vineejay Posted November 21, 2008 Share Posted November 21, 2008 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 1 c. x not equal to 2 d. All of these Modulus Problems Inequality 1. |x^3 - 1| >= 1-x a. -1 2. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) a. -5 b. -5 c. -5 d. -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 Quote Link to comment Share on other sites More sharing options...
12rk34 Posted November 21, 2008 Share Posted November 21, 2008 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 x Hence b. :) Q3: 2. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) =>( x^2 - 7 x + 10) / (x^2 - 6x + 9) => (x-5)(x-2)/(x-3)^2 => (x-5)(x-2) => 2 Hence b. :) Modulus Problems Inequality Q2. |x^3 - 1| >= 1-x Here 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. :rolleyes: Quote Link to comment Share on other sites More sharing options...
ACETARGET Posted November 21, 2008 Share Posted November 21, 2008 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 ) Solve , x>1 , x>-1/2 OR x 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 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 Ans is b Let us know the answer . Quote Link to comment Share on other sites More sharing options...
ACETARGET Posted November 21, 2008 Share Posted November 21, 2008 3. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) Remove modulus , we get following equation to solve ( x^2 - 7 x + 10) / (x^2 - 6x + 9) 0 (x-5) (x-2)/(x-3)^2 0 Solve (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 . Quote Link to comment Share on other sites More sharing options...
john.3030 Posted November 21, 2008 Share Posted November 21, 2008 2. ( x^2 - 7 |x| + 10) / (x^2 - 6x + 9) a. -5 b. -5 c. -5 d. -5 X^2-6x+9=(x-3)^2 > 0. Also X != 3 ------------------(I) ( x^2 - 7 |x| + 10) => ( x^2 - 7x + 10) => (x-2)(x-5) => 2 From (I) & (II), Ans © -5 Quote Link to comment Share on other sites More sharing options...
john.3030 Posted November 21, 2008 Share Posted November 21, 2008 2. |x^3 - 1| >= 1-x a. -1 My appraoch for this problem: From the ans choices put the values. Any other approach? a. -1 |x^3 - 1| = -1.125 and 1-x = 1.5 So optA wrong. b. x c. 0 So ans is D. Quote Link to comment Share on other sites More sharing options...
vineejay Posted November 22, 2008 Author Share Posted November 22, 2008 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. Quote Link to comment Share on other sites More sharing options...
john.3030 Posted November 22, 2008 Share Posted November 22, 2008 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) What is the source of this question. In some sources, some OAs are wrong. Are we missing anything else? Quote Link to comment Share on other sites More sharing options...
thankont Posted November 22, 2008 Share Posted November 22, 2008 Q1) 12rk34 you solved it correctly but I think the answer set is wrong. If it was a) x Quote Link to comment Share on other sites More sharing options...
vineejay Posted November 23, 2008 Author Share Posted November 23, 2008 hey guys..thanks a lot for ur efforts...i thought all my ans are wrong...i guess i should refer to a different book for inequality topic..:) Quote Link to comment Share on other sites More sharing options...
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