[tex]\it 1)\ \ s=1+\Big(\dfrac{1}{2} +\dfrac{1}{2^2} +\dfrac{1}{2^3}+\ ...\ +\dfrac{1}{2^{2009}}\Big) \Rightarrow s>1\ \ \ \ (1)[/tex]
Observăm că s reprezintă suma primilor 2010 termeni ai unei
progresii geometrice, cu rația q= 1/2 și primul termen egal cu 1.
[tex]\it s=\dfrac{1-q^{2010}}{1-q}=\dfrac{1-\dfrac{1}{2^{2010}}}{1-\dfrac{1}{2}}=\dfrac{1-\dfrac{1}{2^{2010}}}{\dfrac{1}{2}}=2\cdot\Big(1-\dfrac{1}{2^{2010}}\Big)=\\ \\ \\ =2-\dfrac{2}{2^{2010}}=2-\dfrac{1}{2^{2009}}<2\ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow s\in(1,\ \ 2)[/tex]
[tex]\it 2)\ \ G_f\cap G_g=P(x,\ y)\\ \\ P(x,\ y)\in G_f \Rightarrow y=f(x) \Rightarrow y=2x-1\ \ \ \ \ \ (1)\\ \\ P(x,\ y)\in G_g \Rightarrow y=g(x) \Rightarrow y=-4x+1\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow\ 2x-1=-4x+1 \Rightarrow 2x+4x=1+1 \Rightarrow 6x=2 \Rightarrow x=\dfrac{2}{6}=\dfrac{1}{3}\ \ \ \ \ (3)\\ \\ (1),\ (3) \Rightarrow y=2\cdot\dfrac{1}{3}-1 =\dfrac{2}{3}-\dfrac{3}{3}=-\dfrac{1}{3}\\ \\ \ G_f\cap G_g=P(\dfrac{1}{3},\ -\dfrac{1}{3})[/tex]
