A high school math problem known function f x 2acos2x b

Let x2 > x1, and x1 and x2 both belong to [0, 2].

f(x2)-f(x1)=-2acos2x2+b+2acos2x1-b=2a(cos2x1-cos2x2)

Since x1 and x2 both belong to [0, 2], 2x1 and 2x2 belong to [0, ].

The cosine function is monotonically decreasing on [0, ], so (cos2x1-cos2x2) > 0

1.When a>0.

f(x2)-f(x1)>0, indicating that the function f(x) is monotonically increasing on [0, 2].

So when x=0, f(x) has a minimum value of -5, i.e., f(0)=-2a+b=-5

When x= 2, f(x) has a maximum value of 1, i.e. f( 2)=b=1

Solution: a=3 b=1 satisfies the condition of the problem.

2.When a>0.

The original function becomes f(x)=b, which is a constant function, which obviously does not satisfy the condition (because the range of the constant function is invariant).

3.When a>0.

f(x2)-f(x1)<0, indicating that the function f(x) is monotonically decreasing on [0, 2].

So when x=0, f(x) has a maximum value of 1, i.e., f(0)=-2a+b=1

When x= 2, f(x) has a minimum value of -5 i.e. f( 2)=b=-5

Solution: a=-3 b=-5 satisfies the conditions of the problem.

To sum up: a = 3 b = 1 or a = -3 b = -5

Solution: There should be a problem with the question, whether f(x)=-acos2x+bx belongs to [0, 2].

then 2x belongs to [0, ].

cos2x belongs to [-1,1].

2cos2x belongs to [-absolute value a, absolute value a].

Easily seen: the absolute value a+b=1 and - the absolute value a+b=-5, so a=3 or -3; b=-2

First, find the monotonicity to determine the relative value of the range when x is 0.

Then substitute the evaluation.

f1(x)=ax+1-2a(the range of f1(x) at x0 is (- 1-a);

When a=0, the range of f1(x) is;

The range of f2(x) when a=2 is [-a2 4,+ a1,a>2.

In summary, the value range of a is (- 0] (2,+

Since x3cosx is an odd function and f(x)=x3cosx+1, if f(a)=2 then x3cosx is equal to 1 f(-a is equal to minus one plus one is 0 la( lalala.

f=2,the power of a "I don't use numbers, it's too troublesome!" >cosa+1=2

The power of a is cosa = 1

f<-a>=-a power cos< -a>+1=-a's power cosa +1=0

First, simplify cos2x=2cos x-1

Original = cos2x + root number 3sin2x + +1=

Then solve 2k < 3+2x<

k -75° i.e. so =

The third question is to be solved on this basis.

Solution: 1, f(-x).

a-2^(-x)]/1+2^(-x)]

a*2^x -1]/[1+2^x]

f(x)=[a+2^x]/[1+2^x]

When -a+2 x = a*2 x -1

i.e. when a=1 and f(-x)=-f(x), then f(x) is an odd function.

If a≠1, then f(x) is a non-odd and non-even function.

2. If f(x) is an odd function, then a=1

f(x)=(1-2^x)/(1+2^x)

1 +[2/（1+2^x)]

2^x＞01+2^x＞1

1/(1+2^x)∈（0,1）

2/(1+2^x)∈(0,2)

f(x)∈(1,1)

This is the value range. arbitrarily take m n, then.

f(m)-f(n)

1+[2/(1+2^m)]+1-[2/(1+2^n)]2*(2^n -2^m)/[1+2^m)(1+2^n)]2^m＞2^n，1+2^m＞0,1+2^n＞0f(m)-f(n)＜0

f(m)＜f(n)

f(x) is a subtractive function.

Have fun!

(1) f(-x)=(x 2+a) (-x)=-f(x) defines the domain (-infinity, 0)u(0, +infinity).

So f(x) is an odd function.

2) f(1)=(1+a) 1=2 solves a=1, so f(x)=(x 2+1) x=x+1 x is the destructor:

When x>0, f(x)min=f(1).

When (0,1), single minus.

When (1,+), Shan Zeng.

Or according to the method of definition.

(1) Define the domain as x is not equal to 0 f(-x)=(x 2+a) (-x)=-f(x), which is an odd function, 2) f(1)=2, a=1

Let x2>x1>1, then f(x2)-f(x1)=x2-x1>0, so it is proved.

Proof: Derived from the original formula: defines the domain as r.

Take x1 and x2 on the definition field, and let x1>x2, f(x1)=(x1-a)(x1-b)2,f(x2)=(x2-a)(x2-b)2, f(x1)-f(x2)=

If you're satisfied, I'll do it again.

(1)f(x)=x 2 2-1+cosx, f'(x)=x-sinx f''(x)=1-cosx 0 is constant, so f'(x)=x-sinx is monotonically increasing on r, and f'(0)=0, so when x>0, f'(x)>0 is constant, so f(x) is increasing the function on (0,+;

2) f(x)=ax 2 2-1+cosx is an increasing function on (0,+, so f'(x)=ax-sinx>0 is constant on (0,+, and f'(0)=0, so there is a positive number, so that f'(x) is an increasing function on (0, ), so when x (0, ), f''(x)>0 is constant, i.e. f''(x)=a-cosx 0, a cosx is true, so a 1

3) From (1) we know that f(x) increases the function on (0,+, and f(0)=0, so when x>0, f(x) > 0

Prove by mathematical induction:

When n=1, 00

Suppose that when n=k, 00, (ak) 2<1 2, cosak<1

Then when n=k+1, a(k+1)=f(ak)= ak) 2 2-1+cosak<1 2-1+1=1 2<1

This means that when n=k+1, the proposition holds.

From , it can be seen that for all positive integers n,0 so 0< a(n+1).

a=1,f(x)=x 2 The derivative of 2-1cosx is x-sinx, and the derivative of this function is 1+cosx, so x-sinx is an increasing function, and when x=0, x-sinx=0, and because it is an increasing function, it is greater than 0 in 0 to infinity, so the original function is increasing in this interval.

Comparing the first question yields a>=1

0< a1<1, so the above equation is less than 0, so an+1