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Difficulty:
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Question Stats:
19% (01:43) correct
81%
(02:25)
wrong
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If \(x, y, \) and \( z \) are three real numbers and \( x \) and \(y \) are different, which of the following must be true?
I. If \( x^2 = y^2 \) , then \( x^2 + y^2 + z^2 > x + y \)
II.If \( |x + z − y| = x + z −y \), then \(|x^2 + z − y| = x^2 + z − y \)
III. \( |x + y − z| ≤ |x| + |y| + |z| \)
A) I only
B) III only
C) I and III only
D) II and III only
E) I, II, and III
I. If \( x^2 = y^2 \) , then \( x^2 + y^2 + z^2 > x + y \)
II.If \( |x + z − y| = x + z −y \), then \(|x^2 + z − y| = x^2 + z − y \)
III. \( |x + y − z| ≤ |x| + |y| + |z| \)
A) I only
B) III only
C) I and III only
D) II and III only
E) I, II, and III
13 Mar 2026, 08:46
Kudos
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is it not B?
I x = 0.5 y = -0.5 z = 0.5 x^2+y^2+z^2 = 0.75 that is less than x+y = 1 false
II. If |x + z − y| = x + z − y, then x+z > y .|x^2 + z − y| = x^2 + z − y .but as in the last option the x could be lower when ^2 false
III. |x + y − z| ≤ |x| + |y| + |z| can work if z is negative or the x and y are negative
I x = 0.5 y = -0.5 z = 0.5 x^2+y^2+z^2 = 0.75 that is less than x+y = 1 false
II. If |x + z − y| = x + z − y, then x+z > y .|x^2 + z − y| = x^2 + z − y .but as in the last option the x could be lower when ^2 false
III. |x + y − z| ≤ |x| + |y| + |z| can work if z is negative or the x and y are negative
13 Mar 2026, 12:21
Kudos
Bookmarks
given x and y are different ,
1. x^2=Y^2 means |x|= |y| but are different, hence have opposite signs.
so certainly x^2+y^2+z^2 a positive value is greater than x+y=0---- MUST TRUE
2. given statement simply can be reiterated as if x+z>y then x^2+z>y
since |a| = a, when a>0
now if we take x=1/2, z=-1 and y=-3/4
so x+z=-1/2 which is greater than -3/4
but x^2+z=1/4 - 1=-3/4 which is not greater than -3/4-------FALSE
3. since in |(x+y-z)| signs are different hence the maximum value of this expression can be the absolute sums of x,y,z-------MUST TRUE
eg |4+2-5|=1 , |-5+4-(-3)|=2 and so on for all real numbers all less than the absolute sum
|2+3-(-3)|=8 which is the maximum value for the given combination.
1. x^2=Y^2 means |x|= |y| but are different, hence have opposite signs.
so certainly x^2+y^2+z^2 a positive value is greater than x+y=0---- MUST TRUE
2. given statement simply can be reiterated as if x+z>y then x^2+z>y
since |a| = a, when a>0
now if we take x=1/2, z=-1 and y=-3/4
so x+z=-1/2 which is greater than -3/4
but x^2+z=1/4 - 1=-3/4 which is not greater than -3/4-------FALSE
3. since in |(x+y-z)| signs are different hence the maximum value of this expression can be the absolute sums of x,y,z-------MUST TRUE
eg |4+2-5|=1 , |-5+4-(-3)|=2 and so on for all real numbers all less than the absolute sum
|2+3-(-3)|=8 which is the maximum value for the given combination.
