Math masters come to a function problem thanks

Since k can be any real number, the original equation does not have a perpendicular x-axis and is not discussed.

Rotate 90 degrees:

After rotation, the straight line is perpendicular to the original line, so the product of the slope is -1 then k'＝-1/k

New equation for straight lines: y=(-1 k)x

When k=0, the new linear equation is x=0

Rotate 45 degrees:

Let the new straight line slope be k'

Then according to the to the to angle formula, there is tan45=(k-k')/(1+kk'So k'=(k-1)/(k+1)

New linear equation: y=(k-1)x (k+1)When k=-1, the new linear equation is x=0

Rotate 30 degrees:

Let the new straight line slope be k'

Then according to the to the to angle formula, there is tan30=(k-k')/(1+kk'So k'=(√3k-1)/(k+√3)

New linear equation: y=(3k-1)x (k+3)When k=- 3, the new linear equation is x=0

90 degrees: When x takes any value, the function value is on the y axis, specific to y=kx, y=0*x

45 degrees: When x is taken as any value, the function is symmetrical to the origin, i.e., y=x30 degrees: using the trigonometric function, the edges of 30 degrees are equal to tan30 degrees * x, so the analytical formula is y=(tan30 degrees * x)x

The straight line y=kx must pass the (0,0) point, that is, the origin, so take the origin as the rotation center, turn 90,45,30 degrees clockwise, and it will pass the (0,0) point.

90 degrees: y=-kx

45 degrees: y=x

30 degrees: This advantage is difficult.

There is no need to turn many corners in this problem, and the sδabc is directly calculated according to the side length ab.

sδabc=√3/4|ab|^2=√3

Since SδABP=SδABC, the distance from the point P to the straight line needs to be: 3 m=5 3 2 or m=-3 3 2

Because p is in the first quadrant, m takes only a positive value. So m=5 3 2

s quadrilateral abpo = s bpo+s abo

m/2+√3/2

m+√3)/2

Let the straight line bp be l, p(m,1 2), b(0,1)kbp=-1 2m

l：(1/2m)x+y-1

Because p is in the second quadrant, a<0

Let the distance from point d to l be d

d=( 3-2m) 1+4m2 under the root number

bp = 1+4m2 2 under the root number

S triangle ABP=1 2*D*BP

3-2m)/4

S triangle abp = s triangle abc = 3

3-2m)/4=√3

m=-3√3/2

Will the question be poorly written? It should be a straight line y=3 3x+1. But this line can't make a regular triangle in the first quadrant.

It is recommended to send the original title picture.

Since a is the intersection point, 2x+3=-x+5 2x=3 x=2 3 y=13 3 a[2 3,13 3] bc The abscissa of the two points is zero, so substituting can obtain b[-3 2,0]c[5,o].

s=[3 2+5] 13 3 1 2=13 2 13 3 1 2=169 12 Okay, you're too talented, I remember when I was a child, I didn't have this condition, I just waited for the morning who wrote it, come on and study hard Upstairs I miscalculated' I just miscalculated, but I changed it.

Solution: (1) y =2x+3 and y=-x+5 together to obtain a(2 3 ,13 3).

y=2x+3 and y=0 (abscissa) together yield: b(-3 2, 0)y=-x+5 and x=0 (ordinate): c(5,0)2)bc=3 2+5= high distance from A to bc=13 3s abc=1 2 times 13 2 times 13 3 times 13 3=269 12

Solution: Because the function is odd, when f(-1)=-1 f(1)=1 because it is an increasing function on [-1 1].

Therefore, the arbitrary function in this interval is in the range of [-1 1] so that g(a)=t*t-2at+t

Treat a as a variable and t as a constant.

There are two scenarios.

1) When t<0 g(a)=t*t-2at+t is a monotonic increasing function of a, as long as g(-1)=t*t+3t>=-1

g(1）=t*t-t>=1

The result is negative.

2) When t>=0 g(a)=t*t-2at+t is a monotonically decreasing function of a.

In this case, only g(1)=t*t-t>=1

The result is a positive value.

The results of (1) and (2) are combined.

Is the question wrong, f(x)<=t*t-2at+t,what is t?

Functions are too abstract I've never been very good at relying on feelings to blind it, friend.

Solution: g(x).

f(x)]^2+f(2x)

9+6log2(x)+[log2(x)]^2+[3+log2(2x)]

9+6log2(x)+[log2(x)]^2+13+7log2(x)+[log2(x)]^2[log2(x)+7/2]^2+3/4

2 Brother x 8, 2 2x 8

2≤x≤41≤log2(x)≤2

The axis of symmetry is log2(x)=-7 2

g(x) monotonically increasing envy group group or answer on [2,4].

The value range is [g(2),g(4)], i.e. [21,31].

I hope the landlord can mine it!

1) When m=-2 9, f(x)=2 x-2 9*3 xf(x+1)<=f(x) is converted to:

2^(x+1)-2/9*3^(x+1)<=2^x-2/9*3^x2^x<=2/9*(2*3^x)

2^(x-2)<=3^(x-2)

3/2)^(x-2)>=1

x-2>=0, i.e., x>=2

2)(4/3)^x>=2^x+m*3^x

i.e. m<=(4 9) x - (2 3) x

Let t=(2 3) x>0, then m<=t 2-t=(t-1 2) 2-1 4=g(t) because t>0, when t=1 2, g(t) takes the minimum value -1 4 so there is m<=-1 4