Question

As intersecções dos gráficos das funções f(x) = 4x + K, com k>0, g(x) = 2 e h(x) = - x+ 5 formam um triângulo com 10 unidades quadradas de área. O valor de k é:

Answer 1

Explicação passo-a-passo:

• $\sf f(x)=g(x)$

$\sf 4x+k=2$

$\sf 4x=2-k$

$\sf x=\dfrac{2-k}{4}$

-> Ponto $\sf A\left(\dfrac{2-k}{4},2\right)$

• $\sf f(x)=h(x)$

$\sf 4x+k=-x+5$

$\sf 4x+x=5-k$

$\sf 5x=5-k$

$\sf x=\dfrac{5-k}{5}$

Substituindo em h(x):

$\sf h\left(\dfrac{5-k}{5}\right)=\dfrac{-5+k}{5}+5$

$\sf h\left(\dfrac{5-k}{5}\right)=\dfrac{-5+k+25}{5}$

$\sf h\left(\dfrac{5-k}{5}\right)=\dfrac{k+20}{5}$

-> Ponto $\sf B\left(\dfrac{5-k}{5},\dfrac{k+20}{5}\right)$

• $\sf g(x)=h(x)$

$\sf -x+5=2$

$\sf x=5-2$

$\sf x=3$

-> Ponto $\sf C(3,2)$

A área do triângulo ABC é dada por:

$\sf A=\dfrac{|det~(D)|}{2}$

sendo:

$\sf D=\left(\begin{array}{ccc} \sf x_{A} &\sf y_{A} &\sf 1 \\ \sf x_{B} &\sf y_{B} &\sf 1 \\ \sf x_{C} &\sf y_{C} & \sf 1\end{array}\right)$

$\sf D=\left(\begin{array}{ccc} \sf \dfrac{2-k}{4} &\sf 2 &\sf 1 \\ \\ \sf \dfrac{5-k}{5} \sf &\sf \dfrac{k+20}{5} &\sf 1 \\ \\ \sf 3 &\sf 2 & \sf 1\end{array}\right)$

$\sf det~(D)=\left(\dfrac{2-k}{4}\right)\cdot\left(\dfrac{k+20}{5}\right)\cdot1+2\cdot1\cdot3+1\cdot\left(\dfrac{5-k}{5}\right)\cdot2-3\cdot\left(\dfrac{k+20}{5}\right)\cdot1-2\cdot1\cdot\left(\dfrac{2-k}{4}\right)-1\cdot\left(\dfrac{5-k}{5}\right)\cdot2$

$\sf det~(D)=\dfrac{40-18k-k^2}{20}+6+\dfrac{10-2k}{5}-\dfrac{3k+60}{5}-\dfrac{2-k}{2}-\dfrac{10-2k}{5}$

$\sf det~(D)=\dfrac{40-18k-k^2+120-12k-240-20+10k}{20}$

$\sf det~(D)=\dfrac{-k^2-20k-100}{20}$

$\sf \dfrac{|\frac{-k^2-20k-100}{20}|}{2}=10$

$\sf \left|\dfrac{-k^2-20k-100}{20}\right|=2\cdot10$

$\sf \left|\dfrac{-k^2-20k-100}{20}\right|=20$

• $\sf \dfrac{-k^2-20k-100}{20}=20$

$\sf -k^2-20k-100=20\cdot20$

$\sf -k^2-20k-100=400$

$\sf k^2+20k+400+100=0$

$\sf k^2+20k+500=0$

$\sf \Delta=20^2-4\cdot1\cdot500$

$\sf \Delta=400-2000$

$\sf \Delta=-1600$

Como $\sf \Delta < 0$, não há raízes reais

• $\sf \dfrac{-k^2-20k+60}{20}=-20$

$\sf -k^2-20k-100=(-20)\cdot20$

$\sf -k^2-20k-100=-400$

$\sf k^2+20k-400+100=0$

$\sf k^2+20k-300=0$

$\sf \Delta=20^2-4\cdot1\cdot(-300)$

$\sf \Delta=400+1200$

$\sf \Delta=1600$

$\sf k=\dfrac{-20\pm\sqrt{1600}}{2\cdot1}=\dfrac{-20\pm40}{2}$

$\sf k'=\dfrac{-20+40}{2}~\Rightarrow~k'=\dfrac{20}{2}~\Rightarrow~\green{k'=10}$

$\sf k"=\dfrac{-20-40}{2}~\Rightarrow~k"=\dfrac{-60}{2}~\Rightarrow~\underbrace{\red{k"=-30}}_{não~serve}$

Logo, $\sf k=10$