Example 25 - Chapter 1 Class 12 Relation and Functions

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 ± √(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 + √(4𝑥^2 + 12𝑥 + 15 − 6))/2 = (−3 + √(4𝑥^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) = (− 𝟑 + √(𝒚 − 𝟔))/𝟐 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 [(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 (x1 – x2) = 0 ⇒ x1 = x2 (x1 + x2 + 3) = 0 ⇒ x1 = – x2 – 3 Since f: N → S So x ∈ N i.e. x is always positive, Hence x1 = – x2 – 3 is not true 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