已知函数f(x)=a•2x2x+2的图象过点(0,2-1).(1)求f(x)的解析式;(2)设P1(x1,y1),P2(x2,y2)为y=f(x)的图象上两个不同点,又点P(xP,yP)满足:OP=1

题目简介

已知函数f(x)=a•2x2x+2的图象过点(0,2-1).(1)求f(x)的解析式;(2)设P1(x1,y1),P2(x2,y2)为y=f(x)的图象上两个不同点,又点P(xP,yP)满足:OP=1

题目详情

已知函数f(x)=
a•2x
2x+
2
的图象过点(0,
2
-1)

(1)求f(x)的解析式;
(2)设P1(x1,y1),P2(x2,y2)为y=f(x)的图象上两个不同点,又点P(xP,yP)满足:
OP
=
1
2
(
OP1
+
OP2
)
,其中O为坐标原点.试问:当xP=
1
2
时,yP是否为定值?若是,求出yP的值,若不是,请说明理由.
题型:解答题难度:中档来源:不详

答案

(1)由题意知class="stub"a
1+
2
=
2
-1

解得a=1,
f(x)=
2x
2x+
2

(2)class="stub"1
2
=xP=class="stub"1
2
(x1+x2)⇒x1+x2=1⇒x2=1-x1
yP=class="stub"1
2
(y1+y2)=class="stub"1
2
(
2x1
2x1+
2
+
2x2
2x2+
2
)=class="stub"1
2
(
2x1
2x1+
2
+
21-x1
21-x1+
2
)

=class="stub"1
2
(
2x1
2x1+
2
+class="stub"2
2+
2
2x1
)=class="stub"1
2
(
2x1
2x1+
2
+
2
2
+2x1
)

=class="stub"1
2
•1=class="stub"1
2

∴yp为定值class="stub"1
2

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