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# Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap

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Math Expert
Joined: 02 Sep 2009
Posts: 57279
Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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25 Apr 2019, 02:11
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Difficulty:

55% (hard)

Question Stats:

66% (02:52) correct 34% (02:12) wrong based on 50 sessions

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Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

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Posts: 232
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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26 Apr 2019, 14:04
Equating points into given equation, we get
|a-π|=1
so a can be approx 4.14 or 2.14

|b-π|=1
so b can be approx 4.14 or 2.14

As a and b are distinct, difference a-b= +-2

Hence, |a-b|= 2

Ans C

Posted from my mobile device
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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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20 Jul 2019, 19:02
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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21 Jul 2019, 01:27
Hey,

Put the Value of x=RootPi in the Equation, You'll get the Values of "y" ( Pi+1 and Pi-1).

Since There are only two solutions to the equation, so above values of "y" will be actually the values of "a" and "b". The order does not matter because of the absolute value sign.

Hope it Helps!
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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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21 Jul 2019, 01:30
DarkHorse2019 wrote:
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7

DarkHorse2019, consider the following solution. I hope this will make sense.

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y + (x^2))^2 = 1$$
square rooting both sides
$$y + (x^2) = ±1$$

$$x^2 = y + 1$$ ...(1)
$$x^2 = y - 1$$ ...(2)

in equation (1)
when $$x = \sqrt{\pi}, y = (15/7)$$ (value of a or b)

in equation (2)
when $$x = \sqrt{\pi}, y = (29/7)$$ (value of b or a)

|a-b| =

|15/7 - 29/7| or |29/7 - 15/7| = |-14/7| or |14/7| = 2
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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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21 Jul 2019, 02:20
vaibhav1221 wrote:
DarkHorse2019 wrote:
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7

DarkHorse2019, consider the following solution. I hope this will make sense.

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y + (x^2))^2 = 1$$
square rooting both sides

$$y + (x^2) = ±1$$

$$x^2 = y + 1$$ ...(1)
$$x^2 = y - 1$$ ...(2)

in equation (1)
when $$x = \sqrt{\pi}, y = (15/7)$$ (value of a or b)

in equation (2)
when $$x = \sqrt{\pi}, y = (29/7)$$ (value of b or a)

|a-b| =

|15/7 - 29/7| or |29/7 - 15/7| = |-14/7| or |14/7| = 2

Shouldn't this be
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y - (x^2))^2 = 1$$
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GMAT 1: 670 Q49 V32
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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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21 Jul 2019, 05:52
DarkHorse2019 wrote:
vaibhav1221 wrote:
DarkHorse2019 wrote:
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7

DarkHorse2019, consider the following solution. I hope this will make sense.

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y + (x^2))^2 = 1$$
square rooting both sides

$$y + (x^2) = ±1$$

$$x^2 = y + 1$$ ...(1)
$$x^2 = y - 1$$ ...(2)

in equation (1)
when $$x = \sqrt{\pi}, y = (15/7)$$ (value of a or b)

in equation (2)
when $$x = \sqrt{\pi}, y = (29/7)$$ (value of b or a)

|a-b| =

|15/7 - 29/7| or |29/7 - 15/7| = |-14/7| or |14/7| = 2

Shouldn't this be
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y - (x^2))^2 = 1$$

Yes, my bad. The procedure is fine though.
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Sky is the limit. 800 is the limit.

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Posts: 9554
Location: Pune, India
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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21 Jul 2019, 21:19
2
1
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

Note that for both points, value of x is given to be $$\sqrt{\pi}$$. Plugging this in, let's get the values of y which are the values of a and b.

Graph:
$$y^2 + \pi^2 = 2\pi*y + 1$$

$$y^2 - 2\pi*y + \pi^2 - 1 = 0$$

$$[y - (\pi + 1)][y - (\pi - 1)] = 0$$

Since |x - y| is not 0 (No option is 0),
$$a = (\pi + 1)$$
$$b = (\pi - 1)$$
(or the other way around)

|a - b| = 2
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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21 Jul 2019, 22:49
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2+x^4-2x^2y = 1$$
$$(y-x^2)^2 =1$$
$$|y-x^2| = 1$$

Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph
=>$$|a-\pi|=1$$
& $$|b-\pi|=1$$
Since they are distinct points
Say$$a = \pi+1$$ and $$b =\pi-1$$
|a-b| =2 since a and b are interchangeable

IMO C
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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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24 Jul 2019, 06:00
Kinshook wrote:
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2+x^4-2x^2y = 1$$
$$(y-x^2)^2 =1$$
$$|y-x^2| = 1$$

Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph
=>$$|a-\pi|=1$$
& $$|b-\pi|=1$$
Since they are distinct points
Say$$a = \pi+1$$ and $$b =\pi-1$$
|a-b| =2 since a and b are interchangeable

IMO C

Hi Kinshook

For - $$|a-\pi|=1$$ and $$|b-\pi|=1$$

How did you assume that |a-\pi| to be 1? It can take any value, right ?
Please suggest and help me with this.
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Joined: 08 Dec 2018
Posts: 11
Location: India
Schools: ISB '20, IIMA PGPX"20
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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24 Jul 2019, 06:09
NamGup wrote:
Kinshook wrote:
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2+x^4-2x^2y = 1$$
$$(y-x^2)^2 =1$$
$$|y-x^2| = 1$$

Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph
=>$$|a-\pi|=1$$
& $$|b-\pi|=1$$
Since they are distinct points
Say$$a = \pi+1$$ and $$b =\pi-1$$
|a-b| =2 since a and b are interchangeable

IMO C

Hi Kinshook

For - $$|a-\pi|=1$$ and $$|b-\pi|=1$$

How did you assume that |a-\pi| to be 1? It can take any value, right ?
Please suggest and help me with this.

Ok.. I understood this point. you substituted the values in the equation.
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap   [#permalink] 24 Jul 2019, 06:09
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