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Gerek kalmadi
Ekrem Emre
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Mar 29, 2016, 12:11:47 PM3/29/16
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17. Soruda A ve B noktalarının apsislerini sırasıyla x1 ve x2 ile gösterirsek x1+x2=8 yazılabilir. Ayrıca şeklin kare olmasından x2-x1=33/4-f(x1) veya 8-2x1=33/4-f(x1) denklemi sağlanmalıdır. Verilenlerden parabolün denklemi (x-4)^2/4 olarak bulunur. Dolayısıyla denklem çözülürse x1=1 ve x2=7 olarak bulunur. Sonuç olarak cevap 7-1=6 olmak üzere 36 dır.
Ekrem Emre
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Mar 29, 2016, 12:11:47 PM3/29/16
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k(x2-4)+4x2+3x-6 şeklinde düzenlenirse paradoks demetinin geçtiği sabit noktalar (-2,4) ve (2,16) olarak bulunur. Dolayısıyla bu noktalardan geçen doğru denklemi y=3x+10 olur.