What is the solution set of (1+|x|)(1+x) 0?

Question

What is the solution set of (1+|x|)(1+x) > 0 ? A. x > -1 B. x < -1 C. x < 0 D. x > 0 E. x > 1

Ben471 · Accepted Answer

What is the solution set of (1+|x|)(1+x)>0 ?

Let a = (1+|x|)(1+x)
since a>0, it implies that 
(1+|x|) is +ve and (1+x) is +ve or 
(1+|x|) is -ve and (1+x) is -ve

(1+|x|) can only be positive,
therefore, (1+x) >0
hence, x>-1

IMO correct answer : A. x>−1

Bunuel · Answer

The first multiple, 1 + |x|, is always positive, thus the product to be positive, the second multiple must also be positive: 1 + x > 0. This gives: x > -1.

Answer: A.

MathRevolution · Answer

=>

Since  1+|x| > 0 , we can divide both sides of the inequality by  1 + |x|  to obtain  1+x > 0  or  x > -1.

Therefore, the answer is A.
Answer: A

gaurav2m · Answer

consider x=0.1, then answer should be D.
plz let me know where I am wrong ?

Bunuel · Answer

(1+|x|)(1+x) > 0 is also true if -1 < x <= 0. This range combined with x > 0 gives x > -1, which is option A.

gaurav2m · Answer

.

Thank you for your reply,

(1+|x|)(1+x) as both of them must be positive for the condition to satisfy, which means x should be greater than 0 but how are you considering answer should x>-1, as for this answer to satisfy isn't we need to take common of two ranges of x say x> 0 and x>-1 then we can say that x>0.

this may be a silly doubt but please help me

Bunuel · Answer

Not sure I can follow what you mean but you can check some negative values from -1 to 0 to see that  (1+|x|)(1+x)  will indeed be more than 0. Also, you can check, more detailed, algebraic solutions above.

Hope it helps.

varmashreekanth · Answer

Hi,
I had a query regarding this question. I understand that the answer is x > -1, but I am getting different answers for different methods of solving this question, so I wanted to understand why one method is wrong and one method is right. The following are the methods in which I tried to solve the question:-

Method 1

We have given that 
 (1+|x|)(1+x) > 0 
=> both (1+|x|) and (1+x) should be of same sign.
=> we know that (1 + |x|) will always be positive. Let us equate it to a positive integer constant K. So therefore the equation becomes:-
 K(1+x)>0
Dividing both sides by positive integer K, we get
 (1+x)>0
=> x > -1
We get the correct answer i.e option A.

Method 2

We have given that 
 (1+|x|)(1+x) > 0 
=> both (1+|x|) and (1+x) should be of same sign.
=> we know that (1 + |x|) will always be positive.

So if we solve the (1+|x|) > 0 and (1+x) > 0 separately, we get :-
(1+x) > 0
=> x > -1 ------------ (1)

(1+|x|) > 0
=> |x| > -1
squaring both sides
=>  x^2 > 1  
=>   x^2-1 > 0  
=>  (x-1)(x+1) > 0  
on solving we get,
 x => (-infinity,-1) & (1, infinity)  ------------------(2)

from equations (1) and (2) we get,
 x>-1  and  x=> (-infinity,-1) & (1, infinity)  ,
so the common area of intersection, is   x=> (1, infinity)  => (x>1)

So in this way, I am getting option E as the answer.

So can anybody tell me what I am doing wrong in method 2? That would be really helpful for me.

Thanks & Regards,
Shreekanth

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What is the solution set of (1+|x|)(1+x) > 0?

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