<p>discuss questions from the math sections. </p>
<p>What were the last 2 grid ins?</p>
<p>Number 18 on the grid-in section was 6. Can’t remember the one before it 
I think there’s 2 48s on that section too</p>
<p>Yes 2 48’s !
Wohoooo.</p>
<p>there were also 2000 in the grid in</p>
<p>Does anyone remember the question for the 2nd 48? The first one was the painting thing but I don’t remember the other one.</p>
<p>Already posted this in the other thread but I’m looking for additional confirmation.
Question 20 of the 25 minute all MC section. It was the one with a shaded region over a parabola and under a line. Did you guys get II and III or just II. Conditions were:
I. xy > 0
II. y > x^2
III. absolute value of x < 2</p>
<p>Seems like it’s II and III.</p>
<p>Also is there an algebraic way to do the question with the two overlapping squares or do we just have to infer from the picture that the shaded region takes up 1/4 the area of the small square.</p>
<p>@MAth112358 The answer to that question is none, not II and III.</p>
<p>@Math112358 I had the same answer. II and III
@Posterguy no. 17 in the grid-in was 2.67</p>
<p>guys, the answer for Question 20 of the 25 minute all MC section. It was the one with a shaded region over a parabola and under a line. I think is was what MUST be true?
I. xy > 0
II. y > x^2
III. lxl < 2
was NONE, here’s why
there are four possible points in the shaded area: (1,1), (-1,1), (2,4), and (-2,4)
- (-1,1) and (-2,4) defies I.
- (2,4) and (-2,4) defies II. because they are EQUAL, NOT greater than or less than 4
- (2,4) and (-2,4) defies III. because l-2l and l2l are EQUAL to 2, NOT greater than or less than 2</p>
<p>Hey guys, was looking over that last question again, here’s what i came up with.
If we assume that the interior shaded region does not contain its boundary, which we can infer from the wording of the question, then II and III are correct; here’s why:</p>
<p>I is obviously incorrect because x can take negative values</p>
<p>II is correct because the region we are looking at rests above y = x^2 which means that all y values in the region are greater than the value of x squared. Remember that the values of x and y are dependent on one another, that is, y = f(x) therefore we can take any point on the curve as (x, f(x)); we cannot just take arbitrary points of y and x and then compare. This condition holds as long as the boundary is not contained in the shaded region, because on the boundary y = x which breaks the condition.
For example, let (x,y) be a point on the curve y = x^2 and let (x,y<em>) be a point in the region bounded by y = x^2 and y = 4 we can take the point x = 0.5. On the boundary, y = x^2 = (0.5)^2 = 0.25. Now this point is on the curve y = x^2 but since the shaded region lies above this curve, the values of y</em> for x = 0.5 are 0.25 < y* < 4.</p>
<p>If the interior does not include the boundary, then x can never reach 2 but is infinitely close to 2 which satisfies condition III.
The answer is II and III unless the boundary is contained in the shaded region in which case the answer is none, but this is far less likely.</p>
<p>Yes I agree with Math112358…and I’m very sure</p>
<p>I had None as my answer since I figured out that I and II were both wrong and the only choice available that didn’t have I or II was “NONE”.</p>
<p>I’m not sure but Math112358, are you trying to say that the curve isn’t part of the shaded area? And what was the value of y where the shading stopped?</p>
<p>I think, if you look it as integrating under a curve, it’s always greater/smaller than or equal to… so wouldn’t that make the curve part of the area?</p>
<p>oh and if x=.25 which would be possible because I’m guessing that the shaded area didnt stop at y=.0625 then y would not be greater than x^2 and that’s how i declared that I and II were false and chose None.</p>
<p>I’m not 100% sure though, but im confident</p>
<p>if x is .25, then x^2 should be .0625, but the curve is not a part of the shaded region, which means when y is .0625, x^2 is less than .0625. so II is correct.</p>
<p>What made you conclude that the curve is not part of the shaded area though?</p>
<p>OKay, so II, y>x^2 , is DEFINITELY WRONG because say you chose the points
(-1,1). x=(-1)^2= +1 AND y=+1. They are EQUAL, not greater than or less than.
Secondly, we can assume that the boundry IS part of the shaded area. Point (x,4.0001) would be OUTSIDE the boundry BUT points (x,3.99999) and (x,4.00) ARE IN THE BOUNDRY.</p>
<p>the question was asking “Which of the following MUST BE true?” right? So, any fallacy would defy any of the three followings.</p>
<p>argh I hate how just one or two stupid/simple mistakes can ruin a score big time :(</p>
<p>Keep in mind that when we find the area under a curve using calculus, we are using a limit to approximate the area. Limits allow us to find solutions which although technically not true, are infinitely close to being true. For example, the difference between x > 2 and x >= 2 can be taken as lim (x->0) 2 - (2 + x).</p>
<p>This is similar to statistics where, if we are looking at a normal distribution, the proportion P(z=x), where x is any number between -1 and 1 inclusive, is taken to be zero because it is so small. For similar reasoning P(z<=x) = P(z<x), because the difference, P(z=x), is zero. The same works for a curve; the boundary of a bounded curve has an infinitely small area and because of this, it doesn’t matter whether we include it or not.</p>
<p>jkjk0202, the point (-1,1) would lie ON the boundary, that is the curve y=x^2. The shaded region for x = -1 would contain the points 1 < y < 4. Therefore, y is greater than x^2</p>
<p>Areas under curves don’t have to be found using limits and approximation you can use integrals and they include the points on the curve…</p>
<p>I just want to know if I got it correct. This bloody question with a couple others could ruin my scores haha :(</p>
<p>so what is the answer then?</p>