Q1·Section I (General (compulsory))·1 of 145Along the yyy-axis f=±1f=\pm1f=±1: no partials, no continuity, no option trueLet f(x,y)=x2+yx2+y2f(x,y)=\dfrac{x^2+y}{\sqrt{x^2+y^2}}f(x,y)=x2+y2x2+y for (x,y)≠(0,0)(x,y)\ne(0,0)(x,y)=(0,0) and f(0,0)=0f(0,0)=0f(0,0)=0. Then: OptionsA∂f∂x\dfrac{\partial f}{\partial x}∂x∂f and ∂f∂y\dfrac{\partial f}{\partial y}∂y∂f exist at (0,0)(0,0)(0,0)Bf(x,y)f(x,y)f(x,y) is continuous at (0,0)(0,0)(0,0)Cf(x,y)f(x,y)f(x,y) is differentiable at (0,0)(0,0)(0,0)Df(x,y)f(x,y)f(x,y) is continuously differentiable at (0,0)(0,0)(0,0)Show answer & discussionShow solution Report a problem with this questionAll questionsQ2 →Keyboard: A–D answer · S solution · ←/→ previous/next