(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.