Senior sophomore function questions, high sophomore function questions

1 by the title: [sin(a 2)-cos(a 2)] 2=sin 2(a 2)+cos 2(a 2)-2sin(a 2)cos(a 2)=1-sina = 8 13

Therefore sina = 5 13

Because 2 a, cosa=-root[(1-(5 13) 2]=-12 13

tan(π-=-tanβ=1/2

i.e. tan = -1 2

tan2 =2tan (1-tan 2 )=-4 3, so tan(a+2 )=(tana+tan2 ) (1-tanatan2 )=(-5 12-4 3) [1-(-5 12)(-4 3)]=-63 16

The first condition is squared at the same time to obtain sina according to the range of cosa second problem, according to the second condition to obtain the value of tan, and then according to the formula of double angle to obtain tan2, from the first question to calculate tana, substituted formula to get the result.

Such a simple question will not be, sad ......

Don't forget to give points!

(1) Derivative f'(x)=-x^2+4ax－3a^2△=4a^2

Find the root x1=-a x2=-3a

Because (0 a 1).

x1>x2

4a^2>0

The opening is upward. The maximum value of the function f(x) is the value obtained when x=x2.

2) When x=1+a, g(x)=2a-1 a*****a 1, when x=1-a, g(x)=-8a+6a-1 -a*****(7-root, number: 17) 16

a (7 + root number 17) 16

Because (7 + root number 17) 16<1

Therefore, the value range of a is (7-root17) 16 a (7+root17) 16

1)f'(x)=3x2-8x-3=(3x+1)(x-3)f'(x) > round jujube 0 is formed at x<-1 3 or x>3.

This is the missing vertical increment interval.

2) From (1) it is known that f(x) decreases on [0,3] and increases on [3,4], and the minimum wax value of f(x) is f(3)=-18 2m

So m<=-9

Problem: Let f(x)=(1-2x) (1+x).

If y=g(x) and y=f-1(1+x).

The image is symmetrical with respect to y=x, then g(2).

Solution 1: Directly find the inverse function of y=f-1(1+x), (the method is very important).

From the definition of the inverse function, it can be seen that y=f-1(1+x) and 1+x=f(y) are the same image, and then exchange x,y can be obtained, and 1+y=f(x) is the inverse function of y=f-1(1+x), so y=g(x)=f(x)-1 can be obtained

So g(2)=f(2)-1=-2

Solution 2, Analysis:

The key to this problem is to solve the problem of y=f-1(1+x), y=f-1(1+x) means to find the inverse function y=f-1(x) first, and then wither to substitute (1+x) into it.

The specific solution is as follows:

The inverse function of y=(1-2x) (1+x) is y=f-1(x)=(1-x) x+2

So y=f-1(1+x)=-x(x+3).

And because y=g(x) and y=f-1(1+x).

The image is symmetrical with respect to y=x, i.e. y=g(x) is y=f-1(1+x)=-x (x+3).

The inverse function of . Then find the inverse function].

So we get y=g(x)=-3x (x+1).

Untie. 1) When the long axis is on the x-axis, then the standard equation of the hyperbola is as follows: x 2 (4k 2)-y 2 (9k 2)=1 and substitute x = -2 and y = 3 2 to obtain: k 2=-1, no solution is not true.

2) When the major axis is on the y-axis, then the standard equation for a=3k, b=2k hyperbola is: y 2 (9k 2)-x 2 (4k 2)=1 substitute x=-2, y=3 2 into k2=1

The grinding range of the hyperbolic standard square is: y 2 9-x 2 4=1

f(x)=丨x 2-4x-3丨.

丨(x-2) 2+1丨.

So when x>=2, f(x) is incremented.

Draw the function image, turn the y<0 part of the graph over, and write it as a symmetric graph about the x-axis, the image is OK

Then look at the graph to find the monotonic interval.

x=0. f（0）=f（-0）=-f（0）∴f（0）=0

Odd functions must pass the origin, which is the basic property of odd functions.

Let y=0 then f(x)=f(x)+f(0).

f（0）=0

Take y=-x, then f(0)=f(x)+f(-x)=0 f(-x)=-f(x).

f(x) is an odd function.

1,x<0 f(x)=x^2-2x-5

0 makes y=-xf(0)=f(x)+f(-x)=0f(-x)=-f(x).

Then why is f(x) an odd function?

1.From the solution set, we can see that the root of equation two is -2, -1 4From Vedic theorem -2+(-1 4)=-b a,-2*(-1 4)=ac.

Since the equation c=-2, so a=-1 4, b=-9 defines the domain as r, then the function image opening is upward and above the x-axis, and the minimum value is greater than 0, and the price is a>0

f(x)=|x-a|-9/x+a x∈[1,6]1)f(x)=|x-1|-9/x+1

x-1-9/x+1

x-9/xf'(x)=1+9 x >0 f(x) monotonically transmits the gods.

2) a∈(1,6)

f(x)=a-x-9/x+a

2a-x-9/x

f'(x)=-1+9 x Bi Blind L.

Intra-interval station x=3

f''(x)=-18/x³<0

f(3) is the maximum value of repentance = 2a-6

i.e. m(a)=2a-6 a (1,6).