modulus
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| X+1| = 2|X-1|
for x > 1
x = 3
for -1 <x < 1
x = 1/3
for x < -1
x = 3 ( discard)
2. says x != 3
so From 1 & 2 x < 1
IMO C
for x > 1
x = 3
for -1 <x < 1
x = 1/3
for x < -1
x = 3 ( discard)
2. says x != 3
so From 1 & 2 x < 1
IMO C
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Question is asking if -1<x<1 ?Vignesh.4384 wrote:Pls help.
(1)
x+1 = 2|x-1|, x = 3
x+1 = 2*(-x+1), x = 1/3
INSUFF
(2)
x not = 3
INSUFF
(1) and (2)
X = 1/3
Answer C
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Key info regarding absolute values
|x| = x if x is +ve
AND
|x| = -x if x is -ve
Stmt 1:
|x+1| = 2|x-1|
Case 1: x is +ve
x+1 = + 2(x-1)
x+1 = 2x -2
x = 3
Case 2: x is -ve
x+1 = - 2(x-1)
x+1 = -2x +2
x = 1/3
2 values... so Insufficient.
Stmt 2:
x <> 3
Insufficient.
Combining the two stmts: x can be equal to 1/3.
So sufficient. Ans C
HT Helps
|x| = x if x is +ve
AND
|x| = -x if x is -ve
Stmt 1:
|x+1| = 2|x-1|
Case 1: x is +ve
x+1 = + 2(x-1)
x+1 = 2x -2
x = 3
Case 2: x is -ve
x+1 = - 2(x-1)
x+1 = -2x +2
x = 1/3
2 values... so Insufficient.
Stmt 2:
x <> 3
Insufficient.
Combining the two stmts: x can be equal to 1/3.
So sufficient. Ans C
HT Helps
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I have a slight confusion here
Case 2: x is -ve
here shouldnt it be -( x+1) = - 2(x-1) instead of x+1 = - 2(x-1) ?
Case 2: x is -ve
here shouldnt it be -( x+1) = - 2(x-1) instead of x+1 = - 2(x-1) ?
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