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x ≠ ±1

by sanju09 » Wed Mar 10, 2010 1:47 am
(∣x∣ + 1)/(∣x∣ - 1) < 4; x ≠ ±1.

What is the solution to the above inequality?
(A) x < -5/3, -1 < x < 1
(B) x < -5/3, -1 < x < 1, x > 5/3
(C) -1 < x < 1, x > 5/3
(D) -5/3 < x < 5/3
(E) 0 < x < 5/3
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by thephoenix » Wed Mar 10, 2010 3:19 am
sanju09 wrote:(∣x∣ + 1)/(∣x∣ - 1) < 4; x ≠ ±1.

What is the solution to the above inequality?
(A) x < -5/3, -1 < x < 1
(B) x < -5/3, -1 < x < 1, x > 5/3
(C) -1 < x < 1, x > 5/3
(D) -5/3 < x < 5/3
(E) 0 < x < 5/3
IMO A

all values of option a satisfies the eqn
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by sanju09 » Wed Mar 10, 2010 3:40 am
thephoenix wrote:
sanju09 wrote:(∣x∣ + 1)/(∣x∣ - 1) < 4; x ≠ ±1.

What is the solution to the above inequality?
(A) x < -5/3, -1 < x < 1
(B) x < -5/3, -1 < x < 1, x > 5/3
(C) -1 < x < 1, x > 5/3
(D) -5/3 < x < 5/3
(E) 0 < x < 5/3
IMO A

all values of option a satisfies the eqn
[spoiler]A ≠ OA[/spoiler]
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by firdaus117 » Wed Mar 10, 2010 3:49 am
Case 1 mod x-1>0
which means x>1 and x<-1
i.e. -1>x>1 -------- (1)
Multiply both sides by mod x-1,
mod x +1 < 4(mod x-1)
3mod x > 5
mod x>5/3
-5/3>x>5/3 ----------(2)
Combining 1 and 2,permissible values of x will lie in interval
-5/3>x>5/3 --------------(3)
Case 2 mod x-1<0
i.e. -1<x<1 -----------------(4)
Multiply both sides by mod x-1,
mod x +1 > 4(modx-1) [the sign of inequality reverses as we are multiplying by a negative quantity]
-5/3<x<5/3 ------------(5)
Combining 4 and 5,
-1<x<1------------------(6)
Finally combining 3 and 6,we get final result as:
[spoiler]x < -5/3, -1 < x < 1, x > 5/3 Option B[/spoiler]
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by sanju09 » Sat Mar 13, 2010 3:31 am
firdaus117 wrote:Case 1 mod x-1>0
which means x>1 and x<-1
i.e. -1>x>1 -------- (1)
Multiply both sides by mod x-1,
mod x +1 < 4(mod x-1)
3mod x > 5
mod x>5/3
-5/3>x>5/3 ----------(2)
Combining 1 and 2,permissible values of x will lie in interval
-5/3>x>5/3 --------------(3)
Case 2 mod x-1<0
i.e. -1<x<1 -----------------(4)
Multiply both sides by mod x-1,
mod x +1 > 4(modx-1) [the sign of inequality reverses as we are multiplying by a negative quantity]
-5/3<x<5/3 ------------(5)
Combining 4 and 5,
-1<x<1------------------(6)
Finally combining 3 and 6,we get final result as:
[spoiler]x < -5/3, -1 < x < 1, x > 5/3 Option B[/spoiler]
Good to see that the OA is matched somehow, the discontinued intervals must not be shown as continuous. Any better explanation to this peculiar problem that only a very few would find on GMAT?
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by firdaus117 » Sat Mar 13, 2010 6:20 am
sanju09 wrote: Good to see that the OA is matched somehow, the discontinued intervals must not be shown as continuous. Any better explanation to this peculiar problem that only a very few would find on GMAT?
Totally agree!But I got too bored typing all that stuff and missed x is not equal to -1 and +1.It's better to solve such question by drawing number line representing different points of discontinuity.
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