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10 Jun 2008, 09:48
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Is IxI< 1?
(1) Ix + 1I = 2Ix – 1I
(2) Ix – 3I ≠ 0
Can any one explain?
(1) Ix + 1I = 2Ix – 1I
(2) Ix – 3I ≠ 0
Can any one explain?
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1) identical to x^2 + 2x - 1 = 4*(x^2-2x+1) -> x = 3, x = 1/3 -> insufficient
2) x != 3 of course insufficient
1&2 sufficient because x = 1/3 is the only solution -> C
2) x != 3 of course insufficient
1&2 sufficient because x = 1/3 is the only solution -> C
12 Jun 2008, 03:22
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Mate just a request I does not look anything like | .. or like \(|\) can you please use the accurate sign next time ?
Lets solve for condition 1.
\(|x+1| = 2|x-1|\)
For every absolute sign that we break up, we are going to see 2 results a + result and a -ive result. Here there are 2 absolute signs that we need to solve for, so we are going to see four roots :
++ , -- , +- and -+
Lets expand one by one :
++
\(x+1 = 2x-2\) this shows that \(x = 3\)
--
\(-x-1 = -2x + 2\) this shows that \(x = 3\)
+-
\(x+1 = -2x+2\) this solves to \(x = 1/3\)
-+
\(-x-1 = 2x-2\) this solves to \(x=1/3\)
So we can gather that x = either 1/3 or 3 so A is insufficient.
B is clearly insufficient.
But when we combine A and B , we can eliminate 3 altogether and are left with x = 1/3 which implies x < 1 hence sufficient.
Answer C.
Lets solve for condition 1.
\(|x+1| = 2|x-1|\)
For every absolute sign that we break up, we are going to see 2 results a + result and a -ive result. Here there are 2 absolute signs that we need to solve for, so we are going to see four roots :
++ , -- , +- and -+
Lets expand one by one :
++
\(x+1 = 2x-2\) this shows that \(x = 3\)
--
\(-x-1 = -2x + 2\) this shows that \(x = 3\)
+-
\(x+1 = -2x+2\) this solves to \(x = 1/3\)
-+
\(-x-1 = 2x-2\) this solves to \(x=1/3\)
So we can gather that x = either 1/3 or 3 so A is insufficient.
B is clearly insufficient.
But when we combine A and B , we can eliminate 3 altogether and are left with x = 1/3 which implies x < 1 hence sufficient.
Answer C.
vivektripathi
