Mathematical value range, a problem of mathematical value range

The known function f(x)=x +1(x>0), f(x)= (-x -4x)+a(x 0); The tangent at points (1,2) has three commonalities with the image of f(x).

point, find the range of values of a.

Solution: f'(x)=2x，f '(1)=2, so the tangent equation for (1,2) is y=2(x-1)+2=2x; Tangent points (1,2) count as a common point.

f(x)= (-x -4x)+a: -x -4x=-x(x+4) 0, i.e., x(x+4) 0, so -4 x 0

The other two common points can only be tangents with f(x) = (-x -4x)+a.

Let 2x= (-x -4x)+a, i.e. have 2x-a= (-x -4x);

square the root number to get 4x -4ax+a =-x -4x;

That is, there is 5x -4(a-1)x+a = 0

Because there are two common points, its discriminant formula =16(a-1) -20a =-4a -32a+16=-4(a +8a-4)>0

i.e. a +8a-4=[a-(-8- 80) 2][a-(-8+ 80) 2]=[a-(-4-2 5)][a-(-4+2 5)]<0

Hence -4-2 5

x 0, f'(x)=2x, it is easy to find the equation of the tangent l l of f(x) at (1,2) as y=2x, and there is only one intersection point between l and the image of f(x).

At x 0, the equation of simultaneous f(x) and l is subtracted y to obtain a quadratic equation (with parameter a) about x

From the problem we know that the equation has two unequal real roots on (- 0), then there is.

0，x1+x2<0，x1x2>0

Simultaneous of these three inequalities can be used to solve the range of a.

In fact, 5 6 2x

17π/6，2x

The interval length of 6 is exactly 2, so the range of sin(2x-6) is exactly [-1,1].

Hello! y=sinx in a certain interval of the single point minus interval is to be remembered!

In [ 2,3 2] is a monotonically decreasing interval of the function, so directly replace x with 2x-6 to find the monotonic decreasing interval of the function!

Thanks for adopting!

Add 9 16 on both sides of the equation at the same time, ie.

x²-3/2x+9/16=k+9/16

x-3/4)²=k+9/16

Because, -1<=x<=1

So -7 4<=x-3 4<=1 4

So 0<=(x-3 4) <=49 16

So 0<=k+9 16<=49 16

9/16<=k<=5/2

This is an inequality with absolute values, and there are two simple models for inequalities with absolute values:

x|0) -aa (a>0) x>a or x<-a

The original problem obviously belongs to the first type, so we can get -15<=15-3a<=15, further solution, -15<=15-3a, solution a<=10

15-3a<=15, solution a>=0

So the original inequality is solved as: 0<=a<=10

Combine the number line, remove the sign of absolute values, the properties of inequalities, and variations.

-15≤15-3a≤15

Subtract 15 from the same and the direction of the unequal sign remains the same.

30≤-3a≤0

Divide by -3 to change direction.

10≥a≥0

Solution: -15 15 - 3a 15

15-15≤15-15-3a≤15-15-30≤-3a≤0

When -3a -30, then a 10

When -3a 0, then a 0

The value range of a is [0,10].

This kind of problem is mainly solved by using the method of "combination of numbers and shapes".

The first question seems to be in the wrong place, because in this case, the value of "m" is an empty set. (In the big exam, the person who wrote the question should not ask such a meaningless question).

It's pretty much the same to change " to ">", or change the range of the range (0,2).

According to the data of your question, I wrote the steps, you can take a look and understand the idea of solving the problem.

This is converted to "x 2 logmx".

Therefore, in the interval (0,2), y=x 2 and y=logmx can be roughly drawn.

The combination of numbers and shapes clearly knows that as long as x belongs to (0,2) and y=x 2 is below y=logmx, it is constant.

So, when x=2, logmx>4, we get (a value).

Then combine the value of the solution with (00, then y<0; when x=1 2).x=2 at y>0

Solution 3 2 < a < 12

a<0, then x=1 2 y<0;x=2 at y>0x= -2 2a (axis of symmetry) >=0

The solution does not exist.

Therefore, the value of the comprehensive A is (3 2,12).

ps: Maybe my calculations will be wrong, but the solution method will definitely not be wrong, take a look at the method carefully, this is a very basic question).

You can draw a sketch and observe it.

1) A+1 0, 3-2a 0, then A+1 3-2A, then 2 3 A 3 2

2) A+1 0, 3-2A 0, clearly inconsistent.

Synthesis one two: 2 3 a 3 2.

x1x2＞1/2

Is it a product, or a separation

x-a)(1-x-a)<1, as a general formula:

x 2-x-(a 2-a-1)>0, from this formula, it can be seen that the image opening is upward and must be greater than 0, so the image has no intersection with the x-axis, then <0 is sufficient, that is:

1+4(a 2-a-1)<0 Heng is established, and the solution:-(1 2) should be detailed enough! Hope it helps!

(x-a)(1-x-a)<1 can be converted to x2+x-a2+a+1>0, i.e., y=x2+x-a2+a+1 has no intersection with the x-axis. Get 1-4(-a2+a+1)<0

1+4a2-4a-4<0 4a2-4a-3<0 (2a-3)(2a+1)<0

（x-a)(1-x-a)<1

Open the brackets so that x-x squared - a+a squared < 1

x square - x-a square + a + a + 1>0

x squared - x+1 4-1 4-a squared + a + 1>0, i.e. (x-1 2) squared - a square + a + 3 4>0

Because (x-1 2) squared" = 0

Therefore, it is necessary to make (x-1 2) squared - a square + a + a + 3 4>0 for any real number constant.

It is -a square + a + 3 4>0, that is, 4a square - 4a-3<0, that is, (2a-3) (2a+1) <0

1/2

x^2-x-a(a-1)>-1

x^2-x-(a^2-a-1)>0

Let f(x)=x 2-x-(a 2-a-1) find the derivative of x and make it equal to 0 to find the minimum.

i.e. f'(x)=2x-1=0

That is, when x=, f(x) is the smallest, and the minimum value is 1 4-1 2-(a 2-a-1) because (x-a)(1-x-a)< 1 is constant for any real number, so 1 4-1 2-(a 2-a-1)>0

Finishing the above equation yields a 2-a-3 4<0

For the left formula, we get (a-1 2) 2-1<0, i.e. (a-1, 2) 2<1, i.e., -1

, we get x 2-x+a+1-a 2>0, that is, the unary quadratic equation is everstable at 0, (-1) 2-4 (a+1-a 2)<0, and the solution is a

x2+x+a2-a-1<0

Then the delta should be less than zero to be established.

delta<0 ji 4a2-4a-5>0 you can solve it yourself sorry the root number is not easy to play!!

Move to -x 2+x+a 2-a-1 0

Because the constant holds, the quadratic function of this opening downward has no intersection with the x-axis.

4a^2-4a-3＜0

2a-3）（2a+1）＜0

1/2＜a＜3/2