1。 f(x)=x^2-ax+2=(x-a/2)^2+2-a^2/4

Let's discuss the scope of a:

(1) when a/2

g(a)=g( 1)=3-a。

(2) When 1

(3) when A/2 >; 5, that is, when a> is at 10, g(a)=g(5)=27-5a.

2.( 1) As can be seen from the meaning of the question, every time the advertising cost increases by 1000 yuan, the sales volume will increase by b (1/2) n.

So sn=n/ 1000 times (b+b * (1/2) n)

=bn/ 1000[ 1+( 1/2)^n]

(2) when a = 10 and b = 4000, the profit f (x) = sn * a = 4000n/1000 [1+(12) n] */kloc-.

=40n[ 1+( 1/2)^n]

To find its maximum value, this requires the derivation of senior three. I won't.

3.( 1) Find | A+B | 2 = A2+B2+2A * B = 4+4+2 * 2 * The cosine of the angle between A and B = 4.

So |a+b|=2, and so is |a-b|.

(2)(a+b)*(a-b)=|a+b||a-b| times cosine of included angle =2*2* root number 3* cosine of included angle.

=a^2-b^2

=0

So the cosine of the included angle is 0, so the included angle is 90 degrees.

Don't forget to add points to solve your problem. If I am a sophomore, let's discuss it together.