Consider the 1st oder equation.?

(i) y = Q(x) is increasing if dy/dx > 0.
Thus, we need (2x+2y) / (1+x^2+y^2) > 0.
Since the denominator is always positive, this reduces to x + y > 0.
Hence, Q is increasing when y > -x.

Similarly, Q is decreasing (dy/dx < 0) when y < -x.

(ii) Q'(x) = dy/dx = (2x+2y) / (1+x^2+y^2).

Since we want to know when Q'(x) = 1, we solve
(2x+2y) / (1+x^2+y^2) = 1
==> 1 + x^2 + y^2 = 2x + 2y
==> (x^2 - 2x + 1) + (y^2 - 2y) = 0
==> (x - 1)^2 + (y - 1)^2 = 1, by completing the square.

So, Q'(x) = 1 at all points on the unit circle with center (1, 1).

(iii) When (x,y) = (1, 1), note that y > -x.
By part (i), this implies that Q(x) is increasing at this point.
(Alternately, plug (1,1) into the DE to find that dy/dx = 4/3 > 0.)

Moreover, since Q is increasing at x = 1, for any x* close to and > 1, Q(x*) > Q(1) = 1.
This new point (x*, Q(x*)) is also in the region y > -x; thus Q is also increasing at x*.
Repeating this argument repeatedly, we see that Q*(x) ≥ 1 for all x ≥ 1.

(Are you sure you also need x < 1 for this last question? That seems much trickier...)

I hope this helps!
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(i) Q is an increasing function where the gradient is positive, i.e.
(2x + 2y)/(1 + x^2 + y^2) > 0 ----> 2x + 2y > 0
[since denominator must be positive for all real x, y]
y > -x

You can do decreasing in the same way.

(ii) Q'(x) = 1 ----> (2x + 2y)/(1 + x^2 + y^2) = 1
2x + 2y = 1 + x^2 + y^2
0 = 1 + x^2 - 2x + y^2 - 2y
1 = 1 + x^2 - 2x + 1 + y^2 - 2y
1 = (x - 1)^2 + (y - 1)^2

Do you recognise what this is?