<prompt> Kwa kila x katika 0,1, fafanuaanzakesi fx 2x, qquadqquad mathrmif quad 0 leq x leq frac12; fx 2-2x , qquad mathrmif quad frac12 x leq 1. endcasesHebu f2x ffx , na fn 1x fnfx kwa kila nambari kamili n geq 2. Kisha nambari ya thamani ya x katika 0,1 ambayo f2005x frac 12 inaweza kuonyeshwa katika fomu pa , ambapo p ni nambari kuu na a ni nambari kamili chanya. Tafuta p a.  <response>: Grafu za y fx na y f2x zimeonyeshwa hapa chini.asyukubwa wa kitengo cm 3;jozi trans 1.8,0;chora0,0--1,0;chora0,0--0,1;chora0,0--12,1--1,0;chora0,12--1,12, iliyokatika;chora1,-0.05--1,0.05;chora-0.05,1--0.05,1;chora-0.05,12--0.05,12;leboy fx, 1,1;lebo0, 0,0, S;lebo1, 1,-0.05, S;lebo0, 0,0, W;lebo1, -0.05,1, W;lebofrac12, -0.05,12, W;kuchorakuhamatrans0,0--1,0;kuchorakuhamatrans0,0--0,1;kuchorakuhamatrans0,0--14,1--12,0--34,1--1,0;kuchorakuhamatrans0,12--1,12,iliyokatika;kuchorakuhamatrans1,-0.05--1,0.05;kuchorakuhamatrans-0.05,1--0.05,1;kuchorakuhamatrans-0.05,12--0.05,12;leboy f2x, 1.2,1 trans;lebo0, 0,0 trans, S;lebo1, 1,-0.05 trans, S;lebo0, 0,0 trans, W;lebo1, -0.05,1 trans, W;lebofrac12, -0.05,12 trans, W;asyKwa n ge 2,fnx fn - 1fx kushotoanzasafuclfn - 12x textif 0 le x le frac12, fn - 12 - 2x textif frac12 le x le 1.mwishosafukulia.Acha gn iwe nambari ya thamani ya x katika 0,1 ambayo fnx frac12 . Kisha fnx frac12 kwa gn - 1 thamani ya x in left 0, frac1101 2 kulia, na gn - 1 thamani za x katika left frac12, 1 kulia.Zaidi ya hayofn kushoto frac12 kulia fn1 0 neq frac12kwa n ge 2. Kwa hivyo, gn 2gn - 1 kwa n yote ge 2. Tangu g1 2, g2005 22005 . Jibu la mwisho ni 2 2005 2007. 