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STEP Siklos Booklet Question 13

Hi there,

I am stuck with Question 13 in the siklos booklet. I have drawn out the diagrams correctly but fail to see (for part A) why the required region is to the left and right of the curve. Also (for part B) why the required region is to the north and south of the curve.

The question is:
Show that
x2 −y2 +x+3y−2=(x−y+2)(x+y−1)
and hence, or otherwise, indicate by means of a sketch the region of the x-y plane for which
x2 − y2 + x + 3y > 2. Sketch also the region of the x-y plane for which
x2 − 4y2 + 3x − 2y < −2.
Give the coordinates of a point for which both inequalities are satisfied or explain why no such point
exists.


Any help would be greatly appreciated.
Reply 1
Avatar for Smaug123
Smaug123
15
Original post by kelsey.phillips
Hi there,

I am stuck with Question 13 in the siklos booklet. I have drawn out the diagrams correctly but fail to see (for part A) why the required region is to the left and right of the curve. Also (for part B) why the required region is to the north and south of the curve.

So is your question that you have two possible regions as your answer and you're not sure which is the correct one? (You divided the plane into two chunks, but you're not sure which chunk is the answer.)

If so, the easiest way is just to grab a point which is in one chunk but not the other, and see whether it satisfies the inequality or not. A point on an axis is usually easiest - the origin is doubly easiest!
Reply 2
Avatar for BabyMaths
BabyMaths
Original post by kelsey.phillips
Hi there,

I am stuck with Question 13 in the siklos booklet. I have drawn out the diagrams correctly but fail to see (for part A) why the required region is to the left and right of the curve. Also (for part B) why the required region is to the north and south of the curve.

The question is:
Show that
x2 −y2 +x+3y−2=(x−y+2)(x+y−1)
and hence, or otherwise, indicate by means of a sketch the region of the x-y plane for which
x2 − y2 + x + 3y > 2. Sketch also the region of the x-y plane for which
x2 − 4y2 + 3x − 2y < −2.
Give the coordinates of a point for which both inequalities are satisfied or explain why no such point
exists.


Any help would be greatly appreciated.


For a point (x,y) below the line x-y+2=0 we have x-y+2>0. If the point is also above x+y-1=0 then we also have x+y-1>0 and so (x-y+2)(x+y-1)>0.
Reply 3
Avatar for ztibor
ztibor
10
Original post by kelsey.phillips
Hi there,

I am stuck with Question 13 in the siklos booklet. I have drawn out the diagrams correctly but fail to see (for part A) why the required region is to the left and right of the curve. Also (for part B) why the required region is to the north and south of the curve.

The question is:
Show that
x2 −y2 +x+3y−2=(x−y+2)(x+y−1)
and hence, or otherwise, indicate by means of a sketch the region of the x-y plane for which
x2 − y2 + x + 3y > 2. Sketch also the region of the x-y plane for which
x2 − 4y2 + 3x − 2y < −2.
Give the coordinates of a point for which both inequalities are satisfied or explain why no such point
exists.


Any help would be greatly appreciated.


For A
the LHS
(x−y)(x+y)+2x+2y−x+y−2=(x−y)(x+y)+2(x+y)−(x−y+2)=(x+y)(x−y+2)−(x−y+2)=(x−y+2)(x+y−1)(x-y)(x+y)+2x+2y-x+y-2=(x-y)(x+y)+2(x+y)-(x-y+2)=(x+y)(x-y+2)-(x-y+2)=(x-y+2)(x+y-1)

For B
x2−y2+x+3y>2→(x−12)2−14−(y−32)2+94>2→\displaystyle x^2-y^2+x+3y>2 \rightarrow \left (x-\frac{1}{2}\right )^2 -\frac{1}{4}-\left (y-\frac{3}{2}\right )^2+\frac{9}{4}>2\rightarrow
→∣x−12∣>∣y−32∣\displaystyle \rightarrow \left |x-\frac{1}{2}\right |>\left |y-\frac{3}{2}\right|

There is four regions here
1. x>=1/2 and y>=3/2 => x+1>y that is the region under the y=x+1 line
2. x<1/2 and y>3/2 => -x+2>y

....
and so on

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