Question

Cum se rezolva exercitiul 2, 3 si 6?
Multumesc pentru ajutor.

Answer 1

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 $\alpha\in ( \frac{\pi}{2} ;\pi)$ deducem ca unghiul este in cadranul doi. In cadranul doi valoarea cosinusului este negativa.
Concluzie:[tex] cos\alpha=- \frac{5}{13} [/tex]