| 落雷 在 2004-10-29 10:46 PM 發表: Let f(x) = (16-k)x^2 +12x -k , where k is a constant. a) Find the discriminant of f(x)=0 . b) Find the condition of k such that f(x)≧0 for all real values of x . a) Δ = b<sup>2</sup> - 4ac = 12<sup>2</sup> - 4(16-k)(-k) = 144 + 64k - 4k<sup>2</sup> b) As f(x) ≧ 0, Δ ≦ 0 and 16-k > 0 144 + 64k - 4k<sup>2</sup>≦0 and k<16 k<sup>2</sup> - 16k - 36 ≧ 0 and k<16 (k - 18)(k + 2) ≧ 0 and k<16 (k ≦ -2 or k ≧ 18) and k<16 ∴k ≦ -2 see if sth wrong @@~ |
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