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On the coordinate plane is point (0, 0) closer to point (u,

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On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 May 2014, 23:50, edited 4 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...


Boss don't post OA .. GIVE A CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\)
Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
\(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\)
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Last edited by x2suresh on 29 Aug 2008, 05:27, edited 1 time in total.
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New post 28 Aug 2008, 08:55
:| hmmm . k .. suresh !! thanks...
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Re: wats this mean? [#permalink]

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New post 28 Aug 2008, 10:33
x2suresh wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...


Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\)
Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
\(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\)


Why will S2 not work? What is the OA?
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Re: wats this mean? [#permalink]

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New post 28 Aug 2008, 10:50
leonidas wrote:
x2suresh wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...


Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = \(sqrt {u^2+v^2 }\)
Distance between (0, 0) and (u, v+1) = \(sqrt {u^2+v^2 +2v+1}\)

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
\(sqrt {u^2+v^2 +2v+1}\) is alwasy less than \(sqrt {u^2+v^2 }\)


Why will S2 not work? What is the OA?


Hey Nemo,

2v+1 --> -ve or +ve.
v<0 v=-1/4 2v+1>0
v=-2 2v+1<0
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New post 28 Aug 2008, 12:26
x2suresh wrote:
leonidas wrote:

Why will S2 not work? What is the OA?


Hey Nemo,

2v+1 --> -ve or +ve.
v<0 v=-1/4 2v+1>0
v=-2 2v+1<0


Got it, didn't try a fraction :(
Thanks X2Suresh
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Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|
Eventually, we can write: v>-0.5

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

Answer is obviously A.
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New post 29 Aug 2008, 02:44
thanks , but
how did u get 2 from 1??
could not quite understand...


walker wrote:
Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|......(1)
Eventually, we can write: v>-0.5....(2)

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

Answer is obviously A.
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New post 29 Aug 2008, 03:13
Full solution:

|v|<|v+1|

1) v<-1: -v<-v-1 --> 0<-1 --> always false
2) -1<=v<=0: -v<v+1 --> v>-0.5 --> -0.5<v<=0
3) v>0: --> v<v+1 --> 0<1 always true.

Therefore, inequality is true when v>-0.5

Fast solution:

large negative v: inequality is false
large positive v: inequality is true
switch point v=-0.5 --> v>-0.5
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New post 29 Aug 2008, 08:11
Duh! I went for D but after reading your solutions I realized I forgot to test fractions.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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New post 13 May 2014, 14:15
An even better way to look at this one:

Is \sqrt{u^2+v^2} > \sqrt{u^2+ (v+1)^2}
\sqrt{u^2+v^2} > \sqrt{u^2+v^2+2v+1}

So \sqrt{u^2+v^2} is equal on both sides so the deciding factor is 2v+1
2v+1>0 or v>-1/2 then the right side of the inequality is the greater one

I sufficiently tells us...u^2= -1 - v...As u^2 can never be negative..we have
1. v > -1
2. v = -1..in both the cases we get the same answer..Suff

II. V can be between 0 and -1/2..or <-1/2 ...Insuff

A it is
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Last edited by JusTLucK04 on 14 May 2014, 00:58, edited 1 time in total.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11


On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?

(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.

(2) \(v<0\). Not sufficient.

Answer: A.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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New post 24 Jun 2014, 21:49
How we could have represented this equation on graph ..... and solved by using graph technique

Please help
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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Bunuel wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11


On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?

(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.

(2) \(v<0\). Not sufficient.

Answer: A.



hi Bunuel,

i could not understand the last step.

v=-1-u^2\leq{-1} --> so the answer to the question is NO.

I did not get how did you infer 'v' lesser than -1 from the last equation??
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

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New post 04 Dec 2014, 03:40
arshu27 wrote:
Bunuel wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11


On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?

(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.

(2) \(v<0\). Not sufficient.

Answer: A.



hi Bunuel,

i could not understand the last step.

v=-1-u^2\leq{-1} --> so the answer to the question is NO.

I did not get how did you infer 'v' lesser than -1 from the last equation??


First of all please read Writing Mathematical Formulas on the Forum.

As for your question, we need to find whether \(v>-\frac{1}{2}\).

(1) gives \(v=-1-u^2\). Since u^2 (the square of a number) must be non-negative, then we have that \(v=-1-(nonnegative)\leq{-1}\), therefore v is NOT greater than -1/2, so we have a definite NO answer to the question.

Hope it's clear.
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