯﷮2𝑎﷯ Putting values x = − 12 ± ﷮ 12﷮2﷯ − 4 4﷯ 15 − 𝑦﷯﷯﷮2 4﷯﷯ x = − 12 ± ﷮144 − 16 15 − 𝑦﷯﷯﷮8﷯ = − 12 ± ﷮16(9 −(15 − 𝑦)﷯﷮8﷯ = − 12 ± ﷮16(9 −15 + 𝑦)﷯﷮8﷯ = − 12 ± ﷮16(𝑦 − 6)﷯﷮8﷯ = − 12 ± ﷮16﷯ ﷮𝑦 − 6﷯﷮8﷯ = − 12 ± ﷮ 4﷮2﷯﷯ ﷮𝑦 − 6﷯﷮8﷯ = − 12 ± 4 ﷮𝑦 − 6﷯﷮8﷯ = 4 − 3 ± ﷮𝑦 − 6﷯﷯﷮8﷯ = − 3 ± ﷮𝑦 − 6﷯﷮2﷯ So, x = − 3 + ﷮𝑦 − 6﷯﷮2﷯ or − 3 − ﷮𝑦 − 6﷯﷮2﷯ As x ∈ N , So, x is a positive real number x cannot be equal to − 3 − ﷮𝑦 − 6﷯﷮2﷯ Hence, x = − 𝟑 + ﷮𝒚 − 𝟔﷯﷮𝟐﷯ Hence, x = − 3 + ﷮𝑦 − 6﷯﷮2﷯ Since f: N → S So y ∈ S i.e. y ∈ Range of f For every y in range of f, there is a pre-image x in N Hence, 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 = − 3 + ﷮𝑦 − 6﷯﷮2﷯ Let g(y) = − 3 + ﷮𝑦 − 6﷯﷮2﷯ where g: S → N So, inverse of f = f–1 = − 𝟑 + ﷮𝒚 − 𝟔﷯﷮𝟐﷯