High School Math Problems, Detailed Explanations! Thank you 10

Solution: y x 2-3x-4=x 2-3x+9 4-25 4=(x-3 2) 2-25 4

Define the domain as 0,m

Then the function value is maximum when x=0.

That is, the maximum value of y = (0-3 2) 2-25 4=9 4-25 4=-4 and the range is 25 4, 4

That is, when x = m, the function is smallest and y is minimum = -25 4, i.e., -25 4 = < (m-3 2) 2-25 4< = -40 = < (m-3 2) 2< = 9 4

i.e. m>=3 2 (1).

i.e. (m-3 2) 2<=9 4

m-3 2>=-3 2 and m-3 2<=3 20<=m<=3 (2).

So: 3 2<=m<=3

The value range of m is [-3, 2,3].

When the axis of symmetry is x=-3 2, at this time y=min. -25 4x>-3 2, the function increases monotonically, and when x=3, y=-4, so x is between -3 2 and -3.

Let's start with the answer being a closed interval of 3 2 to 3.

Draw well, explain that you draw the following figure, the minimum range is -25 4, this is the value of the vertex of the function, so m is greater than or equal to the axis of symmetry x=3 2, the range is less than -4, and when y=-4, x is equal to 0 or 3, so m is less than or equal to 3

First, break down the factor, find the zero point, and draw the graph.

Let the function f(x)=x-a x-1, the set m=, p=, and if m is a true subset of p, then the range of the real number a is .

Solution]: f(x)=(x-a) (x-1).

m=f(x)=(x-a)/(x-1)<0

When a>1, x (1,a).

When a=1, x=empty set.

When a<1, x (a,1).

f'(x)=(a-1)/(x-1)^2

p=f'(x)>0

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

When a>1, x r

When a=1, x=empty set.

When a=1, x=empty set.

m is a true subset of p.

So: a>1

Question 1, question 2, question 3, question 4, question 5, search the whole network.

The problem is known that the function f(x)=|x+a|+|2x-1|(a r) (when a=1, find the solution set of inequality f(x) 2;

If the solution set of f(x) 2x contains [12].

1] to find the range of values of a

Analysis (1) Through classification discussion, the absolute value sign in the absolute value function is removed and transformed into a piecewise function, and the solution set of the inequality f(x) 0 can be obtained.

2) From the meaning of the question, the inequality can be reduced to |x+a|+2x-1 2x, i.e. |x+a|1, the solution is -a-1 x -a+1, which is contained by the solution set of f(x) 2x [12].

1], available.

a−1≤12

a+1 1 can be solved to obtain the value range of a

Answer(1) When a=1, the inequality f(x) 2 can be reduced to |x+1|+|2x-1|2, when x 12

, the inequality is 3x 2 and the solution is x 23

Therefore, the solution set of the inequality f(x) 2 is x 23;When -1 x 12

, the inequality is 2-x 2, and the solution is x 0, so the solution set of the inequality f(x) 2 is -1 x 0;

When x -1, the inequality is -3x 2, and the solution is x 23, so x -1;

In summary, the set of solutions to the original inequality is;

2) Because the solution set of f(x) 2x contains [12].

1], the inequality can be reduced to |x+a|+2x-1 2x, i.e. |x+a|1, the solution is -a-1 x -a+1, which is known.

a−1≤12

A+1 1 gives 32

a 0, so the range of values for a is [ 32

1）a=-4 b=-16

Drawing method Draw an image of g(x) first: If f(x)<=g(x) (both are absolute values), then the expression of f g must be the same.

2) f(x)>=m+2)x-m-15] to 2x 2-(m+6)x+m-1> scramble=0

The coefficient of the quadratic term is 2, so the opening is upward, there is a minimum value, which is the axis of symmetry, and the corresponding chain of values is the minimum value, and the minimum value at this time is calculated, and the other "=0" solves a range of m.

2x 2-(m+6)x+m-1>=0 is reduced to the equation (1-x)m>=-2x 2+6x-1 about m

m>=-2(1-x)-1+3 (1-x) is the checkpoint function, because x is greater than 2, so -2(1-x)-1+3 (1-x)>=2*3-1=-7 is m>=-7

Compound m>=-7 with the range of one m found above to find the range of m.

1 a=-2 b=-8

2m is less than or equal to -5 (derivative).

3 exists.

First of all, it should be noted that if the whole of positive integers is taken as the sample space, then the space is infinite and does not belong to the classical generalization. But the last digit of the square of a positive integer depends only on the last digit of the positive integer, and the last digit of the positive integer is 0, 1, 2 ,...Any number in 9, now taking any positive integer means that these ten numbers are equally possible. Therefore, take the sample space as the desired event as a {1 , 9 }, so take p(a)=2 10=1 5.

The last digit of the square of a positive integer depends on the last digit of the corrected integer, when the last digit is 1 and 9, the square can be 1, and the last digit has ten cases, therefore.

The probability should be 2 10=

(1 square is 1) 2 square is 4 3 square is 9 4 square is 6 5 square is 5 6 square is 6 7 square is 9 8 square is 4 (9 square is 1) 10 square is 0

Regardless of the number of digits in the following numbers, the single digit obtained is the same as the above, and the probability that the last digit is 1 is: two parentheses above.

The last numbers are:

It can be seen that only when the mantissa is 1 or 9, the end of the square is 1

Then the probability is 2 10=

Let the base salary constant be c and the bonus be xv.

i=c+xv

7000=c+100000x

8000=c+200000x

Solve the equation to obtain: x=, c=60000.

1)i=6000+

2)6000=（1-40%）i

i=10000

v=（i-6000）/

The profit for the month was 400,000.