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21 Nov 2019, 01:07
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
51% (02:15) correct
49%
(02:07)
wrong
based on 183
sessions
History
Date
Time
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Not Attempted Yet
Which quadrant, if any, contains no point (x, y) that satisfies the inequality \(y ≥ (x - 3)^2 - 1\)?
(A) I
(B) II
(C) III
(D) IV
(E) All quadrants contain at least one point that satisfies the given inequality.
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21 Nov 2019, 11:14
Kudos
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Bunuel
Explanation: y ≥ (x - 3)^2 - 1
Solving above equation, we get y ≥ (x-2)(x-4)
So, Putting y=0, We get two values of x=2,4
by putting x=0, we get y ≥ 8
To find out which area is covered by the graph put the Coordinate (0,0) in the original question y ≥ (x - 3)^2 - 1
We get: 0≥8 Which is False.
So (0,0) does not lie in the area covered by the graph, Therefore the equation covers the area above the line.
Thus 4th Quadrant does not contain any point that satisfies the inequality. ( Rest 3 Quadrants will have a few points that would satisfy the inequality).
Answer is D.
Hoping other will reply, so we can have good discussion over this question
Please provide kudos, if you find my explanation good enough 
24 Nov 2019, 16:40
Kudos
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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
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
