Question

(U. F. Viçosa-MG) Seja a função real f tal que:


$f(x + 2) = f(x) + \frac{5}{6}$ e $f(0) = \frac{5}{4}$


Pode-se afirmar que f(12) vale:

a) 77/6

b) 25/4

c) 65/6

d) 53/4

e) 19/12


Resposta: b) por quê?

Answer 1

Seja $f(0)=\dfrac{5}{4}$ tome a função $f(x+2)=f(x)+\dfrac{5}{6}$ e vá tomando valores pra x até chegar em $f(12)$

$f(x+2)=f(x)+\dfrac{5}{6} \quad\mbox{para x=0}\\\\f(0+2)=\overbrace{f(0)}^{\frac{5}{4}}+\dfrac{5}{6}\\\\f(2)=\dfrac{5}{4}+\dfrac{5}{6}\\\\f(2)=\dfrac{5\cdot6+5\cdot4}{4\cdot6}\\\\f(2)=\dfrac{50}{24}\quad\mbox{simplificando por 2}\\\\f(2)=\dfrac{25}{12}$

A partir daqui vou fazer um pouco mais rapido se não vai ficar mt longo, basta so repetir o processo

$f(2+2)=\overbrace{f(2)}^{\frac{25}{12}}+\dfrac{5}{6} \quad\mbox{para x=2}\\\\f(4)=\dfrac{25}{12}+\dfrac{5}{6}\\\\f(4)=\dfrac{35}{12}$

.......................................................

$f(4+2)=\overbrace{f(4)}^{\frac{35}{12}}+\dfrac{5}{6} \quad\mbox{para x=4}\\\\f(6)=\dfrac{35}{12}+\dfrac{5}{6}\\\\f(6)=\dfrac{15}{4}$

.......................................................

$f(6+2)=\overbrace{f(6)}^{\frac{15}{4}}+\dfrac{5}{6} \quad\mbox{para x=6}\\\\f(8)=\dfrac{15}{4}+\dfrac{5}{6}\\\\f(8)=\dfrac{55}{12}$

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$f(8+2)=\overbrace{f(8)}^{\frac{55}{12}}+\dfrac{5}{6} \quad\mbox{para x=8}\\\\f(10)=\dfrac{55}{12}+\dfrac{5}{6}\\\\f(10)=\dfrac{65}{12}$

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$f(10+2)=\overbrace{f(10)}^{\frac{65}{12}}+\dfrac{5}{6} \quad\mbox{para x=10}\\\\f(12)=\dfrac{65}{12}+\dfrac{5}{6}\\\\\boxed{\boxed{f(12)=\dfrac{25}{4}}}$

Letra B