If the inequality about x 2 x 8 x 4 a 0 has a solution in 1 x 4, then the range of values of the rea

This question should be simple with the separation parameter method.

The inequality 2x-8x-4-a0 has a solution within 1 x 4.

Meaning: in the interval (1,4) there is an x value that makes a<2x-8x-4 hold.

Only the maximum value of a< (2x -8x-4) is required.

Let f(x)=2x -8x-4=2(x-2) -12When x=1 or x=4, the maximum value is -10, but this maximum value cannot be taken.

When x=2, the minimum value is -12

Therefore, the value range of f(x) is -12 and the value range of f(x) is -10 <, so the value range of a is a -10

Converted to y=2x -8x-4 in the range of 1 x 4, it is found to be (-12,-4) If there is a solution, a is here, i.e. a<-4

If the inequality has a solution within 1 x 4, it is satisfied.

1.There are two solutions to inequalities.

f(1) =0 ,2-8-4-a =0,a =-10f(4) =0, 32-32-4-a =0,a =-4 =0, 64+8(4+a) 0,a -122, there is a solution to the inequality.

0, a = -12, axis of symmetry x = 2 on (0, 4).

To sum up, the value range of the real number a is -12 =a =-10

Ling Jinhong slips y=x 2

The original question is 8y 2+8(a-2)y-a+5 0 to any y>=0 constant.

1. δ 0 solution obtains 1 2 = 3 or the dispersion a0 is 5-a>0, that is, a

Shirt void cavity t = x 2

The inequality can be reduced to 8t 2+8(a-2)t-a+5>0 on t>=0.

Let f(t)=8t 2+8(a-2)t-a+5 function opening upward, if t>=0 or shirt, f(t)>0 is constant.

f（0）>0 x=-8(a-2)/(2*8)0 (2-a)/2

Personally, I think the best way to do this problem is to combine numbers and shapes.

x^2-2<-|x-a|

Both figures of the left and right equations are easier to draw -|x-a|It is an inverted V shape.

The first tipping point is -|x-a|At (0,-2)y-(-2)=-1(x-0)=>y=-x-2, when y=0, x=-2, i.e., a=-2

So a>-2

The second tipping point is -|x-a|One of the branches is tangent to the function y=x 2-2.

y'=2x=1 =>x=1 2 tangent (1 2,-7 4)y+7 4=x-1 2 => y=0 =.x=9 4 i.e. a=9 4>-2 think it's good Graphics, the landlord can draw it himself.

Good luck with your studies.

x^2+|x-a|<2 Considering that x-a does not know whether it is greater than or less than or equal to 0, put x 2+|x-a|<2 into.

When x-a is greater than or equal to 0.

x^2+x-a-2<0

At this point, there should be at least one correct solution.

√b^2-4ac﹚/-2a≥0

Because the parabolic axis of symmetry x=-1 2 0

So just when x=0 x 2 + x-a-2 0 is fine.

The solution is a -2 at this time

When x-a is less than 0.

x^2-x+a-2<0

So δ = b 2-4ac -2a 0

The axis of symmetry at this point is x=1 2 0

So as long as δ 0 there must be a correct solution.

Solution a 9 4

Sum up. 2＜a≤9/4

.Look... Okay, no.,,I don't understand and ask again.。。

2x^2-8x-4-a>0

a<2x^2-8x-4

A should be less than 1 so y=2x 2-8x-4=2(x-2) 2-12, and when x=2, y has a minimum value y=-4

So a<-4

Revise the answer upstairs.

2x 2-8x-4-a>0 has a solution.

Only the maximum value needs to be greater than 0

Let y=2x 2-8x-4=2(x-2) 2-12, when x tends to 4, y tends to defeat the maximum value of -4

So a<-4

2x^2-8x-4-a>0

a<2x^2-8x-4

a should be less than 10 x=4 yo x=1 y<0 no solution.

After the discussion and grasp the argument, take the union set a<-4 buduima

Revise the upstairs answers.

2x 2-8x-4-a>0 has a solution.

Only the maximum value needs to be greater than 0

Let y=2x 2-8x-4=2(x-2) 2-12, and when x tends to 4, y tends to be the most finger-carrying value -4

So a<-4

Let y=2x 2-8x-4-a be parabolic because the parabola opening is upwards and there is no solution at 10 x=4 yo x=1 y<0.

After discussion, take the union set A<-4 Buduima

2x²-8x-4>a

2x²-8x-4

2(x-2)²+4

1a then a is less than his minimum.

So a<-4

The original inequality 2x2-8x-4-a 0 can be reduced to: a 2x2-8x-4, as long as a is less than the maximum value of y=2x2-8x-4 in 1 x 4, y=2x2-8x-4=2(x-2)2-12 y=2x2-8x-4 in 1 x 4 the maximum value is -4 then there is: a -4

So the answer is: a -4

A classic way to find the range of values of parameters like this is to come up with parameters!

Solution: The original inequality is deformed to give a 2x 2-8x-4 (1 x 4).

Let y 2x 2-8x-4The axis of symmetry is a function of the straight line x 2 y to obtain a minimum value of -12 at the absence of a macro infiltration x 2 and a maximum value of -4 at x 4 and y a can be regarded as a constant function, it is to have an intersection with y 2x 2-8x-4 (1 x 4), then the straight line y a must be between the minimum and maximum values.

i.e. -12 a 4

If you have any questions, please ask!