Ritu, I solved the question by actually plotting the function in the three-dimensional plane. You'll see that every point in the plane in a pre-image, so the entire plane is the range.
It's rather a silly method, but it works
Its not exactly in terms of z, ritu. I drew a graph like this one below, where the upper red area is the range and the lower area is the domain.
The dots in the doamin have corresponding coloured dots in the range.
You an see that the pattern is such that for evry value in the domain, the range becomes the entire red area (though I havent shown the other 4 quadrants).
We will say that the range is entire plane if we can get as output anything in the plane by using some input combination in the domain. Lets say I want to generate a vector (a, b) in plane then what must be the value of (x, y) such that f(x, y) = (a, b). Well its easy, solving the system
x - y = a
x + y = b
x = (a+b)/2
y = (b-a)/2
Thus, if we want to generate the output (a, b) in the co-domain there is an input combination in the domain ((a+b)/2, (b-a)/2) and that will do the job for us. Hence, any point in the plane is mapped by some point in the domain. And therefore the range is the entire plane.
for ques no. 39,
it is the case of omission of a relevant variable from our model which leads to model specification error..
so definitely there is a bias component in our final model.
the new model will give the biased estimates of the coefficients of d true model,
and this bias can be either downward or upward...
but it is given that,
a2>0 and K,L >0 and both K and L r complements which means both L and K r moving together.
which confirms that d biasing will be upwards.
(u can refer to D.N.GUJARATI ECONOMETRICS BOOK,chapter no.7)