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

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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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New post 25 Apr 2019, 02:11
1
10
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

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

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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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New post 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

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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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New post 20 Jul 2019, 19:02
Dear Moderators Gladiator59, VeritasKarishma generis
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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New post 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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New post 21 Jul 2019, 01:30
DarkHorse2019 wrote:
Dear Moderators Gladiator59, VeritasKarishma generis
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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New post 21 Jul 2019, 02:20
vaibhav1221 wrote:
DarkHorse2019 wrote:
Dear Moderators Gladiator59, VeritasKarishma generis
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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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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New post 21 Jul 2019, 05:52
DarkHorse2019 wrote:
vaibhav1221 wrote:
DarkHorse2019 wrote:
Dear Moderators Gladiator59, VeritasKarishma generis
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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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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New post 21 Jul 2019, 21:19
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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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New post 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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New post 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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Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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New post 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.
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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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