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x-y plane

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by shashank.ism » Mon Feb 08, 2010 7:02 am
Consider the following two curves in the x-y plane:
y = x3 + x2 + 5
y = x2 + x + 5
Which of the following statements is true for -2 < x < 2?

a) The two curves intersect once
b) The two curves intersect twice
c) The two curves do not intersect
d) The two curves intersect thrice
e) None of the above
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by harsh.champ » Mon Feb 08, 2010 7:05 am
shashank.ism wrote:Consider the following two curves in the x-y plane:
y = x3 + x2 + 5
y = x2 + x + 5
Which of the following statements is true for -2 < x < 2?

a) The two curves intersect once
b) The two curves intersect twice
c) The two curves do not intersect
d) The two curves intersect thrice
e) None of the above
First ,finding the points of intersection:-

x3 + x2 + 5 = x2 + x + 5
=>x3 - x = 0
=>x2(x-1) = 0.
Hence, 2 points. B
Last edited by harsh.champ on Mon Feb 08, 2010 8:04 am, edited 1 time in total.
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by Brent@GMATPrepNow » Mon Feb 08, 2010 7:57 am
harsh.champ wrote:
shashank.ism wrote:Consider the following two curves in the x-y plane:
y = x3 + x2 + 5
y = x2 + x + 5
Which of the following statements is true for -2 < x < 2?

a) The two curves intersect once
b) The two curves intersect twice
c) The two curves do not intersect
d) The two curves intersect thrice
e) None of the above
First ,finding the points of intersection:-

x3 + x2 + 5 = x2 + x + 5
=>x3 - x = 0
=>x2(x-1) = 0.
Hence, 2 points. B
This question is out of scope.
That said, the above solution is good up to the point where we get x^3 - x = 0
Factor: x(x^2 - 1) = 0
Factor: x(x+1)(x-1) = 0
x=-1, 0, 1
Three points of intersection (D)
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by harsh.champ » Mon Feb 08, 2010 8:07 am
Brent Hanneson wrote:
harsh.champ wrote:
shashank.ism wrote:Consider the following two curves in the x-y plane:
y = x3 + x2 + 5
y = x2 + x + 5
Which of the following statements is true for -2 < x < 2?

a) The two curves intersect once
b) The two curves intersect twice
c) The two curves do not intersect
d) The two curves intersect thrice
e) None of the above
First ,finding the points of intersection:-

x3 + x2 + 5 = x2 + x + 5
=>x3 - x = 0
=>x2(x-1) = 0.
Hence, 2 points. B
This question is out of scope.
That said, the above solution is good up to the point where we get x^3 - x = 0
Factor: x(x^2 - 1) = 0
Factor: x(x+1)(x-1) = 0
x=-1, 0, 1
Three points of intersection (D)
Thanks Brent for pointing out my mistake.
I did like this :-
>x3 - x = 0
=>x2(x-1) = 0.

whereas it should be x(x2 - 1 ) = 0,hence ,we get 3 pts. of intersection(0,1,-1)

And thus the ans. will be D.
A very silly mistake though but can make a lot of difference in the exam.

Thanks once again :)
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