Question

31 . Se considera expresia E(x) = (2x+3)^2 - (2x-1)(4-x)-2(x+2)^2+3x-2 , unde x este un numar real .
a) Aratati ca E(x) = 4x^2-2x+3,pentruorice numar real x

b)Detrminati valorile intregi ale lui n , pentru care E(n) mai mic sau egal 2(n+4)+3

Answer 1

$a)~E(x)=(2x+3)^2-(2x-1)(4-x)-2(x+2)^2+3x-2\\\\ E(x)=(2x)^2+2 \cdot 2x \cdot 3+3^2-\Big(8x-2x^2-4+x\Big)-2\Big(x^2+2\cdot x \cdot 2+2^2\Big)+\\\\+3x-2\\\\ E(x)=4x^2+12x+9-\Big(9x-2x^2-4\Big)-2\Big(x^2+4x+4\Big)+3x-2\\\\ E(x)=4x^2+12x+9-9x+2x^2+4-2x^2-8x-8+3x-2\\\\ E(x)=4x^2-2x+3$

$\displaystyle b)~E(n)\leq 2(n+4)+3;~E(n)=4n^2-2n+3\\\\ E(n)\leq 2(n+4)+3\Rightarrow 4n^2-2n+3\leq 2n+8+3 \Rightarrow \\\\ \Rightarrow 4n^2-2n+3-2n-8-3\leq 0 \Rightarrow 4n^2-4n-8\leq 0\Big|:4 \Rightarrow \\\\ \Rightarrow n^2-n-2\leq 0\\\\n^2-n-2=0\\\\ \Delta=(-1)^2-4 \cdot 1 \cdot (-2)=1+8=9 > 0\\\\ n_1=\frac{-(-1)-\sqrt{9} }{2\cdot 1}=\frac{1-3}{2} =\frac{-2}{2} =-1 \\\\ n_2=\frac{-(-1)+\sqrt{9} }{2\cdot 1}=\frac{1+3}{2} =\frac{4}{2} =2$

$\displaystyle \left\begin{array}{ccc}\cfrac{n}{~~~~~~~~~~~~~~~} \\n^2-n-2\end{array}\right|\frac{-\infty~~~~~~~~~~~~~~~~~~~~-1~~~~~~~~~~~~~~~~~~~~~2~~~~~~~~~~~~~~~~~~~~~~\infty}{++++++++++0-------0+++++++++}$

$n^2-n-2\leq 0 \Rightarrow -1\leq n\leq 2\Rightarrow n\in \{-1;0;1;2\}$