Tough Inequality Questions

[vineejay]

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

[12rk34]

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:

[ACETARGET]

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 .

[ACETARGET]

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 .

[john.3030]

 

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

[john.3030]

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.

[vineejay]

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.

[john.3030]

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?

[thankont]

Q1) 12rk34 you solved it correctly but I think the answer set is wrong. If it was a) x

[vineejay]

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..:)