MTH202 Assignment Solution: Question 1 Let a function f: R®R be defined by f(x) =2x3 + 7 Show that f is bijective. Solution: f is one-to-one Let f(x1) = 2 x13 + 7 and f(x2) = 2 x23 + 7 By the definition we know that, f(x1) = f(x2) for x1 , x2 Î R 2 x13 + 7 = 2 x23 + 7 2 x13 - 2 x23 = 7 – 7 2( x13 - x23 ) = 0 x13 - x23 = 0 (x1 - x2)( x12 + 2x1 x2 + x22) = 0 x1 - x2 = 0 ; x12 + 2x1 x2 + x22 = 0 x1 = x2 (The second equation gives no real solution) Accordingly f is one-to-one. f is onto f(x) =2x3 + 7 let y Î R. We search for x Î R such that f(x) = y 2x3 + 7 = y x3 = OR x = f(x) = f [ ] f(x) = 2 + 7 = y – 7 + 7 f(x) = y Thus f is a Bijective Question 2; Find the sum of the series whose nth term is . Solution MTH202 Assignment Solution: Question 1 Let a function f: R®R be defined by f(x) =2x3 + 7 Show that f is bijective. Solution: f is one-to-one Let f(x1) = 2 x13 + 7 and f(x2) = 2 x23 + 7 By the definition we know that, f(x1) = f(x2) for x1 , x2 Î R 2 x13 + 7 = 2 x23 + 7 2 x13 - 2 x23 = 7 – 7 2( x13 - x23 ) = 0 x13 - x23 = 0 (x1 - x2)( x12 + 2x1 x2 + x22) = 0 x1 - x2 = 0 ; x12 + 2x1 x2 + x22 = 0 x1 = x2 (The second equation gives no real solution) Accordingly f is one-to-one. f is onto f(x) =2x3 + 7 let y Î R. We search for x Î R such that f(x) = y 2x3 + 7 = y x3 = OR x = f(x) = f [ ] f(x) = 2 + 7 = y – 7 + 7 f(x) = y Thus f is a Bijective Question 2; Find the sum of the series whose nth term is . Solution MTH202 Assignment Solution: Question 1 Let a function f: R®R be defined by f(x) =2x3 + 7 Show that f is bijective. Solution: f is one-to-one Let f(x1) = 2 x13 + 7 and f(x2) = 2 x23 + 7 By the definition we know that, f(x1) = f(x2) for x1 , x2 Î R 2 x13 + 7 = 2 x23 + 7 2 x13 - 2 x23 = 7 – 7 2( x13 - x23 ) = 0 x13 - x23 = 0 (x1 - x2)( x12 + 2x1 x2 + x22) = 0 x1 - x2 = 0 ; x12 + 2x1 x2 + x22 = 0 x1 = x2 (The second equation gives no real solution) Accordingly f is one-to-one. f is onto f(x) =2x3 + 7 let y Î R. We search for x Î R such that f(x) = y 2x3 + 7 = y x3 = OR x = f(x) = f [ ] f(x) = 2 + 7 = y – 7 + 7 f(x) = y Thus f is a Bijective
Question 2; Find the sum of the series whose nth term is . Solution
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