wuhanmin
幼苗
共回答了21个问题采纳率:81% 举报
y = x - k√(x^2 - 1),设 √(x^2 - 1) = t (t ≥ 0),x = √(t^2 + 1)
则 y = √(t^2 + 1) - kt (t≥0),令 t = tanu ( 0≤u<Pi/2),则:
y = secu - ktanu = 1/cosu - k(sinu/cosu) = (1-ksinu)/cosu
∴ ksinu + y cosu = 1
∴ √(k^2 + y^2)sin(u+φ) = 1,tanφ = y/k
∴ sin(u+φ) = 1/√(k^2 + y^2) ≤ 1
∴ y ≥ √(1 - k^2 )
补充:x →+∞,√(x^2 - 1) → x,y → (1-k)x → +∞ (1-k为正数定值),
可知y无上界.
∴ y的值域是 [√(1 - k^2 ),+∞]
1年前
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