Y = cosx + √3sinx
y' = -sinx + √3cosx = 0
√3cosx = sinx
tgx = √3
x = π/3 + πk
-π/2 < π/3 + πk < π/2
-π/2 - π/3 < πk < π/2 - π/3
-5π/6 < πk < π/6
-5/6 < k < 1/6
k = 0, x=π/3 - максимум
y(π/3) = cos(π/3) + √3*sin(π/3) = 0.5 + √3*√3/2 = 2
y(-π/2) = cos(-π/2) + √3*sin(-π/2) = 0 - √3*1 = -√3
y(π/2) = cos(π/2) + √3*sin(π/2) = √3
Наибольшее значение при x=π/3, y=2
Наименьшее значение при x= -π/2, y= -√3
-12x= -2,4-3,6
-12x= -6
x= -6/-12
x=0,5
( 6x - 13)² = ( 6x - 11)²
36x² - 156x + 169 = 36x² - 132x + 121
36x² - 36x² - 156x + 132x = 121 - 169
- 24x = - 48
24x = 48
x = 2
Я вектора напишу просто буквами
b(b+xa)=0
b+xa=(2;x+1)
b(2;x+1)=0
2*2+x+1=0
x=-5