函数f(x)=x2(0<x<1)的图象如图,其在点M(t,f(t))处的切线为l,l与x

函数f(x)x2(0<x<1)的图象如图,其在点M(tf(t))处的切线为llx轴和x1分别交于PQ,点N(1,0),设PQN的面积Sg(t)

(1)g(t)的表达式;

(2)g(t)在区间(mn)上单调递增,求n的最大值;

(3)PQN的面积为b时的点M恰有两个,求b的取值范围.

答案

 (1)设点M(tt2),由f(x)x2(0<x<1),得f′(x)2x

过点M的切线PQ的斜率k2t.

切线PQ的方程为y2txt2.

y0,得x,取x1,得y2tt2

P(0)Q(1,2tt2)

Sg(t)(1)(2tt2)t3t2t.

(2)(1)得,g(t)(t34t24t)

g′(t)(3t28t4)

g′(t)0,解得t1t22()

t(0)时,g′(t)>0g(t)为增函数.

t(1)时,g′(t)<0g(t)为减函数.

g(t)在区间(mn)上单调递增,

n的最大值为.

(3)t时,g(t)有极大值,也就是(0,1)上的最大值为.

g(0)0g(1).

要使PQN的面积为b时点M恰好有两个,

<S<.

b的取值范围为()

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