5. Determine whether each of these functions is a bijection from R to R. (a) /()=2x-10…
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Question “5. Determine whether each of these functions is a bijection from R to R. (a) /()=2x-10…”
5. Determine whether each of these functions is a bijection from R to R. (a) /()=2x-10 (b)/(a) – 4 +4 (c)/(w) – (x + 1)/(x+2) (d)/(a) 7+1
Answer
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Bijection function the fi>R and onto function a is one (1): 20-10 checking one f(0) =2X2 -10 f(0)=2012-10 putting for =f (62) 221-10 -222 =222 1702 Hence f (-2) =f(x2)=x2=X2 =f(x2) =f(x2)=X2 =f(x2) =f(x2)=X2 =f(x2) =f(x2) =f(x2) =x2=X2=X2=X2=X2=X2=X2=X2=X2=X2=X2=x2=X2=X2=X2=X2=X2=222 =222 I checked onto let f (x) = such that the YER 21-10-y 21 = ng CamScanner scans 10 T +5 cs
Checking for y=f (x) by putting value of ja in fco? Ja f(x+5) =2(45)-10 =y +10.20 =y. Thus, for every JER, f(x]=y. Thus, f is onto exijtXER such that it is one and is bijection. Both can be Scanned using CamScanner
(6) f@2 – 4a? +4 Checking one f(0) =4x +4 f(2)=402 +4 putting $ (27) =f(+2). 4×7 +4 =402 +4 4a2=4×2 image 22 =X 0.757 =-72 because I doesn’t have unique It isn’t one.one. +4 = 8 it is not only one function. It is function bijection Scanned using CamScanner
CSF.ROR: 8W = (2322Xt2 proving one) f(0) =Xgt2 X2+2 f (02), = x2d1 X2+2 autting F(x) =f(x2) + 1 -X2 +2 xit2 2+2 (2242)-(242) (X2d 2) (X2 + 2) – (x2+2) Xada – 1 – 2 Kaj + 2 . X2+2 1,12 = 1242 dig az one one firmetion. CamScanner – f = cs Scanned
checking onto. Let f(x=y such that f(x2=242 that YER = x+2x+2x+2x+2 G + 1)–(x+2)xt 2 – 3-7) = niz *+2 = 1: 1-2+27 1-9 Scanned using CamScanner
Now check y = fa(x)= putton?x+1 x+2a valme?y-172-1 42 Joy 24-1 1-7 2y -] 12y -] = 12 sy.
CARA PECA] = 2deg+1. Checking one one three, flag = (10) =x x+2 + 1 putting flag) =f(x2) =f(x2) ocz+ 1 =x +7X2=X2 Therefore flag) =f(a), so X =x2 function is only one function. Check f(x onto =y so that TGR F(x) is Y x-1 09. CamScanner scan
Now, cheekily, y = f (x), and putting an avalue of 2x=JJ-2 = (69-7-7)+7 ygt.5.7.-3+7 so that f. is onto function, one asa and one is onto bijtetion. L Bijection . +602=6+) lis +2) CS Scanner (2-83 + 2 CamScanner 465 NO YES YES. |
Conclusion
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