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M22-11

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Bunuel
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M22-11 [#permalink] New post   16 Sep 2014, 01:16
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33% (02:11) correct 67% (01:55) wrong based on 123 sessions
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In the coordinate plane, is the point \((0, 0)\) closer to the point \((u, v)\) than to the point \((u, v + 1)\)?


(1) \(v + u^2 = -1\)

(2) \(v < 0\)
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Bunuel
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Re M22-11 [#permalink] New post   16 Sep 2014, 01:16
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Official Solution:


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

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

Essentially, 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 \(\sqrt{(u-0)^2+(v-0)^2} \lt \sqrt{(u-0)^2+(v+1-0)^2}\)?

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

Is \(u^2+v^2 \lt u^2+v^2+2v+1\)?

Is \(v \gt -\frac{1}{2}\)?

(1) \(v + u^2 = -1\).

Rearrange: \(v=-1-u^2\). Now, since \(u^2 \ge 0 \), then \(v=-1-u^2 \le -1\), so the answer to the question is NO. Sufficient.

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


Answer: A
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Re: M22-11 [#permalink] New post   15 Jul 2020, 04:56
I think this is a high-quality question and I agree with explanation.
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Re: M22-11 [#permalink] New post   09 Aug 2020, 20:59
coreyander wrote:
I think this is a high-quality question and I agree with explanation.

can you graphically explain the soln ? both statement tells v is negative , I am not able to visualise the situation in which v+1 cant be nearer to origin than v (as v is negative )
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Re: M22-11 [#permalink] New post   24 Sep 2020, 07:06
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ayushkumar22941 wrote:
coreyander wrote:
I think this is a high-quality question and I agree with explanation.

can you graphically explain the soln ? both statement tells v is negative , I am not able to visualise the situation in which v+1 cant be nearer to origin than v (as v is negative )


Hi,

I tried the same method of approach as you had applied and i too got the answer wrong ,The reason is as follows:-

the cases work if v>-1/2 (i,e if v is +ve then v+1 will be further from the origin and if V is -ve v+1 will be closer to the origin) -----(1)

But an interesting thing happens from at 0<v<=-1/2

if v= -1/4 then v+1 will be farther away from the origin Contratry to what is written in bracket of (1)

hence the solution by Bunnuel is correct


Thanks
hope it helps!
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Bunuel
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Re: M22-11 [#permalink] New post   05 May 2023, 12:13
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M22-11 [#permalink] New post   27 May 2023, 06:16
Hi Bunuel - i solved the problem a different way.

CHeck this out - prompt one says that 'v' has to be a negative number less than or equal to (-1) because 'u' squrared has to be zero or a positive number. so if '1' is added to negative 1 then that in all situations becomes closer to zero (0,0).

the second prompt just says that v is negative. now if v is less than -.5 then we are good. but if v is -.1 then if you add a positive 1 to that number it becomesfurther away from (0,0).

I basically picturised the problem instead of applying the distance formula. is this a good way to atempt this problem?
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Bunuel
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Re: M22-11 [#permalink] New post   27 May 2023, 08:43
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hloonker wrote:
Hi Bunuel - i solved the problem a different way.

CHeck this out - prompt one says that 'v' has to be a negative number less than or equal to (-1) because 'u' squrared has to be zero or a positive number. so if '1' is added to negative 1 then that in all situations becomes closer to zero (0,0).

the second prompt just says that v is negative. now if v is less than -.5 then we are good. but if v is -.1 then if you add a positive 1 to that number it becomesfurther away from (0,0).

I basically picturised the problem instead of applying the distance formula. is this a good way to atempt this problem?


Yes, your reasoning, if I understood it correctly, is right.
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Re: M22-11 [#permalink]
27 May 2023, 08:43
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Bunuel
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