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Question 37 (iii) (B) If the track of the final race (for the biker 𝑏_1) follows the curve 𝑥^2=4𝑦; ( where 0≤𝑥≤20√2. & 0≤𝑦≤200), then state whether the track represents a one-one and onto function or not. (Justify).Given 𝑥^2=4𝑦 𝑦=𝑥^2/4 Let 𝑦=𝒇(𝒙)=𝒙^𝟐/𝟒 In [0, 20√2]→[0, 200] 𝒇(𝒙)=𝒙^𝟐/𝟒 Checking one-one 𝒇(𝒙_𝟏 )=〖𝒙_𝟏〗^𝟐/𝟒 𝒇(𝒙_𝟐 )=〖𝒙_𝟐〗^𝟐/𝟒 Putting 𝒇(𝒙_𝟏 )= 𝒇(𝒙_𝟐 ) 〖𝒙_𝟏〗^𝟐/𝟒= 〖𝒙_𝟐〗^𝟐/𝟒 〖𝒙_𝟏〗^𝟐 = 〖𝒙_𝟐〗^𝟐 x1 = x2 or x1 = –x2 But x1 = –x2 is not possible in our domain [0, 20√2] Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 So, x1 = x2 when f (x1) = f (x2) ∴ f(x) is one-one Check onto 𝒇(𝒙)=𝒙^𝟐/𝟒 Let f(x) = y , such that y ∈ R 𝑥^2/4=𝑦 𝑥^2=4𝑦 𝑥=±2√𝑦 Since 0≤𝑦≤200 and where 0≤𝑥≤20√2 For all values of y in [0, 200], there is a corresponding value of x in [0, 20√2] Thus, for every y in [0, 200], there exists a pre-image in [0, 20√2] ∴ f(x) is onto Thus, for every y in [0, 200], there exists a pre-image in [0, 20√2] ∴ f(x) is onto Let f(x) = y , such that y ∈ R x2 = y x = ±√𝑦 Note that y is a real number, so it can be negative also Putting y = −3 x = ±√((−3)) 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑎𝑠 𝑟𝑜𝑜𝑡 𝑜𝑓 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑎𝑙 Hence, x is not real So, f is not onto

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.