2)f(1)=2·1-1
f(2)=2·2-1
...........
f(2012)=2·2012-1
S=2(1+2+...+2012)-2012=2·(1+2012)·2012/2-2012=
=2013·2012-2012=2012(2013-1)=2012²=4048144
3)[tex]2^{x^2+x+0,5}=2^2\cdot 2^{\frac{1}{2}
[/tex]
[tex]2^{x^2+x+0,5}=2^{\frac{5}{2}}\\
x^2+x+0,5=\frac{5}{2}\\
2x^2+2x-4=0|:2\\
x^2+x-2=0\\
\triangle=b^2-4ac=9\\
x_{1/2}= \frac{-1\pm3}{2} \\
x_1= 1 \\
x_2=-2[/tex]
6)Folosind teorema fundamentala a trigonometriei obtinem:
[tex]sin^2\alpha+cos^2\alpha=1\\
( \frac{12}{13})^2 +cos^2\alpha=1\\
cos^2\alpha=1- \frac{144}{169} \\
cos\alpha=\pm \frac{5}{13} [/tex]
Deoarece [tex]\alpha\in ( \frac{\pi}{2} ;\pi)[/tex] deducem ca unghiul este in cadranul doi. In cadranul doi valoarea cosinusului este negativa.
Concluzie:[tex]
cos\alpha=- \frac{5}{13} [/tex]
