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Thread: Tough Inequality Questions

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    Tough Inequality Questions

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    Inequality is one topic iam really weak at...pls help out with these ans

    1. (x^2 + 4x + 4) / (2x^2 -x-1) >0

    a. x < -2 b. x >1 c. x not equal to 2 d. All of these

    Modulus Problems Inequality

    1. |x^3 - 1| >= 1-x

    a. -1 < x < 0 b. x < -1 c. 0 < x d. All except a

    2. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) < 0

    a. -5<= x <= -2 ; 2< x< 5
    b. -5 < x , -2; 2 < x < 5
    c. -5 < x < -2 ; 2 < x < 3 ; 3 < x < 5
    d. -5 < x < -2 ; 3 < x < 5

    The difficult part in inequality for me is figuring out when to include the value which makes the eqn zero and when not to

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    An Urch Guru Pundit Swami Sage 12rk34's Avatar
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    Q1: (x^2 + 4x + 4) / (2x^2 -x-1) >0
    => (x+2)^2/(2x+1)(x-1) >0
    =>(2x+1)(x-1) >0
    => x >-1/2 , x > 1 => x >1
    OR, x < -1/2 , x < 1 => x < -1/2
    Hence b.
    Q3:
    2. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) < 0
    =>( x^2 - 7 x + 10) / (x^2 - 6x + 9) < 0 or (x^2 + 7 x + 10) / (x^2 - 6x + 9) < 0
    => (x-5)(x-2)/(x-3)^2 < 0 or (x+5)(x+2)/(x-3)^2 < 0
    => (x-5)(x-2) < 0 or (x+5)(x+2) < 0
    => 2 < x < 5 or -5 < x <-2
    Hence b.
    Modulus Problems Inequality

    Q2. |x^3 - 1| >= 1-x

    Here x <-1 and x >0 both satisfy this inequality, so (d) will be the answer.
    But if we solve this problem, it gives imaginary values of x which is not possible in normal GMAT problems. Hence it seems difficult to arrive at real values of x by following simple mathematical steps.
    Last edited by 12rk34; 11-21-2008 at 01:20 PM.

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    1.
    (x^2+4x+4)/(2x^2-x-1) >0

    (x+2)^2 / (x-1) (2x+1)>0
    (x-1) (2x+1) >0 , as (x+2)^2 is +ve whatever the value of x

    so either (x-1) >0 and (2x-1) >0 OR (x-1 ) <0 and (2x-1) <0

    Solve , x>1 , x>-1/2 OR x<1 , x<-1/2

    so x>1 , Choice B.


    2. I think catch here is to put value of X from choice , from the equation it seems the values change when x is -1 or 1 , so values should be put accordingly

    say X <-1

    LHS = +VE (value is more than 1 )

    RHS = +ve (value is more than 1 but less than LHS)

    LHS >= RHS

    Try other values like -1<x<0 and so on

    Ans is b

    Let us know the answer .

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    3.
    ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) < 0

    Remove modulus , we get following equation to solve

    ( x^2 - 7 x + 10) / (x^2 - 6x + 9) < 0 OR ( x^2 - 7 x + 10) / (x^2 - 6x + 9) >0

    (x-5) (x-2)/(x-3)^2 <0 OR (x-5) (x-2)/(x-3)^2 >0

    Solve

    (x-5) (x-2)<0 OR (x-5) (x-2) >0 ,,,, (x cannot be equal to 3 --- basis this choice a and b eliminated )

    IMO D


    NOTE : I differ from 12rk34 in solving this inequality .Please assist .
    Last edited by ACETARGET; 11-21-2008 at 01:59 PM. Reason: Solution for Q3 added.

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    Quote Originally Posted by vineejay View Post

    2. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) < 0

    a. -5<= x <= -2 ; 2< x< 5
    b. -5 < x , -2; 2 < x < 5
    c. -5 < x < -2 ; 2 < x < 3 ; 3 < x < 5
    d. -5 < x < -2 ; 3 < x < 5
    X^2-6x+9=(x-3)^2 > 0. Also X != 3 ------------------(I)

    ( x^2 - 7 |x| + 10) < 0
    => ( x^2 - 7x + 10) < 0 or ( x^2 + 7x + 10) < 0
    => (x-2)(x-5) < 0 or (x+2)(x+5) < 0
    => 2<x<5 or -5<x<-2 -----------------------------(II)
    From (I) & (II), Ans (C) -5 < x < -2 ; 2 < x < 3 ; 3 < x < 5

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    Quote Originally Posted by vineejay View Post
    2. |x^3 - 1| >= 1-x

    a. -1 < x < 0 b. x < -1 c. 0 < x d. All except a
    My appraoch for this problem: From the ans choices put the values. Any other approach?

    a. -1<x<0. Say x=-1/2 => |x^3 - 1| = -1.125 and 1-x = 1.5 So optA wrong.
    b. x<-1, say x=-2. The inequality holds true.
    c. 0<x, Say x=1,2,3 etc. The inequality holds true.

    So ans is D.

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    Now the OAs

    1. d. 12rk..i don't know how it is d..but thats the ans

    Modulus Problems

    1. d

    2. d

    I have'nt understood , so Ques 2, why can't we take C as the answer as if we substitute between 2 and 3 in the inequlality, it satisfies the eqn.

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    Quote Originally Posted by vineejay View Post
    Now the OAs

    1. d. 12rk..i don't know how it is d..but thats the ans

    Modulus Problems

    1. d

    2. d

    I have'nt understood , so Ques 2, why can't we take C as the answer as if we substitute between 2 and 3 in the inequlality, it satisfies the eqn.
    Yes. let x=5/2. it satisfies the equation ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) < 0

    What is the source of this question. In some sources, some OAs are wrong. Are we missing anything else?

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    An Urch Guru Pundit Swami Sage
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    Q1) 12rk34 you solved it correctly but I think the answer set is wrong. If it was a) x<-1/2 b) as it stands c) x not equal to -2 (to avoid 0) then answer would be d.

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    hey guys..thanks a lot for thy efforts...i thought all my ans are wrong...i guess i should refer to a different book for inequality topic..

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