Consider the number of zeros of the function fx x2 2x a 1 a R .

Let f(x)=x -2x-a-1=0

In the equation f(x)=0, =(-2) -4*1*(-a-1)=4+4a+4=4a+8

When 0, i.e., 4a+8 0,a -2, the equation f(x)=0 has two different real solutions, and the number of zeros is 2

When =0, i.e., 4a+8=0, a=-2, the equation f(x)=0 has a real solution, and the number of zeros is 1

When 0, i.e., 4a+8 0, a -2, the equation f(x)=0 has no real solution, and the number of zeros is 0

In summary, when a -2, the function f(x) has 2 zeros;

When a=-2, the function f(x) has 1 zero;

When a -2, the function f(x) has no zero point.

The correct answer is 6, but in my opinion, x=0 is a cosx2=0, and there is also x2=root 2 root 2= because it is in the interval 0, 4, so you can also take a multiply by 3=, and when multiplied by 5, it will be more than 4, which is 270 degrees, and I would like to ask how to take 6.

The correct solution is the solution: let f(x)=0, we get x=0 or cosx2=0x=0 or x2=k+

k∈zx∈[0，4]

k=0，1，2，3，4

There are 6 solutions to the equation.

The number of zeros of the function f(x)=xcosx2 in the interval [0,4] is 6.

When a<-2, there is no zero point.

When a=-2, a zero point.

When x>-2, two zeros.

Answer: f(x)=x (1 2)-(1 2) x=0x (1 2)=(1 2) xg(x)=x (1 2)>=0, the definition domain is x> rock meng=0, the monotonically increasing function h(x)=(1 2) x>0, the sense domain of the Dingdan jujube year is r, and the monotonically decreasing function is: g(x)=h(x) There is 1 intersection in the first quadrant so:

f(x)=x (1 2)-(1 2) x the number of zeros is 1.

x+2=0, y=0, is the zero point of the lift.

ax+1=0,ax=-1

When a=0, there is no solution to the reputation.

a≠ when Bibi 0, x=- a is the zero point.

f(x) = lnx-ax^2(a>0)

Define the domain x 0

Derivative: f (x) = 1 x-2ax = (1-2ax x) = x

Monotonic increase interval: (0,1 (2a)).

Monotonic reduction interval: (1 (2a),+

When x approaches 0, f(x) approaches -; When x approaches +, f(x) approaches - maximum f(1 (2a)) = ln[1 (2a)]-a*1 (2a) = -1 2ln(2a)-1 2 = -1 2[ln(2a)+1].

When the maximum f(1 (2a)) 0, there is no zero point, at this time:

0＜2a＜1/e

a＜1/(2e)

When the maximum value f(1 (2a))=0, a zero point is at this time, at this time:

a=1/(2e)

When the maximum f(1 (2a)) 0, two zeros, at this time:

a＞1/(2e)

Answer: f(x)=x (1 2)-(1 2) x=0x (1 2)=(1 2) x

g(x)=x (1 2)>=0, the domain is defined as x>=0, the monotonically increasing function h(x)=(1 2) x>0 is defined as r, the monotonically decreasing function is so: g(x)=h(x) has 1 intersection in the first quadrant, so: f(x)=x (1 2)-(1 2) x the number of zeros is 1.

Solution: |2^x-1|=a

The absolute value is 0, when a 0, x has no solution, i.e. 0 zeros when a 0, 2 x-1=0, x=0, there is 1 zero point, when 0 a 1, 2 x-1=-a or 2 x-1=a, x=log2(1-a) or x=log2(1+a), there are 2 zeros.

When a 1, 2 x-1 0, |2^x-1|=2 x-1=a, x=log2(1+a), with 1 zero.

To sum up: when a 0: 0 zeros.

When a 0: 1 zero.

When 0 a 1: 2 zeros.

When a 1: 1 zero.

When a<0 does not.

When a=0, there are two.

When a>0, there are four.

∵f（1）f（2）＜0

f(x) has only one number of zeros on (1,2). Choose C

by f(x)=x3-x2-x+1=0

We get x2(x-1)-(x-1)=(x-1)(x2-1)=(x-1)2(x+1)=0, and we get x=1 or x=-1, so there are two identical zeros 1 on [0,2], so the answer is: 2

2^x-1|=a

The absolute value is 0, when a 0, x has no solution, i.e. 0 zeros.

When a 0, 2 x-1=0, x=0, and only the width has 1 zero.

When 0 a 1, 2 x-1=-a or 2 x-1=a, x=log2(1-a) or the fool x=log2(1+a), has 2 zeros.

When a belt mountain pie 1, 2 x-1 0, |2^x-1|=2 x-1=a, x=log2(1+a), with 1 zero.

To sum up: when a 0: 0 zeros.

When a 0: 1 zero.

When 0 a 1: 2 zeros.

When a 1: 1 zero.