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1 < = x < =0, -1 < = y < = 1, z > = -2, which

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1 < = x < =0, -1 < = y < = 1, z > = -2, which [#permalink] New post 20 Feb 2004, 04:51
-1 < = x < =0, -1 < = y < = 1, z > = -2, which of the following could have the largest value?

a. x^2 + y ^2
b. x^2 - z^2
c y^2 - z ^2
d. x^2 + y^2 -z^2
e. 2 *( x-z)*(x+z) + 2*y^2
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 [#permalink] New post 20 Feb 2004, 11:24
Another vote for A.
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 [#permalink] New post 20 Feb 2004, 14:43
agree with kpadma. Answer should be E.
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 [#permalink] New post 25 Feb 2004, 09:16
Another vote for E. The greatest possible value for the expression in E is +infinity.

The greatest value for A is just 2.

Batliwala,

I have seen that you are positing a lot of good question these days. Most of them may not need to be confirmed with official naswer. However, there are few questions where official answer is required.

It would be great if you could publish official answer where there is a diagreement among the members.
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 [#permalink] New post 25 Feb 2004, 12:50
Please note that 2*(X-Z)*(X+Z)+2Y^2=2X^2-2Z^2+2Y^2 so no matter if Z>= -2 or <=-2 it is always negative, as with the example of gmatblast it should be -inf.Also think it should be A)
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 [#permalink] New post 26 Feb 2004, 05:53
correct ans is E.

substituting values for x, y, and z and limiting conditions E. has the greatest value =4.
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 [#permalink] New post 26 Feb 2004, 07:41
batliwala wrote:
correct ans is E.

substituting values for x, y, and z and limiting conditions E. has the greatest value =4.


Well I agree with the answer E but I think the value of the expression in E will be +inf.

Here is how I thought of it:

2 *( x-z)*(x+z) + 2*y^2

To get the maximum value of this expression, we need to select the values as follws:

Maximum possible value of (x-z)
Maximum possible value of (x+z)
Maximum possible value of y


To get the Maximum possible value of (x-z):

Take the maximum possible value of x = 0
Take the minimum possible value of z = -2

Hence, maximum possible value of (x - z) = [0-(-2)] = 2

To get the Maximum possible value of (x+z):

Take the maximum possible value of x = 0
Take the maximum possible value of z = +inf

Hence, the maximum possible value of (x + z) = +inf

To get the Maximum possible value of y:

Take the maximum possible value of y = 1

So the maximum possible value of the given expression = 2 (2) (+inf) + 2(1) = +inf.

What am I missing?




Maximum possible value for x = 0
Minimum possible value for z = +inf
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 [#permalink] New post 28 Feb 2004, 15:03
gmatblast wrote:
batliwala wrote:
correct ans is E.

substituting values for x, y, and z and limiting conditions E. has the greatest value =4.


Well I agree with the answer E but I think the value of the expression in E will be +inf.

Here is how I thought of it:

2 *( x-z)*(x+z) + 2*y^2

To get the maximum value of this expression, we need to select the values as follws:

Maximum possible value of (x-z)
Maximum possible value of (x+z)
Maximum possible value of y


To get the Maximum possible value of (x-z):

Take the maximum possible value of x = 0
Take the minimum possible value of z = -2

Hence, maximum possible value of (x - z) = [0-(-2)] = 2

To get the Maximum possible value of (x+z):

Take the maximum possible value of x = 0
Take the maximum possible value of z = +inf

Hence, the maximum possible value of (x + z) = +inf

To get the Maximum possible value of y:

Take the maximum possible value of y = 1

So the maximum possible value of the given expression = 2 (2) (+inf) + 2(1) = +inf.

What am I missing?




Maximum possible value for x = 0
Minimum possible value for z = +inf


You can't have z equal -2 and +inf in the same equation.
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 [#permalink] New post 28 Feb 2004, 15:59
kpadma wrote:
Ans: E


I'm changing my answer to A.

A simple question and I scrowed up!!
God, Show mercy on my Soul in the GMAT exam !!!

-1 < = x < =0, ==> 0 =< X^2 =< 1
-1 < = y < = 1, ==> 0 =< Y^2 =< 1
z > = -2 ==> 4 =< Z^2 =< +INF

a. x^2 + y ^2 ==> 1 + 1 = 2
b. x^2 - z^2 ==> 1 - 4 = -3
c y^2 - z ^2 ==> 1 - 4 = -3
d. x^2 + y^2 -z^2 ==> 1 + 1 - 4 = -2
e. 2 *( x-z)*(x+z) + 2*y^2
2 ( X^2 + Y^2 - Z^2) ==> 2( 1 + 1 - 4) = -4

So, the answer is A.

Dear Batliwala, the unoffical ETS question banker,

What is the official answer?
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 [#permalink] New post 28 Feb 2004, 18:26
YES. Kpadma!! I think the answer should be A.

Such a simple problem and got screwed up. Thanks for the follow up.
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 [#permalink] New post 02 Mar 2004, 06:16
Here's the unofficial answer.
The correct choice is E.
x^2 < = 1, Y^2 <=1, Z^2 >=0

Evaluating the maximum value of the answer choices:
a. 1+1 = 2
b. 1-0 =1
c. 1- 0 = 1
d. 1+1 -0 = 2
e. 2(x^2-Z^2) + 2y^2 = 2(1-0)+ 2*1 = 4.

aayi maala wanchwa.
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 [#permalink] New post 02 Mar 2004, 07:20
batliwala wrote:
Here's the unofficial answer.
The correct choice is E.
x^2 < = 1, Y^2 <=1, Z^2 >=0

Evaluating the maximum value of the answer choices:
a. 1+1 = 2
b. 1-0 =1
c. 1- 0 = 1
d. 1+1 -0 = 2
e. 2(x^2-Z^2) + 2y^2 = 2(1-0)+ 2*1 = 4.

aayi maala wanchwa.


Yes, In deed, Man, Simple problems tend to confuse me! :oops:
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The Answer is E [#permalink] New post 29 Dec 2005, 22:41
***********************************************************
Step 1: 2 *( x-z)*(x+z) + 2*y^2 = 2* (x^2 - z^2) + 2*y^2
Using : (a+b)*(a-b) = a^2-b^2
Step 2: 2* (x^2 - z^2) + 2*y^2 = 2(x^2+y^2-z^2)
Taking 2 as common

Step 3: 2(x^2+y^2-z^2) = 2( (-1^2) + (1^2) - (0^2))
Selecting maximum values for x and y and minimum value for z (Here sign doesnt make any diffrence, since we are squaing the values :-D)
i.e. x = -1; y = 1 or -1; z = 0

Step 4: 2( (-1^2) + (1^2) - (0^2)) = 4
Solving :)
************************************************************
Note here the maximum value the option A can give is 2 since it is x^2 + y^2 and max values for x and y (ignoring the sign) is 1
The Answer is E   [#permalink] 29 Dec 2005, 22:41
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