设二元函数z=(x2+y2)earctan(y/x)，则函数f(k)=lim(x,y)→(0,0)2z/xy(x，y)沿直线y=

设二元函数z=(x²+y²)e^arctan(y/x)，则函数f(k)=lim_{(x,y)→(0,0)}²z/xy(x，y)沿直线y=kx(-∞﹤k﹤+∞)的最大值为()
A、1
B、-e^π/2
C、-(π/2)
【正确答案】：A
【题目解析】：由∂z/∂x=2xe^arctan(y/x)+(x²+y²)e^arctan(y/x)[-(y/x²)][1+(y/x)²]=(2x-y)e^arctan(y/x)得∂²z/∂x∂y=-e^arctan(y/x)+(2x-y)e^arctan(y/x)+1/x/[1+(y/x)²]=[(x²-xy-y²)/(x²+y²)]e^arctan(y/x)所以，f(k)=(1-k-k²)/(1+k²)e^arctank(-∞﹤k﹤+∞)． 由f′(k)={(1+k²)(-1-2k)-(1-k-k²)•2k/(1+k²)²+(1-k-k²)/(1+k²)•[1/(1+k²)]}e^arctank=[-5k/(1+k²)²]e^arctank{﹥0,k﹤0, =0,k=0, ﹤0,k﹥0.于是f(0)=1是f(k)的最大值。