Question

Ex 35 va rog muuuultttt!!!

Answer 1

Răspuns:

Explicație pas cu pas:

35.

E(x) = (x + 4)^2 - 2(2 - x)(2x - 3) - (x - 3)^2 - 2(3x + 8)

= x^2 + 8x + 16 - 2(4x - 6 - 2x^2 + 3x) - x^2 + 6x - 9 - 6x - 16

= x^2 + 8x + 16 - 14x + 4x^2 + 12 - x^2 + 6x - 9 - 6x - 16

= 4x^2 - 6x + 3

___________

4n^2 - 6n + 3 ≤ -2n + 11

4n^2 - 6n + 2n + 3 - 11 ≤ 0

4n^2 - 4n - 8 ≤ 0

n^2 - n - 2 ≤ 0

Δ = 1 + 8 = 9

n1 = (1 + 3)/2 = 2

n2 = (1 - 3)/2 = -1

n ∈ {-1; 0; 1; 2}

Answer 2

$35.~~E(x)=(x+4)^2-2(2-x)(2x-3)-(x-3)^2-2(3x+8),~x\in\mathbb{R}\\\\ a)~E(x)=4x^2-6x+3\\\\ E(x)=(x+4)^2-2(2-x)(2x-3)-(x-3)^2-2(3x+8)\\\\ E(x)=x^2+2\cdot x \cdot 4+4^2-2(2 \cdot 2x-2 \cdot 3-2x^2+3x)-\\\\-(x^2-2\cdot x \cdot 3+3^2)-6x-16\\\\ E(x)=x^2+8x+16-2(4x-6-2x^2+3x)-(x^2-6x+9)-6x-16\\\\ E(x)=x^2+8x+16-8x+12+4x^2-6x-x^2+6x-9-6x-16\\\\ E(x)=4x^2-6x+3$

$b)~E(n)\leq -2n+11 \\\\ E(n)=4n^2-6n+3\\\\ E(n)\leq -2n+11 \Rightarrow 4n^2-6n+3\leq -2n+11 \Rightarrow \\\\\Rightarrow 4n^2-6n+2n+3-11\leq 0 \Rightarrow 4n^2-4n-8\leq 0 \Big|:4 \Rightarrow n^2-n-2\leq 0 \Rightarrow \\\\ \Rightarrow (n+1)(n-2)\leq 0\\\\ (n+1)(n-2)= 0 \Rightarrow n+1=0 \Rightarrow n=-1\\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow n-2=0 \Rightarrow n=2$

$~~~~~~~~~n~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~~~~~~-1~~~~~~~~~~~~~~~~~~~2~~~~~~~~~~~~~~~~\infty\\(n+1)(n-2)~~+++++++++++~~0~-----~~0~+++++++\\\\\\ (n+1)(n-2)\leq 0 \Rightarrow -1\leq n\leq 2 \Rightarrow n\in\{-1,0,1,2\}$