Neferteena wrote:
y>= (x-3)^2 -1
I quadrant -> x +ve, y +ve
Ex: (20,5)
20 >= (5-3)^2-1
20>=3
satisfied. Hence we can conclude that it lies in first quadrant.
II Quadrant -> x -ve, y +ve
Ex: (20, -1)
50>=(-1-3)^2-1
50>= 15
Satisfied.
III Quadrant -> x -ve, y -ve
Ex: (-1,-1)
-1>=(-1-3)^2 -1
-1>=15
Not satisfied. As x is within a square term, it will always be positive, y will remain negative. Hence there will be no points in this quadrant.
IV Quadrant -> x +ve, y -ve
(3,-1)
-1>= (3-3)^2-1
-1>=-1
Satisfied.
Ans: C
Your approach is not always true.
If you plug in (-1,2) in Quadrant II,
the inequality becomes 2>=15, which is not satisfied.
Similarly, if you plug in (2,-1) in Quadrant IV,
the inequality becomes -1>=0, which again is not satisfied
Statistics : Posted by attiyaehsan • on 21 Nov 2019, 01:07 • Replies 7 • Views 6213







