It is known that in △ABC, AB=AC, BD⊥AC, BD= 1/2AB.

Find the degree of ∠BAC

Solution: Let BD⊥AC intersect with straight line AC at point D.

(1) When point D is on alternating current.

∫BD = 1/2AB

∴∠BAC=30

(2) When point D is on the extension line of CA,

∫BD = 1/2AB

∴∠BAD=30

∴∠BAC= 150

f'(x)=2x-m/x，

h'(x)=2x- 1，

Let f'(x)=0 and get m = 2x2; x=√m/2，

Let h'(x)=0 and x= 1/2.

In order to satisfy that f(x) and h(x) have the same monotonicity in the common area, the extreme points of the two functions must be the same, that is,

√m/2= 1/2, so m= 1/2.

K(x)=-2lnx+x-a=0, let two zeros be X 1 ≥ 1, X2 ≤ 3, a =-2LNX1=-2LNX2+x2;

Let g (x1) =-2lnx1+x1,Y (x2) =-2LNX2+x2,

g '(x 1)=-2/x 1+ 1，(x 1 ≥ 1)，g(x 1)≥g(2)=-2ln 2+2；

Y'(x2)=-2/x2+ 1，(x2≤3)，y(x2)≤y(3)=-2ln 3+3；

So there is -2ln2+2≤a≤-2ln3+3.