# Example 25

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 ± 𝑏2 − 4𝑎𝑐2𝑎 Putting values x = − 12 ± 122 − 4 4 15 − 𝑦2 4 x = − 12 ± 144 − 16 15 − 𝑦8 = − 12 ± 16(9 −(15 − 𝑦)8 = − 12 ± 16(9 −15 + 𝑦)8 = − 12 ± 16(𝑦 − 6)8 = − 12 ± 16 𝑦 − 68 = − 12 ± 42 𝑦 − 68 = − 12 ± 4 𝑦 − 68 = 4 − 3 ± 𝑦 − 68 = − 3 ± 𝑦 − 62 So, x = − 3 + 𝑦 − 62 or − 3 − 𝑦 − 62 As x ∈ N , So, x is a positive real number x cannot be equal to − 3 − 𝑦 − 62 Hence, x = − 𝟑 + 𝒚 − 𝟔𝟐 Let g(y) = − 3 + 𝑦 − 62 where g: S → N Step 2: gof = g(f(x)) = g(4x2 + 12x + 15) = −3 + 4 𝑥2 + 12𝑥 + 15 − 62 = −3 + 4 𝑥2 +12𝑥 + 92 = −3 + (2𝑥)2+ 32 +2 2𝑥 ×32 = −3 + 2𝑥 + 322 = −3 + 2𝑥 +32 = 2𝑥2 = x Hence, gof = x = IN Step 3: fog = f(g(x)) = f( − 3 + 𝑦 − 62) = 4 − 3 + 𝑦 − 622 + 12 − 3 + 𝑦 − 62 + 15 = 4 −3 + 𝑦 − 624 + 6 −3 + 𝑦 −6 + 15 = −3 + 𝑦 −62– 18 + 6 𝑦 −6 + 15 = (–3)2 + 𝑦 −62– 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 + 𝒚𝟑 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] = 04 (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 ± 𝑏2 − 4𝑎𝑐2𝑎 Putting values x = − 12 ± 122 − 4 4 15 − 𝑦2 4 x = − 12 ± 144 − 16 15 − 𝑦8 = − 12 ± 16(9 −(15 − 𝑦)8 = − 12 ± 16(9 −15 + 𝑦)8 = − 12 ± 16(𝑦 − 6)8 = − 12 ± 16 𝑦 − 68 = − 12 ± 42 𝑦 − 68 = − 12 ± 4 𝑦 − 68 = 4 − 3 ± 𝑦 − 68 = − 3 ± 𝑦 − 62 So, x = − 3 + 𝑦 − 62 or − 3 − 𝑦 − 62 As x ∈ N , So, x is a positive real number x cannot be equal to − 3 − 𝑦 − 62 Hence, x = − 𝟑 + 𝒚 − 𝟔𝟐 Hence, x = − 3 + 𝑦 − 62 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 − 62 Let g(y) = − 3 + 𝑦 − 62 where g: S → N So, inverse of f = f–1 = − 𝟑 + 𝒚 − 𝟔𝟐

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Example 23 Important

Example 25 Important You are here

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Class 12

Important Question for exams Class 12

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.