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.