In the xy-plane shown, the shaded region consists of all points that

Given: In the xy-plane shown, the shaded region consists of all points that lie above the graph of y=x^2 - 4x and below the the x-axis.

Asked: Does the point (a,b) (not shown) lie in the shaded region if b<0?

(1) 0 < a < 4
Value of b is unknown
NOT SUFFICIENT

(2) a^2 - 4a < b
0 > b > a^2 - 4a
Since b is inside the bound of 0 and a^2 - 4a
SUFFICIENT

IMO B

An excellent official question by GMAT.
We have y=x^2-4x

To find : Does point (a,b) lie in shaded region, given b<0.
statement 1 : 0<x<4
Consider a square having vertices at (0,0),(0,-4),(4,0),(4,-4)(Just for better visualization)
Now if take point (1,1), it will be under the shaded region.
But point (1,-4) will be outside the shaded region and under sqaure.
I choose (1,-4) as -4 can be the min value of b.
Thus statement 1 is not sufficient.

Statement 2: a^2-4a<b
There is general rule for the curve I am quoting below, if we need to find whether (a,b) lies inside or outside or on the curve.
For any curve y=x^2-4x ,
if b>a^2-4a (point (a,b) will lie above the curve)
if b=a^2-4a (point(a,b) will lie on the curve)
if b<a^2-4a (point (a,b) will lie below the curve)

Now as per question we know that b<0
and b>a^2-4a, hence point(a,b) will lie inside shaded region
Statement 2 is sufficient.

Hope it will help

In the xy-plane shown, the shaded region consists of all points that lie above the graph of y = x^2– 4x and below the x-axis. Does the point (a,b) (not shown) lie in the shaded region if b < 0 ?

(1) 0 < a < 4

We don't know what b is. We know b < 0 from the question stem, but we can can a yes and no answer.

if (1,-2) then
\(1^2 - 4(1) < -2\); YES

if (1,-4) then
\(1^2 - 4(1) > -2;\) NO

INSUFFICIENT.

(2) \(a^2 – 4a < b\)

Since \(b > a^2 - 4a\), b will be located above the graph. And since \(b < 0\), point (a,b) MUST be in the shaded region. SUFFICIENT.

Answer is B.

I have my reservations on statement 2. I am sharing my thought process, not sure what exactly I am missing here:-

Statement (2): a^2 - 4a < b

Now b <0, therefore the above inequality translates as:-

a^2 - 4a < 0
i.e. a*(a-4) < 0
i.e. a>0 and a<4 OR a<0 and a>4 (this case is not possible, a cannot be negative and positive together)
i.e. 0<a<4
this is exactly the same as statement (1). In this case shouldn't the answer be E?

Regards

Hi Abheek - One thing you may be missing is that by changing the inequality from a^-4a<b to a^2-4a < 0 --> you have pretty much moved from a subset of values to a super set of possibilities. A was insufficient because there was no clear answer in this super set where all we know is that 0<a<4 - we were getting both YES to question and NO to question. However statement 2, is actually a subset of those values where b (value of y) is greater than a^2-4a (the value of y on the parabola given when x=a). Within this subset the question answers "YES" only to the asked question - i.e. Yes b lies in the shaded region.

Does this help. I am working on a video to share my approach on this problem.

Please see below my approach to this problem. The crux of the solution is on the first three inferences that are outlined in the video. I hope this is helpful.

chetan2
I would be so appreciative if you could kindly explain "since the equation b=a2−4ab=a2−4a is that of the the line..
a^2-4a< b will be inside the parabola and a^2-4a>b will be outside it.." this further. I am not sure how you can draw the conclusion that it is inside or outside of the line.

Hey woohoo921,
Thanks for posting your query.


Let me help you understand how Chetan2u wrote the relationships he did. I will go in a slightly different way and hope that this makes things perfectly clear to you.


UNDERSTAND THE REGION:
First, try to understand the shaded region completely. What kind of points make up this region?

Observe that this region consists of all points that lie above the parabola but below the x-axis. To understand how the coordinates of these points will be, we will analyze each part of the definition of the region one by one:

• Above the parabola
  
  • The equation of the parabola is \(y = x^2 – 4x\). All points that lie on the parabola will exactly satisfy this equation. For example, some points on the parabola are: (x, y) = (0, 0), (1, -3), and (2, -4).
  • Now, look at (0, 0) carefully. We know that this point lies on the parabola.
    
    • Where then would lie (0, 1)? You would say that since this point is exactly 1 unit above (0, 0), it will lie above the parabola. And that is correct.
    • How about (0, 2)? Same story again! This is also above (0, 0) and hence, lies in the region above the parabola.
    • In fact, every point (0, y), where y > 0 [where 0 is the y-coordinate of (0, 0)], will lie above (0, 0) and hence, above the parabola.
  • Similarly, keep in mind the point (1, -3) that lies on the parabola. Then, all points (1, y), for which y > -3 [where -3 is the y-coordinate of (1, -3)], will lie above (1, -3) and hence, above the parabola.
  
  
  Essentially, all points that have a y-coordinate greater than the y value you get using the equation of the parabola (\(y = x^2 – 4x\)) will lie above the parabola.
  Hence, for all these points, \(y > x^2 – 4x\). -------------(1)
  
  
• Below the x-axis
  This one is easier to understand.
  
  • The equation of the x-axis is y = 0.
  • And all points that lie below the x-axis have y smaller than 0.
  In other words, all points that lie below the x-axis satisfy y < 0. ---------(2)

To conclude, we will combine (1) and (2). So, all points (x, y) that lie in the shaded region satisfy \(x^2 – 4x < y < 0\). -------(3)


CONNECTION WITH (a, b):
Now, let’s tie up all we learned above with the main question. -

• From (3), if a point (x, y) lies in the shaded region, it satisfies \(x^2 – 4x < y < 0\).
• So, if (a, b) has to lie in the shaded region, it must satisfy \(a^2 – 4a < b < 0\).

The question already tells us that b < 0, so all we need to check is whether \(a^2 – 4a < b\).


And that is precisely what was confusing you!


I hope this makes everything crystal clear. 😊


Best,
Shweta
Quant Product Creator, e-GMAT

Hi, I am unable to understand this- 'the moment a2−4a>0, a>5 or a<0 on x axis, and this point will be outside the parabola..'
Can you please elaborate?
Looking forward to hearing from you. Thanks!

First go through these three posts:
https://www.gmatclub.com/forum/veritas- ... he-graphs/
https://www.gmatclub.com/forum/veritas- ... s-part-ii/
https://www.gmatclub.com/forum/veritas- ... -part-iii/

They explain the concept of how the xy co-ordinate plane is divided into two parts by a line. The parabola does the same. It divides the plane into two regions:
y > x^2 - 4x
y < x^2 - 4x

Stmnt 2 gives us that (a, b) lies in the region y > x^2 - 4x which is the region inside the parabola (if you want to verify, check (2, 1). It satisfies y > x^2 - 4x.
Since the question stem tells us that b < 0, so we are looking at a point below the x axis. The region inside the parabola below x axis is the shaded region. So the point must lie on it.[/quote]

Hi VeritasKarishma!! I understand that point "a" lies inside the parabola. But how do we ensure that "b" does too? it just says b<0. it could lie anywhere below the x-axis. how do we assume that it lies inside the parabola only below the x-axis?

Also, we derive this from st 2 "0<a<4", but even statement one says this!! So how come statement 1 isn't sufficiency then (I know i must be missing something very silly)[/quote]

Note that (a, b) specifies a single point (a point whose x co-ordinate is 'a' and y co-ordinate is 'b'). So (2, 6) specifies a single point. Only a or only b can't specify a point.
Now the entire plane is divided into two parts - one lying inside the parabola and the other outside.
The one lying inside the parabola is x^2 - 4x < y and the one outside is x^2 - 4x > y. The actual parabola is x^2 - 4x = y.

The shaded region is inside the parabola so all points (a, b) inside it satisfy a^2 - 4a < b (the inequality given by statement 2)

When b is negative (the y coordinate of the point is negative), this will be the shaded region.

So (a, b) are points such as (2, -1).[/quote]

Hi,
I am unable to understand how the 'outside the parabola' and inside the parabola is being derived.
Would it always be the case for every equation that when-
1. expression with y> expression with x: outside the parabola and
2. expression with y< expression with x: inside the parabola?
The parabola is in the 4th quadrant in this qs. What if it were in the first quadrant? Would the greater than/ less than y and inside/ outside parabola relation still be the same?

This is how I solved this question:

Axis of symmetry of the given graph passes through the point = \(\frac{0+4}{2}\) = 2

In order to find the y-coordinate of the vertex of the graph we will put the value of the point of axis of symmetry into the given equation:

\(y = x^2– 4x\)
\(y = 2^2– 4*2\)
\(y = -4\)

(1) 0 < a < 4

This is not sufficient as we do not know exact value of b.

(2) \(a^2 – 4a < b\)

\(a(a – 4) < b\)

Since b is negative we can take the given inequality as: \(a(a – 4) < 0\)

For this to be true we will have 2 cases:

(a) a<0 & a-4>0 => a<0 & a>4 (Not a valid case)

(b) a>0 & a-4<0 => a>0 & a<4 => 0<a<4

We can now take the extreme cases: a=0.1 and a=3

i) a=0.1

\(b>0.1(0.1 – 4)\)
\(b>-0.39\)

so, both a=0.1 & -0.39<b<0 falls within the shaded region (y=-4).

ii) a=3

\(b>3(3 – 4)\)
\(b>-3\)

so, both a=3 & -3<b<0 falls within the shaded region (y=-4). Therefore, sufficient.

chetan2u KarishmaB egmat Could you please confirm whether my approach is correct?

I am not sure of the logic of this approach. The value of a is not independent in statement 2. It depends on the value of b. So say if b = -3, then the range of values of a will be different, not 0 to 4. Of course, when we try numbers, they will satisfy because the statement is enough to answer. Can we establish that the statement will also work for all numbers based on 2 cases? I would worry about it.

Here is how you can be sure of the logic. The Veritas links I gave above don't work anymore so I will brief my thought process here.

The parabola is y = x^2 - 4x

It divides the co-ordinate plane into two regions - one inside the parabola and one outside.
For one of these regions, y > x^2 - 4x and for the other y < x^2 - 4x. Which is which, let's find out.

The point (2, -1) obviously lies inside the parabola within our shaded region. What relation of x and y holds for it?
\(x^2 - 4x = 2^2 - 8 = -4\)
\(y = -1\)
\(y > x^2 - 4x\) here
So every point inside the parabola satisfies y > x^2 - 4x and every point outside it satisfies y < x^2 - 4x.

Now, if we have a point (a, b) such that b > a^2 - 4a, it must lie inside the parabola and within the shaded region (because b < 0).

Answer (B)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.