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

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Ex 1.3 , 9 (Method 1) Consider f: R+ → [-5, 𕔴 given by f(x) = 9x2 + 6x – 5. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = ﷐﷐﷐﷮y +6﷯﷯ − 1﷮3﷯ . f(x) = 9x2 + 6x – 5 Step 1: Let f(x) = y y = 9x2 + 6x – 5 0 = 9x2 + 6x – 5 – y 9x2 + 6x – 5 – y = 0 9x2 + 6x – (5 + y) = 0 x = ﷐−𝑏 ± ﷐﷮﷐𝑏﷮2﷯ − 4𝑎𝑐﷯﷮2𝑎﷯ Putting values x = ﷐− 6 ± ﷐﷮﷐6﷮2﷯ − 4﷐9﷯ ﷐−﷐5 + 𝑦﷯﷯﷯﷮2﷐9﷯﷯ x = ﷐− 6 ± ﷐﷮36 + 4﷐9﷯﷐5 + 𝑦﷯﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36 + 36(5 + 𝑦)﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36﷐1 + ﷐5 + 𝑦﷯﷯﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36﷐6 + 𝑦﷯﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36﷯﷐﷮(6 + 𝑦)﷯﷮18﷯ = ﷐− 6 ± ﷐﷮﷐6﷮2﷯﷯﷐﷮(6 + 𝑦)﷯﷮18﷯ = ﷐− 6 ± 6 ﷐﷮(6 + 𝑦)﷯﷮18﷯ = ﷐6﷐− 1 ± ﷐﷮﷐6 + 𝑦﷯﷯﷯﷮18﷯ = ﷐− 1 ± ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ So, x = ﷐− 1 − ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ or ﷐− 1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ As x ∈ R+ , i.e., x is a positive real number x cannot be equal to ﷐−1 − ﷐﷮﷐6 + 𝑦﷯ ﷯﷮3﷯ Hence, x = ﷐−𝟏 + ﷐﷮﷐𝟔 + 𝒚﷯﷯﷮𝟑﷯ Let g(y) = ﷐−𝟏 + ﷐﷮﷐𝟔 + 𝒚﷯﷯﷮𝟑﷯ where g: [–5,𕔴 → R+ Step 2: gof = g(f(x)) = g(9x2 + 6x – 5) = ﷐−1 + ﷐﷮﷐6 + 9﷐𝑥﷮2﷯ + 6𝑥 − 5﷯﷯﷮3﷯ = ﷐−1 + ﷐﷮﷐9﷐𝑥﷮2﷯ + 6𝑥 + 1﷯﷯﷮3﷯ = ﷐−1 + ﷐﷮﷐﷐(3𝑥)﷮2﷯+ ﷐1﷮2﷯ +2﷐3𝑥﷯ × 2﷯﷯﷮3﷯ = ﷐−1 + ﷐﷮﷐﷐3𝑥 + 1﷯﷮2﷯﷯﷮3﷯ = ﷐−1 + 3𝑥 + 1﷮3﷯ = ﷐3𝑥﷮3﷯ = x Hence, gof = x = IR+ Step 3: fog = f(g(x)) = f(﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯) = 9﷐﷐﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯﷯﷮2﷯ + 6﷐﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯﷯ – 5 = 9﷐﷐﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷯﷮2﷯﷮9﷯ + 2﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷯ – 5 = ﷐﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷯﷮2﷯– 2 + 2﷐﷮﷐6 + 𝑦﷯﷯ – 5 = (–1)2 + ﷐﷐﷐﷮﷐6 + 𝑦﷯﷯﷯﷮2﷯– 2﷐﷮﷐6 + 𝑦﷯﷯ – 2 + 2﷐﷮﷐6 + 𝑦﷯﷯ – 5 = 1 + (6 + y) – 2 – 5 = 7 – 7 + y = y Hence, fog = y = I[–5,𕔴 Since, gof = IR+ & fog = I[–5,𕔴 f is invertible, and inverse of f = g(y) = ﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ So, f–1 = ﷐−𝟏 + ﷐﷮﷐𝟔 + 𝒚﷯﷯﷮𝟑﷯ Ex 1.3 , 9 (Method 2) Consider f: R+ → [-5, 𕔴 given by f(x) = 9x2 + 6x – 5. Show that f is invertible with the inverse f−1 of given f by f-1 (y) = ﷐﷐﷐﷮y +6﷯﷯ − 1﷮3﷯ . f(x) = 9x2 + 6x – 5 f is invertible if it is one-one and onto Checking one-one f (x1) = 9(x1)2 + 6x1 – 5 f (x2) = 9(x2)2 + 6x2 – 5 Putting f (x1) = f (x2) 9(x1)2 + 6x1 – 5 = 9(x2)2 + 6x2 – 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = – 5 + 5 9(x1)2 – 9(x2)2 + 6x1 – 6x2 = 0 9[(x1)2 – (x2)2 ]+ 6[x1 – x2] = 0 9[(x1 – x2) (x1 + x2) ]+ 6[x1 – x2] = 0 3(x1 – x2) [3(x1 + x2) + 2] = 0 (x1 – x2) [3x1 + 3x2 + 2] = ﷐0﷮3﷯ (x1 – x2) [3x1 + 3x2 + 2] = 0 Hence, if f (x1) = f (x2) , then x1 = x2 ∴ f is one-one Check onto f(x) = 9x2 + 6x – 5 Let f(x) = y such that y ∈ [-5, 𕔴 Putting in equation y = 9x2 + 6x – 5 0 = 9x2 + 6x – 5 – y 9x2 + 6x – 5 – y = 0 9x2 + 6x – (5 + y) = 0 x = ﷐−𝑏 ± ﷐﷮﷐𝑏﷮2﷯ − 4𝑎𝑐﷯﷮2𝑎﷯ Putting values x = ﷐− 6 ± ﷐﷮﷐6﷮2﷯ − 4﷐9﷯ ﷐−﷐5 + 𝑦﷯﷯﷯﷮2﷐9﷯﷯ x = ﷐− 6 ± ﷐﷮36 + 4﷐9﷯﷐5 + 𝑦﷯﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36 + 36(5 + 𝑦)﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36﷐1 + ﷐5 + 𝑦﷯﷯﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36﷐6 + 𝑦﷯﷯﷮18﷯ = ﷐− 6 ± ﷐﷮36﷯﷐﷮(6 + 𝑦)﷯﷮18﷯ = ﷐− 6 ± ﷐﷮﷐6﷮2﷯﷯﷐﷮(6 + 𝑦)﷯﷮18﷯ = ﷐− 6 ± 6 ﷐﷮(6 + 𝑦)﷯﷮18﷯ = ﷐6﷐− 1 ± ﷐﷮﷐6 + 𝑦﷯﷯﷯﷮18﷯ = ﷐− 1 ± ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ So, x = ﷐− 1 − ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ or ﷐− 1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ As x ∈ R+ , i.e., x is a positive real number x cannot be equal to ﷐−1 − ﷐﷮﷐6 + 𝑦﷯ ﷯﷮3﷯ Hence, x = ﷐−𝟏 + ﷐﷮﷐𝟔 + 𝒚﷯﷯﷮𝟑﷯ Since f: R+ → [-5,𕔴 So y ∈ [-5,𕔴 i.e. y is greater than or equal to −5 i.e. y ≥ -5 y + 5 ≥ 0 ⇒Hence the value inside root is positive Hence ﷐﷮𝑦+6﷯ ≥ 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 = ﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ Let g(y) = ﷐−1 + ﷐﷮﷐6 + 𝑦﷯﷯﷮3﷯ where g: [–5,𕔴 → R+ So, inverse of f = f–1 = ﷐−𝟏 + ﷐﷮﷐𝟔 + 𝒚﷯﷯﷮𝟑﷯