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Bunuel
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 25 Apr 2019, 02:11
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C
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65% (02:53) correct 35% (02:39) wrong based on 126 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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C
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 26 Apr 2019, 14:04
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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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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 20 Jul 2019, 19:02
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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 21 Jul 2019, 01:27
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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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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 21 Jul 2019, 01:30
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DarkHorse2019
Dear Moderators Gladiator59, VeritasKarishma generis
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7
Show more

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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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 21 Jul 2019, 02:20
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vaibhav1221
DarkHorse2019
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
Show more

Shouldn't this be
\(y^2 + (x^2)^2 - 2x^2 y = 1\)
\((y - (x^2))^2 = 1\)
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 21 Jul 2019, 05:52
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vaibhav1221
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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
\(y^2 + (x^2)^2 - 2x^2 y = 1\)
\((y - (x^2))^2 = 1\)
Show more

Yes, my bad. The procedure is fine though.
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 21 Jul 2019, 21:19
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Bunuel
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) \(1 + \sqrt{\pi}\)
Show more

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 21 Jul 2019, 22:49
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Bunuel
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) \(1 + \sqrt{\pi}\)
Show more

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

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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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 24 Jul 2019, 06:00
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Kinshook
Bunuel
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
Show more


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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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 24 Jul 2019, 06:09
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Kinshook
Bunuel
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.
Show more


Ok.. I understood this point. you substituted the values in the equation.
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Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap 30 Oct 2024, 11:40
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