QUESTION 2Solve the problem.Let f(x,y) = In (x² + y2). Which one of the following curves is a simple closed path in the domain of the function f? Hint: Think about when the vector field isdefined for all values of (x,y) inside the region and also on the boundary.The ellipse +24= 1The unit circle (x - 5)² + y² = 1The unit circle x² + y² = 1The rectangle with vertices at (-3, 2), (3, 2), (3, -2), and (-3, -2)

A function in two variables is given as fx, y = lnx2+y2. Our aim is to find the correct option for the given curve, which is a simple closed path in the function's domain.The given function is fx, y = lnx2+y2. This function is defined for all (x, y) points except where x2+y2=0. This implies this function f(x, y) is not defined at (0, 0) as ln(0) is not defined. (EXPLANATION FOR CORRECT ONE) Consider the unit circle x-52+y2=1. As this is a unit circle having a center at point (5, 0) and radius 1, the point (0, 0) will never lie inside or on the boundary of the unit circle. Thus, it can be concluded that the function is defined at all points of the curve x-52+y2=1 inside and on the boundary of the curve, and this is a simple closed path in the domain of the function. Thus, the correct option is "The unit circle x-52+y2=1"(EXPLANATION FOR INCORRECT) Consider the ellipse x2a2+y2b2=1. As it is known that point (0, 0) lies inside the region of the given ellipse, the function will not be defined for all the points inside the ellipse. Thus, it is not the simple closed path for the function's domain. For the unit circle x2+y2=1, the point (0, 0) lies inside the curve, this is the incorrect option for the given function. Similarly, if one draws the rectangle having the vertices (-3, 2), (3, 2), (3, -2), and (-3, -2), it can be seen that point (0, 0) will lie inside the region on the rectangle, thus, it not the simple closed path where the function is defined. Thus, all the options are incorrect except "The unit circle x-52+y2=1".