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16 Apr 2013, 23:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
70% (01:20) correct
30%
(01:20)
wrong
based on 695
sessions
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16 Apr 2013, 23:32
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\(|y|\geq{y^3}\)
Case y>0
\(y\geq{y^3}=y-y^3\geq{0}=y(1-y^2)\geq{0}\) this is +ve if \(y\geq{0}\) and \(1-y^2\geq{0}\) (\(-1\leq{x}\leq{1}\)) so the only interval in which this is positive is \(1\geq{y}\geq{0}\) ( you have to intersect \(y\geq{0}\) with \(-1\leq{x}\leq{1}\) AND with the rage we are considering y>0 and take the common part)
Case y<0
\(-y\geq{y^3}=-y-y^3\geq{0}=-y(1+y^2)\geq{0}\) this is +ve if \(-y\geq{0}=y\leq{0}\) and \(1+y^2\geq{0}=y^2\geq{-1}=\) always true => always positive. Intersect those ranges and find out that if y<0 the function is always +ve.
Take case 1 and 2: interval \(1\geq{y}\geq{0}+y\leq{0} = y\leq{1}\)
The question now:is \(y\leq{1}\)?
(1) y < 1 Sufficient
(2) y < 0 Sufficient
Let me know if it's clear
Case y>0
\(y\geq{y^3}=y-y^3\geq{0}=y(1-y^2)\geq{0}\) this is +ve if \(y\geq{0}\) and \(1-y^2\geq{0}\) (\(-1\leq{x}\leq{1}\)) so the only interval in which this is positive is \(1\geq{y}\geq{0}\) ( you have to intersect \(y\geq{0}\) with \(-1\leq{x}\leq{1}\) AND with the rage we are considering y>0 and take the common part)
Case y<0
\(-y\geq{y^3}=-y-y^3\geq{0}=-y(1+y^2)\geq{0}\) this is +ve if \(-y\geq{0}=y\leq{0}\) and \(1+y^2\geq{0}=y^2\geq{-1}=\) always true => always positive. Intersect those ranges and find out that if y<0 the function is always +ve.
Take case 1 and 2: interval \(1\geq{y}\geq{0}+y\leq{0} = y\leq{1}\)
The question now:is \(y\leq{1}\)?
(1) y < 1 Sufficient
(2) y < 0 Sufficient
Let me know if it's clear
17 Apr 2013, 00:01
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Is y^3 ≤ |y|?
You can use simple reasoning instead of algebra. Let's see for which range(s) y^3 ≤ |y| holds true.
When y ≤ 0, then LHS = y^3 ≤ 0 and RHS = |y| >= 0, thus in this case y^3 ≤ |y|.
When 0 < y ≤ 1, then also y^3 ≤ |y|.
When y>1, then obviously y^3 > |y|.
So, we have that y^3 ≤ |y| holds true when y ≤ 1.
(1) y < 1. Sufficient.
(2) y <0. Sufficient.
Answer: D.
Hope it's clear.
You can use simple reasoning instead of algebra. Let's see for which range(s) y^3 ≤ |y| holds true.
When y ≤ 0, then LHS = y^3 ≤ 0 and RHS = |y| >= 0, thus in this case y^3 ≤ |y|.
When 0 < y ≤ 1, then also y^3 ≤ |y|.
When y>1, then obviously y^3 > |y|.
So, we have that y^3 ≤ |y| holds true when y ≤ 1.
(1) y < 1. Sufficient.
(2) y <0. Sufficient.
Answer: D.
Hope it's clear.
