Является ли число 28, 4 членом арифметической прогрессии если а1=3, 2 ; а5=4, 8 если да, то определите номер этого члена n.

4,8= 3,2 + d* (5-1)

4d=1,6

d=0,4

 

28,4= 3,2 + 0,4*(n-1)

25,2= 0,4n - 0,4

25, 6= 0,4n

n=64

это 64 член арифметической прогрессии. 

The given equation can be re-written as sin

2

4x−2sin4xcos

4

x+cos

2

x=0

Add and subtract cos

8

x

∴(sin4x−cos

4

x)

2

+cos

2

x(1−cos

6

x)=0

Since both the terms are +ive (cos

6

x≤1), above is possible only when each term is zero for the same value of x.

sin4x−cos

4

x=0 .(1)

and cos

2

x(1−cos

6

x)=0 .(2)

From (2) cosx=0 or cos

2

x=1

∵z

3

=1⇒z=1 only

as other values will not be real.

Case I: If cosx=0 i.e., x=(n+

2

1

)π, then from (1)

sin4(n+

2

1

)π+0=0

or sin(4n+2)π=0 which is true.

∴x=(n+

2

1

)π (3)

Case II: When cos

2

x=1 i.e., sinx=0

∴x=rπ then from (1), sin4rπ−1=0 or −1=0 which is not true. Hence the only solution is given by (3).

1. y=(x-4)^2*(x-1)

y'=2x - 8

y'=0, 2x - 8 =0

          x = 4

y(4)=3

y(1.5)=3.125

y(4.5)=0.875

 

ответ 3,125

 

2.  y=2x^3+2x^2-10x+1

y'= 6x^2 + 4x - 10

y'=0, 6x^2 + 4x -10=0

d=256

x1 = (-4 -4)/2=-4

x2=0

 

y(0)= 1

y(-4)= 201

y(-1)=-2+2+10+1=11

y(2)=16+8-20+1=5

ответ 201

 

y=x+16/x

y'= 1-16/x^2

y'=0 1-16/x^2=0

16/x^2=1

16=x^2

x1 = 4

x2 = -4

 

y(4)=8

y(-4)=-8

y(1)=17

y(8)= 10

ответ -8

 

y=6+27x-x^3

y'=27-3x^2

y'=0, 27-3x^2=0

3x^2=27

x^2=9

x1=3

x2=-3

y(3)=60

y(-3)=102

y(4)=50

ответ 50