It is known that vectors a and b satisfy a = b =1and ka+b = √ 3a-kb (k > 0), so f (k) = a.b.

(1) Find f(k)=a? B (represented by k)

(2) when k > 0, f(k)≥x? -2tx- 1/2 holds for any t∈[- 1, 1], and sets the value range of real number X.

Solution: (1) Both sides are squares. ka+b =√3. One thousand bytes. Obtained:

k？ Answer? +2ka？ b+b？ =3(a？ -2ka？ b+k？ b？ )，∵?a ? = ? b ? = 1，∴a？ =b？ = 1, replaced by:

8ka？ b=2k？ +2, so f(k)=a? b=(k？ +1)/4k = (k/4)+(1/4k) ≧ 2 √ (116) =1/2 if and only if k/4 = 65438.

That is, when k= 1, take the equal sign.

(2) In order to make f(k)≥x? -2t- 1/2 holds for any t∈[- 1, 1], and the minimum value of f(k) must satisfy the inequality:

1/2≥x？ -2tx- 1/2, that is, the inequality x? -2tx- 1≦0 holds for any t∈[- 1, 1].

That's (x-t)? -t？ -1≦0, so there is (x-t)? ≦t？ + 1; Because-1≦t≦ 1 so t? ≦ 1, so t? + 1≦2;

That's (x-t)? ≦2，-√2≦x-t≦√2，t-√2≦x≦t+√2；

When t=- 1, there is-1-√ 2 ≦ x ≦-1+√ 2 .......... ①; When t= 1, there is1-√ 2 ≦ x ≦1+√ 2 .......... ②;

① ∩ ② = {x ︱1-√ 2 ≦ x ≦-1+√ 2}, which is the value range of X.