High School Math Comprehensive! Big brothers and sisters can help

=a2-4b

1.The equation has no real solution and δ< 0

At this point a2<4b

When a=0, b=1 or 2

When a=1, b=1 or 2

A=2, b=2

There are 5 cases in total, so the probability that the equation has no real solution is 5 c14?c13=5/12

2.The equation has only one real number solution δ=0

At this point a2<4b

When a=0, b=0

A=2, b=1

There are 2 cases in total, so the probability that the equation has only one real solution is 2 c14?c13=2/12=1/6

3.The equation has two real number solutions δ> 0

At this point a2<4b

When a=1, b=0

A=2, b=0

When a=3, b=0,1,2

There are 5 cases in total, so the probability that the equation has two real solutions is 5 c14?c13=5/12

The expected number of roots is 0*5 12+1*1 6+2*5 12=1

The condition for having a solid root is δ>=0, where b <=a2 4

The full value of a and b is the area of the rectangle = 2 * 3 = 6

The area of y <=x2 4 in this rectangle is (2 2)3 12+(3-2 2)*2=6-8 3 2

So the probability of having a real root is (6-8 3 2) 6=1-4 9 2

Topic 1 will help you do it.

First of all, how did you get this picture?

Keeping in mind a formula, this type of problem can be thought of as a large pyramid and a small pyramid. The square ratio of the height of the pyramid is equal to the ratio of the base area of the pyramid and applies to any pyramid and cone. So 150 54 = the square of h (h-14).

150 54 = the square of h (h-14), and the equation can be solved.

The first method (more complicated, but easy to understand).

The function f(x) is defined as an increasing function in r and f(1-ax-x 2) is less than f(2-a) to any 0=0

interval 0=0 then a<1

Therefore, 00 is -2-2 2=1 is a<-2

The interval 0=0 is a<3

Hence a<-2

Therefore, in summary, a<1

The second method is the function f(x) defined as an increasing function in r and f(1-ax-x 2) is less than f(2-a) to any 0==0

a<(x 2+1) (1-x) at 0=0

Combine the image of the quadratic function f(x)=x2+ax+1-a, with the opening pointing upwards, and the axis of symmetry is x=-a2

That is, when x<-a2, f(x) decreases monotonically, and when x>-a2, f(x) increases monotonically.

Categorize the value of a and discuss it.

i) When a>0, -a 2<0, so [0,1] is the increasing interval of f(x), as long as f(0)>0 is satisfied, there is x 2+ax+1-a>0 for any x [0,1], so f(0)=1-a>0, and the solution is a<1

ii) When 0 -a 2 1 is -2 a 0, in order for f(x) to satisfy f(x) on [0,1], > only f(-a 2) > 0 (f(-a 2) is the minimum value of the function)) i.e., (-a 2) 2+a*(-a 2)+1-a>0 is solved to -2-2 21, i.e., a<-2, f(x) decreases monotonically on [0,1], and to make f(x) >0, only f(1)>0, i.e., 1 2+a*1+1-a=2>0, obviously, the former formula is true for any a<-2.

Combining the above three cases, 0 can be obtained

The equation for the ellipse g is x 2 4 + y 2 3 = 1

The equation for the circle c can be reduced to (x+1 2) 2+y 2=9 4 to connect pb,pc,bc to get the right triangle pbc, in the right triangle pbc, pb 2=pc 2-bc 2=pc 2-9 4, so when pc takes the maximum, pb takes the maximum, and because c is on the x-axis, when p is also on the x-axis, pc is the maximum, at this time, pc=5 2

Therefore, the maximum value of PB is the root number [(5 2) 2-9 4] = 2

Solution: According to the title, the (2AT+B) side can be used to obtain:

4A square T square + 4ABT + B square.

Substituting t back to the original equation yields:

AT square + BT + C = 0

In the above equation, multiply 4a on the left and right sides together, and get:

4A square T square + 4ABT + 4AC = 0

To compare the size of two numbers, it can be used as a difference method, i.e.:

Subtract square b-4ac from square (2AT+B) and compare the difference with 0 to get:

2AT+B)-(B-square-4AC).

4A square T square + 4ABT + B square - B square + 4AC

4A square T square + 4ABT + 4AC

And we have already calculated it earlier:

4A square T square + 4ABT + 4AC = 0

Therefore, it can be seen that the difference between the two numbers is 0, so: the two numbers are equal, choose a.

First, (2at+b) square = 4a 2*t 2+4abt+b 2 secondly, ax square + bx square + c = 0, multiplied by 4a4a 2*x 2+4abx+4ac=0 on both sides

Therefore, 4a 2*t 2+4abt+b 2=-4ac+b 2 choose a

A (2at+b) square - (b square - 4ac) is equal to 4a (at square + bt + c), because t is the sum of the equation, so (at square + bt + c) is equal to 0, i.e. the two equations are equal.