高三测试-函数及导数测试题-2013-8-1(6)
时间:2025-02-22
时间:2025-02-22
对任意向量a=(x1,y1)∈V,b=(x2,y2)∈V,以及任意λ∈R,均有f[λa+(1-λ)b]=λf(a)+(1-λ)f(b),则称映射f具有性质P.
现给出如下映射:
①f1:V→R,f1(m)=x-y,m=(x,y)∈V; ②f2:V→R,f2(m)=x2+y,m=(x,y)∈V; ③f3:V→R,f3(m)=x+y+1,m=(x,y)∈V.
其中,具有性质P的映射的序号为________.(写出所有具有性质P的映射的序号)
解析:a=(x1,y1),b=(x2,y2).
f1[λa+(1-λ)b]=f1[λx1+(1-λ)x2,λy1+(1-λ)y2]=λx1+(1-λ)x2-λy1
-(1-λ)y2.
λf1(a)+(1-λ)f1(b) =λ(x1-y1)+(1-λ)(x2-y2) =λx1-λy1+(1-λ)x2-(1-λ)y2 =λx1+(1-λ)x2-λy1-(1-λ)y2. ∴f1具有性质P
f2[λa+(1-λ)b]=f2[λx1+(1-λ)x2,λy1+(1-λ)y2]=[λx1+(1-λ)x2]2+λy1+(1-λ)y2
222λf2(a)+(1-λ)f2(b)=λ(x21+y1)+(1-λ)(x2+y2)=λx1+(1-λ)x2+λy1+(1
-λ)y2
≠f2[λa+(1-λ)b] ∴f2不具有性质P
f3[λa+(1-λ)b]=λx1+(1-λ)x2+λy1+(1-λ)y2+λf3(a)+(1-λ)f3(b) =λ(x1+y1+1)+(1-λ)(x2+y2+1) =λx1+(1-λ)x2+λy1+(1-λ)y2+1
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