f'( 1)=2-a f( 1)=0
The tangent equation at (1, f( 1)): y=(2-a)(x- 1)
The intersection of simultaneous solution and x 2+y 2 = 1: x 1.
( 1,0)((3-4a+a^2)/(5-4a+a^2),2(-2+a)/(5-4a+a^2)
4/(5-4 * A+A 2) = 2 Solution: a= 1 or a=3.
(2) 1+ 1/x^2-a/x>; 0 x^2-ax+ 1>; 0 △=a^2-4<; = 0 | a | & lt= 2∴-2 & lt; = a & lt=2