So a-b = 2 (1)

F '(x)= 2ax+b- 1 slope k=-2a+b at the point.

This line is perpendicular to the line with x-3y=0.

Then -2a+b=3 ②

A=-5 b=-7 comes from ① ② solution, so f(x)=-5x? -7 times

(2)f(x)=-5x？ -7x monotonically increases in the interval [m, m+ 1].

First of all, f(x) is a parabola with a downward opening, and the symmetry axis is x=-0.7.

Increasing (-∞, -0.7) in the interval.

[m, m+ 1] If it is within this interval, then m+ 1≤-0.7 gives the solution of m≤- 1.7.

2.( 1) Pass Cosine Theorem

cosB=(a？ +c？ -B? )/2ac≥(2ac-b？ )/2ac= 1/2

And B∈(0, π), so B(0, π/3)

(2)y= (sin? B+cos？ B+2sinBcosB)/(sinB+cosB)

=(sinB+cosB)？ /(sin b+ cosB)= sin b+ cosB =√2 sin(b+π/4)

0 & ltB& lt； =60

45 & ltb+45 & lt； = 105

So sin(B+π/4)∈(√2/2, 1)

So y∈( 1, √2)

3.f(x)=lnx+a( 1-a)x？ -2( 1-a)x

Domain x>0

f '(x)= 1/x+2a( 1-a)x-2( 1-a)=[ 1+2a( 1-a)x？ -2( 1-a)x]/x

Molecules greater than 0

Discuss denominator g(x)=2a( 1-a)x? -2( 1-a)x+ 1

A<0 (1-a) < 0 parabolic downward opening 2a.

δ= 4( 1-a)？ -8a( 1-a)= 12a？ - 16a+4 = 4(a- 1)(3a+ 1)

①0 & gt； When a≥- 1/3, that is, δ≤ 0, g(x)≤0 is a constant, so f(x) is a decreasing function.

②a & lt； - 1/3 δ >; 0 equation has two real roots x 1, x2 (x 1

x 1 = 2( 1-a)-√( 12a？ - 16a+4)/4a( 1-a)= 1/2a-√(3a？ -4a+ 1)/2a( 1-a)

x2= 1/2a+√(3a？ -4a+ 1)/2a( 1-a)

F(x) is a decreasing function on (-∞, x 1) and (x2,+∞).

F(x) is the increasing function on (x 1, x2).

4.( 1)f(x)=x？ -9x？ /2+6x-a

f'(x)=3x？ -9x+6 is greater than m, which holds.

f'(x)=3x？ The minimum value of -9x+6 is at the symmetry axis x=3/2.

f'(x)min=3×9/4-9×3/2+6=-3/4

As long as m is not greater than the minimum value of f'(x), inequality is a constant.

So the maximum value of m is -3/4.

(2) g(x)=x? -9x？ /2+6x=a has only one real root.

g'(x)=3x？ -9x+6=3(x- 1)(x-2)

So f(X) increases at (-∞, 1); To 1, take 5/2 of the maximum value.

Decreasing in the interval of (1, 2); Take the minimum value of 2 o'clock direction as 2.

It is increasing at the speed of (2, +∞).

Y=a represents a straight line parallel to the x axis.

Draw a rough picture of y=f(x), and you can see that

The condition that there is only one real root is a≥5/2 or a≤2.

5.f(x)=2cos2x+sin？ x = 2 cos2x+( 1-cos2x)/2 = 3 cos2x/2+ 1/2

cos2x=cos(π/6)=√3/2

So the original formula =(3√3+2)/4.