Question

Exercițiile 19 si 20 va rog e urgent !​

Answer 1

$\it 19)\ f(x) = (1-3a)x+1\\ \\ T(-a,\ 2a^2+a)\in Gf \Rightarrow f(-a) = 2a^2+a \Rightarrow (1-3a)\cdot(-a)+1=2a^2+a \Rightarrow\\ \\ \Rightarrow -a + 3a^2 +1 = 2a^2+a \Rightarrow -a + 3a^2 +1 - 2a^2-a =0 \Rightarrow a^2-2a+1=0\Rightarrow\\ \\ \Rightarrow (a-1)^2 =0\Rightarrow a-1=0\Rightarrow a=1$

Pentru a = 1, funcția devine:

$\it f(x) = (1-3)x +1 = -2x+1\\ \\ f(-5)=-2\cdot(-5)+1= 10+1=11$

$\it 20)\ h(x)= (2a-1)x+1\\ \\ M(2a,\ 3a^2+4)\in Gh \Rightarrow h(2a) = 3a^2+4 \Rightarrow (2a-1)\cdot2a+1=3a^2+4 \Rightarrow\\ \\ \Rightarrow4a^2-2a+1-3a^2-4=0 \Rightarrow a^2-2a-3=0\Rightarrow a^2+a-3a-3=0\Rightarrow\\ \\ \Rightarrow a(a+1)-3(a+1)=0 \Rightarrow (a+1)(a-3)=0\Rightarrow \begin{cases}\it a+1=0 \Rightarrowa=-1\\ \\ \it sau\\ \\ \it a-3=0\Rightarrow a=3\end{cases}$

$\it I)\ a=-1 \Rightarrow h(x) = -3x+1 \Rightarrow h(a)=h(-1)=4\\ \\ \sqrt{h(a)}=\sqrt4=2$

$\it II)\ a=3 \Rightarrow h(x) = 5x+1 \Rightarrow h(a)=h(3)=16\\ \\ \sqrt{h(a)}=\sqrt{16}=4$