②中,y=(sin x)'+cos x=2cos x∈[-2,2];
③中,y=sin x+(cos x)'=0.
答案:B
7. 设曲线y=ex在点(0,1)处的切线与曲线y=1/x(x>0)上点P处的切线垂直,则点P的坐标为 .
答案: (1,1)
8.曲线y=1/x和y=x2在它们交点处的两条切线与x轴所围成的三角形的面积是 .
解析:由{■(y=1/x "," @y=x^2 "," )┤得交点A的坐标为(1,1).
由y=x2,得y'=2x,
∴y=x2在点A(1,1)处的切线方程为y-1=2(x-1),即y=2x-1.
由y=1/x,得y'=-1/x^2 ,
∴y=1/x在点A(1,1)处的切线方程为y-1=-(x-1),即y=-x+2.
如图,S△=S阴=1/2×3/2×1=3/4.
答案:3/4
9.已知曲线C:f(x)=x3-x+2,则经过点P(1,2)的曲线C的切线方程是 .
解析:设经过点P(1,2)的直线与曲线C相切于点(x0,y0),则由f'(x)=3x2-1,得在点(x0,y0)处的切线的斜率k=f'(x0)=3x_0^2-1,所以在点(x0,y0)处的切线的方程为y-y0=(3x_0^2-1)(x-x0).
又因为点(x0,y0)与点P(1,2)均在曲线C和切线上,所以有{■(y_0=x_0^3 "-" x_0+2"," @2"-" y_0="(" 3x_0^2 "-" 1")(" 1"-" x_0 ")," )┤
联立消去y0,得x0-x_0^3=(3x_0^2-1)(1-x0),
解得x0=1或x0=-1/2,
于是k=2或-1/4.
所以所求切线方程为y=2x或y=-1/4x+9/4.
答案:y=2x或y=-1/4x+9/4
10.已知f(x)=cos x,g(x)=x,求当f'(x)+g'(x)≤0时x的值.
分析:先求出f'(x),g'(x),再解三角不等式.
解:∵f(x)=cos x,g(x)=x,
∴f'(x)=(cos x)'=-sin x,
g'(x)=x'=1,
由f'(x)+g'(x)≤0,得-sin x+1≤0,
即sin x≥1,但sin x∈[-1,1],
∴sin x=1.∴x=2kπ+π/2(k∈Z).
11.求过曲线y=f(x)=cos x上一点P(π/3 "," 1/2)且与过这点的切线垂直的直线方程.
解:∵y=f(x)=cos x,∴f'(x)=-sin x.
∴f'(π/3)=-sinπ/3=-√3/2.
∴过点P且与过这点的切线垂直的直线的斜率为2/√3 .∴所求的直线方程为y-1/2=2/√3 (x"-" π/3),
即2x-√3y-2π/3+√3/2=0.
12.已知曲线C:y=2x2,点A(0,-2)及点B(3,a),从点A观察点B,若视线不被曲线C挡住,求实数a的取值范围.