设曲线Y=ax^2与Y=lnx相切求a(要有解题步骤)还有一题:当x>1时f(x)=ln(x^2+a^2),x≤1时f(x)=sinb(x-1)如果f(x)在点x=1处可导,求a,b

Question

设曲线Y=ax^2与Y=lnx相切求a(要有解题步骤)
还有一题:当x>1时f(x)=ln(x^2+a^2),x≤1时f(x)=sinb(x-1)如果f(x)在点x=1处可导,求a,b

Answer 1

Y=ax^2 ,y'=2ax ； Y=lnx ,x'=1/x
切点设为(x,y),则 ax^2=lnx ,2ax=1/x
显然x>0 ,2ax=1/x >0
两式相除得 x/2=x*lnx
x=e^(1/2)
x>1时 f'(x)=2x/(x^2+a^2); x1+时）f'(x)
可导一定连续,于是 (x->1-时）f(x) =（x->1+时）f(x) = f(1)
将x=1代入有 2/(1+a^2) =b
ln(1+a^2) =0
得 a=0 ,b=2