解答:解:①中,f′(x)=g′(x)+2x.
∵y=g(x)在點(diǎn)(1,g(1))處的切線方程為y=2x+1,
∴g′(1)=2,∴f′(1)=g′(1)+2×1=2+2=4,
∴曲線y=f(x)在點(diǎn)(1,f(1))處切線斜率為4,
故①錯(cuò)誤.
②中,不等式(a-3)x
2<(4a-2)x即(x
2-4x)a-3x
2+2x<0,
令g(a)=(x
2-4x)a-3x
2+2x,a∈(0,1)
由題意可得g(a)<0在a∈(0,1)恒成立,結(jié)合一次函數(shù)的單調(diào)性可得
,即
,解不等式組可得x≤-1或x≥
,
∴x的取值范圍是
(-∞,-1]∪[,+∞),
故②正確;
③中,∵變量X與Y相對(duì)應(yīng)的一組數(shù)據(jù)為(10,1),(11.3,2),(11.8,3),(12.5,4),(13,5),
=
11.72,
==3,
∴這組數(shù)據(jù)的相關(guān)系數(shù)是r=
=0.3755,
變量U與V相對(duì)應(yīng)的一組數(shù)據(jù)為 (10,5),(11.3,4),(11.8,3),(12.5,2),(13,1)
==3,
∴這組數(shù)據(jù)的相關(guān)系數(shù)是-0.3755,
∴第一組數(shù)據(jù)的相關(guān)系數(shù)大于零,第二組數(shù)據(jù)的相關(guān)系數(shù)小于零,即r
2<0<r
1,
故③正確;
④中,由對(duì)照數(shù)據(jù),計(jì)算得
4 |
|
i=1 |
xi2=86,
=
=4.5,
==3.5,
4 |
|
i=1 |
xiyi=66.5,
4=63,
42=81,
∴求得回歸方程的系數(shù)為b=0.7,a=0.35,
∴所求線性回歸方程為y=0.7x+0.35,
故④錯(cuò)誤;
故選C.