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If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?

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If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 04:05
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Difficulty:

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If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 06:49
2
1 Kudos for an excellent question.
IMO graphical approach can save lot of time in this question.

We just need to figure out whether point of minima of ||x|-0.34|+0.66=y lies within the circle or not, and whether local maxima at x=0 lies within circle or not.
\(x^2+y^2=1\) is a curve of a circle of radius 1.

At x=0, ||x|-0.34|+0.66= 1, which lies on the circle.

Minima of \(||x|-0.34|+0.66=y\) occurs when |x|-0.34=0 or x=+0.34 and -0.34, and y=0.66, lie inside the circle.

Now we are good to draw the curve.

There are 3 points where both curves meet.

D



Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



______
Smth similar I met in my GMAT exam, thus make sure you are familiar with such questions :)
Kudos are always appreciated :tongue_opt2

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If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 06:52
3
Such tasks are easy to solve graphically: solutions to the system of equation are points of intersection of graphs of each equation.

Remember: When a module sign is "layed" on F(x), graphical part of F(x), which lays below the X-axis, is reflected over the X-axis. This operation gives you the graph of |F(x)|.

Thus, the given task is solved with consequent graphics:

Image

As already discussed, to obtain ||x|-0.34| graph, we reflect "negative" part of |x|-0.34 over X-axis. Then shift ||x|-0.34| by 0.66 upwards to obtain ||x|-0.34|+0.66. Than we draw the plot of x^2+y^2=1

Finally, one can see 3 points of intersection -> 3 solutions to the system -> x can take 3 values -> answer is D
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 07:38
This question is a tough one :?
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 08:42
Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



______
Smth similar I met in my GMAT exam, thus make sure you are familiar with such questions :)
Kudos are always appreciated :tongue_opt2


nick1816
Bunuel
What is wrong with the given solution in the image:
here x=0 or 1 0r -1

but i consider case last one in pic (from d)
here i get x=.84 and y =-.52
then also x^2+y^2=1 then y are we nopt considering such cases
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 09:23
There are couple of problems with this approach for this particular question.
1. This approach can take a lot of time. Whenever you see second order equations, go for the graphical approach.

2. You didn't find the domain for x, for each equation.
For example-
||x|-0.34|+0.66=y
y=x-0.34+0.66, when x≥0.34
y=0.34-x+0.66, when \(0.34 ≥x ≥0\)
y=0.34+x+0.66, when 0≥x≥-0.34
y=0.34-x+0.66, when x≤-0.34

Now you can solve each case, and look whether the solution lies in the domain or not.




vanam52923 wrote:
Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



______
Smth similar I met in my GMAT exam, thus make sure you are familiar with such questions :)
Kudos are always appreciated :tongue_opt2


nick1816
Bunuel
What is wrong with the given solution in the image:
here x=0 or 1 0r -1

but i consider case last one in pic (from d)
here i get x=.84 and y =-.52
then also x^2+y^2=1 then y are we nopt considering such cases
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 10:08
nick1816 wrote:
There are couple of problems with this approach for this particular question.
1. This approach can take a lot of time. Whenever you see second order equations, go for the graphical approach.

2. You didn't find the domain for x, for each equation.
For example-
||x|-0.34|+0.66=y
y=x-0.34+0.66, when x≥0.34
y=0.34-x+0.66, when \(0.34 ≥x ≥0\)
y=0.34+x+0.66, when 0≥x≥-0.34
y=0.34-x+0.66, when x≤-0.34

Now you can solve each case, and look whether the solution lies in the domain or not.




vanam52923 wrote:
Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



______
Smth similar I met in my GMAT exam, thus make sure you are familiar with such questions :)
Kudos are always appreciated :tongue_opt2


nick1816
Bunuel
What is wrong with the given solution in the image:
here x=0 or 1 0r -1

but i consider case last one in pic (from d)
here i get x=.84 and y =-.52
then also x^2+y^2=1 then y are we nopt considering such cases

My bad ,thanks a lot :)
But are such questions asked on GMAT?
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 27 Aug 2019, 10:20
nick1816 VeritasKarishma Bunuel chondro48 Is there an algebra way to solve this question?



nick1816 wrote:
1 Kudos for an excellent question.
IMO graphical approach can save lot of time in this question.

We just need to figure out whether point of minima of ||x|-0.34|+0.66=y lies within the circle or not, and whether local maxima at x=0 lies within circle or not.
\(x^2+y^2=1\) is a curve of a circle of radius 1.

At x=0, ||x|-0.34|+0.66= 1, which lies on the circle.

Minima of \(||x|-0.34|+0.66=y\) occurs when |x|-0.34=0 or x=+0.34 and -0.34, and y=0.66, lie inside the circle.

Now we are good to draw the curve.

There are 3 points where both curves meet.

D



Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



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Kudos are always appreciated :tongue_opt2

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If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post Updated on: 28 Aug 2019, 00:39
Kinshook wrote:
nick1816 VeritasKarishma Bunuel chondro48 Is there an algebra way to solve this question?



nick1816 wrote:
1 Kudos for an excellent question.
IMO graphical approach can save lot of time in this question.

We just need to figure out whether point of minima of ||x|-0.34|+0.66=y lies within the circle or not, and whether local maxima at x=0 lies within circle or not.
\(x^2+y^2=1\) is a curve of a circle of radius 1.

At x=0, ||x|-0.34|+0.66= 1, which lies on the circle.

Minima of \(||x|-0.34|+0.66=y\) occurs when |x|-0.34=0 or x=+0.34 and -0.34, and y=0.66, lie inside the circle.

Now we are good to draw the curve.

There are 3 points where both curves meet.

D



Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



Kudos are always appreciated :tongue_opt2


Hi Kinshook nick1816 VeritasKarishma and Bunuel,

Yes, there is. Below is a framework to do so. Correct me, experts, if I make mistake.

Note that using algebra, the solution is laborious and time consuming.

Transform \(x^2=1-y^2\) to |x|=√(1-y^2). Substituting |x| with √(1-y^2) into the second equation, we get:
|√(1-y^2)-0.34|=y-0.66.

If √(1-y^2) > 0.34, then √(1-y^2)-0.34 = y-0.66 --> √(1-y^2)= y-0.34. After squaring both sides and performing tedious calculation, we will find that there are two solution points available when √(1-y^2) > 0.34.

If √(1-y^2) <= 0.34, then 0.34-√(1-y^2) = y-0.66 --> √(1-y^2)=1-y. It is obvious that y=1 satisfies the equation and correspondingly |x|=√(1-y^2)=0. Thus, there is one solution point (0,1) when √(1-y^2) <= 0.34.

Finally, there are 3 values that x can take.

Final answer is (D)

Originally posted by chondro48 on 27 Aug 2019, 17:04.
Last edited by chondro48 on 28 Aug 2019, 00:39, edited 1 time in total.
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 28 Aug 2019, 00:36
Kinshook wrote:
nick1816 VeritasKarishma Bunuel chondro48 Is there an algebra way to solve this question?



nick1816 wrote:
1 Kudos for an excellent question.
IMO graphical approach can save lot of time in this question.

We just need to figure out whether point of minima of ||x|-0.34|+0.66=y lies within the circle or not, and whether local maxima at x=0 lies within circle or not.
\(x^2+y^2=1\) is a curve of a circle of radius 1.

At x=0, ||x|-0.34|+0.66= 1, which lies on the circle.

Minima of \(||x|-0.34|+0.66=y\) occurs when |x|-0.34=0 or x=+0.34 and -0.34, and y=0.66, lie inside the circle.

Now we are good to draw the curve.

There are 3 points where both curves meet.

D



Philipp98 wrote:
If \(x^2+y^2=1\) and \(||x|-0.34|+0.66=y\), than how many values can \(x\) take?

A. 0
B. 1
C. 2
D. 3
E. 4



______
Smth similar I met in my GMAT exam, thus make sure you are familiar with such questions :)
Kudos are always appreciated :tongue_opt2


You can solve it using the usual method of definition of absolute values (as done by others here) but if you are unable to solve it graphically, it is probably a good idea to skip the question. With algebra, it will be far too time consuming and error prone.
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 29 Aug 2019, 11:01
nick1816

The explanation given by you to find out the value of y at the equation's minima is crystal clear. We have arrived at y = 0.66 at minima i.e x = +/- 0.34.

Now, how do we figure if y = 0.66 falls inside the circle?

Ofcourse we can substitute x = 0.34 in the circle equation to find out the value of y. And if the value of y > 0.66, we can deduce that 0.66 is inside the circle. However, this is a time consuming process that requires squaring and square-roots

Is there are short cut to find if 0.66 falls inside the circle?

VeritasKarishma Bunuel Philipp98 - Please help me with this.
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Re: If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 29 Aug 2019, 11:12
1
0.34≃1/3
0.66≃2/3
\((\frac{1}{3})^2+(\frac{2}{3})^2\)= \(\frac{1}{9}+\frac{4}{9}\)=\(\frac{5}{9}\)<1

GMATaspirant641 wrote:
nick1816

The explanation given by you to find out the value of y at the equation's minima is crystal clear. We have arrived at y = 0.66 at minima i.e x = +/- 0.34.

Now, how do we figure if y = 0.66 falls inside the circle?

Ofcourse we can substitute x = 0.34 in the circle equation to find out the value of y. And if the value of y > 0.66, we can deduce that 0.66 is inside the circle. However, this is a time consuming process that requires squaring and square-roots

Is there are short cut to find if 0.66 falls inside the circle?

VeritasKarishma Bunuel Philipp98 - Please help me with this.
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If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?  [#permalink]

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New post 29 Aug 2019, 23:13
GMATaspirant641 wrote:
nick1816

The explanation given by you to find out the value of y at the equation's minima is crystal clear. We have arrived at y = 0.66 at minima i.e x = +/- 0.34.

Now, how do we figure if y = 0.66 falls inside the circle?

Ofcourse we can substitute x = 0.34 in the circle equation to find out the value of y. And if the value of y > 0.66, we can deduce that 0.66 is inside the circle. However, this is a time consuming process that requires squaring and square-roots

Is there are short cut to find if 0.66 falls inside the circle?

VeritasKarishma Bunuel Philipp98 - Please help me with this.


Check the solution given by Philipp98 above. Notice how the graph of ||x| - 0.34| + 0.66 = y is drawn.
Also note that each green line has slope 1 or -1. Since the graph intersects the circle at (0, 1), the line of slope 1 will be inside the circle (45 degrees line). Hence you know that it will intersect the circle in 3 points.

Check out these two posts on how to draw absolute value graphs:

https://www.veritasprep.com/blog/2011/0 ... h-to-mods/
https://www.veritasprep.com/blog/2014/1 ... solutions/
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If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?   [#permalink] 29 Aug 2019, 23:13
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