7.(北師大版第54頁A組第6題)值域
變式1: 解:作出函數(shù)的圖象,容易發(fā)現(xiàn)在上是增函數(shù),在上是減函數(shù),求出,,,注意到函數(shù)定義不包含,所以函數(shù)值域是.
變式2:解:∵
y= cos2x+sinx=-2sin2x+sinx+1,令t= sinx Î [-1,1],
則y=-2t2+t+1,其中tÎ [-1,1],
∴y Î [-2, ],即原函數(shù)的值域是[-2, ].
變式3: 解:(I) ∵ f (1 + x)
= f (1-x),
∴
- = 1,
又方程
f (x) = x 有等根 Û a x 2 + (b-1) x = 0 有等根,
∴
△= (b-1) 2 = 0 Þ b = 1 Þ a = -,
∴
f (x) = -x 2 + x.
(II) ∵ f (x)
為開口向下的拋物線,對稱軸為 x = 1,
1° 當(dāng) m≥1 時,f (x) 在 [m,n] 上是減函數(shù),
∴ 3m = f (x)min = f (n) = -n 2 + n (*),
3n = f (x)max = f (m) = -m 2 + m,
兩式相減得:3 (m-n) = -(n 2-m 2)
+ (n-m),
∵ 1≤m < n,上式除以 m-n 得:m + n = 8,
代入 (*) 化簡得:n 2-8n + 48 = 0 無實數(shù)解.
2° 當(dāng) n≤1 時,f (x) 在 [m,n] 上是增函數(shù),
∴ 3m = f (x)min = f (m) = -m 2 + m,
3n = f (x)max = f
(n) = -n 2 +
n,
∴ m = -4,n = 0.
3° 當(dāng) m≤1≤n 時,對稱軸 x = 1 Î [m,n],
∴ 3n = f (x)max
= f (1) = Þ n = 與 n≥1 矛盾.
綜合上述知,存在
m = -4、n = 0 滿足條件.