x & gt0

(1)c=0 f(x)=lnx easy to own 1 zero x= 1.

(2)c & gt； 0

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

f’(x)>0 1 & gt； 2cx^2 x & lt； Root number (1/2c)

f '(x)& lt； 0 1 & lt； 2cx^2 x & gt； Root number (1/2c)

F' (x) = 0 1 = 2cx 2x = radical sign (1/2c)

At (0, 1/2c), f (x) monotonically increases ... f (root sign 1/2c)) is the maximum.

F (root number 1/2c))=ln (root number (1/2c))-1/2 = (1/2) (ln (1/2c)-/kloc-.

When ln (1/2c) < 1,1/2c < E, c> 1/2e, f(x) has no zero.

When ln (1/2c) = 1, 1/2c = E and C = 1/2e, F (x) has 1 zeros.

When ln (1/2c) >: 1, 1/2c >; e，0 & ltc & lt 1/2e f(x)>0

Then prove that f (x) has a negative value (you don't need the following method after learning to find the limit), or you can directly construct an x to make f(x).

In the interval of (0, 1), f (x) is less than 0, and there are 1 zeros.

Here, we can use inequality to construct x and directly find the limit.

inequality

X> radical 1/2c, inequality lnx-CX 2.

(building root number1/2c >; Root number e f (1) =-c

(3)

C<0 f'(x)>0 function monotonically increases.

f( 1)=-c & gt； 0

1. build x

0 & lte^c<； 1

f(e^c)=c-c(e^c)^2=c( 1-(e^c)^2)<； 0

So there is a zero point.

two

0 & ltc & lt 1/2e

Let x1> x2

ln(x 1x2)=c(x 1^2+x2^2)

ln(x 1/x2)=c(x 1-x2)

(x 1x2)=e^(c(x 1^2+x2^2))>； 1

ln(x 1/x2)/ln(x 1x2)=(x 1^2-x2^2)/(x 1^2+x2^2)

Suppose x 1x2.

ln(x 1/x2)& lt； =(x 1^2-x2^2)/(x 1^2+x2^2)

Let t = x1/x2 t > 1

lnt & gt(t^2- 1)/(t^2+ 1)

Let g (t) = lnt-(t2-1)/(t2+1).

g'(t)=(t^2- 1)^2/t>； 0 grams (tons) is increasing. ...

g( 1)=0 g(t)>0

Contradict with the hypothesis

So x1x 2 >; E dezheng