If x is a real number, then the quantity (1 − |x|)(1 + x) is positive

Question

If x is a real number, then the quantity (1 − |x|)(1 + x) is positive if and only if

(A) |x| < 1

(B) |x| > 1

(C) x < −1 or − 1 < x < 1

(D) x < 1

(E) x < −1

Timebomb · Answer

chetan2u  can you confirm if the OA is correct? Shouldnt the OA be D?

globaldesi · Answer

Answer should not be D since x<1 means lets take -2
That will 1+x negative.
However still C is also not correct. 
 Bunuel  can you comment

Timebomb · Answer

globaldesi 
if you take x=-2 then 1+x is negative and so is (1-|x|). making the expression positive.

cosimo · Answer

If X=-1 the expression is not positive but it's equal to zero, so C is the correct answer.

chetan2u · Answer

You could do it two ways.. (I) Algebraic/inequality way.. We are told (1 − |x|)(1 + x) >0 Two cases.. A) Both (1 − |x|) and (1 + x) are positive.. 1-|x|>0, or |x|<1, that is -10, or x>-1 OVERLAP -11, that is -1>x or x>1 1+x<0, or x<-1 OVERLAP x<-1 Thus answer is x<-1 and -1 1 ..... at 2, (1 − |x|)(1 + x)=(1-|2|)(1+2)=(-1)(3)-31..NO (C) x < −1 or − 1 < x < 1...YES (D) x < 1 .......Clearly at -1, (1 − |x|)(1 + x)=(1-|-1|)(1-1)=0.. no (E) x < −1........At x=0, (1 − |x|)(1 + x)=(1-|0|)(1+0)=(1)(0)=1.. positive but NOT included in this choice Hence C

KarishmaB · Answer

"If and only if" means we are looking for the EXACT range.

We know 
|x| = x if x>= 0 and 
|x| = -x if x < 0

(1 − |x|)(1 + x)
If x is negative, this becomes ( 1 + x)^2
This MUST be 0 or positive. When x = -1, it is 0, else it is positive.

So this is positive when x < 0 but x is not -1

Look at the options. 
Only option (C) covers this range and eliminates -1 so answer must be (C).

Note that (C) covers the range of 0 to 1 also which we must be getting when we assume x to be positive. We don't need to check that because we are looking for exact range and only (C) eliminates -1 as a value and covers everything else we obtained above.

Kinshook · Answer

Asked: If x is a real number, then the quantity (1 − |x|)(1 + x) is positive if and only if

(1-|x|)(1+x) > 0

Case 1: x<0
(1+x)ˆ2>0
\(x \neq -1\)
x < -1 & -1<x<0 (1)

Case 2: x>=0
(1-x)(1+x)>0
1 - xˆ2>0
xˆ2 <1
0<=x<1 (2)

Combining results (1) & (2)
x<-1 or -1<x<1

(A) |x| < 1
NOT NECESSARILY TRUE

(B) |x| > 1
NOT NECESSARILY TRUE

(C) x < −1 or − 1 < x < 1
TRUE

(D) x < 1
NOT NECESSARILY TRUE

(E) x < −1
NOT NECESSARILY TRUE

IMO C

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If x is a real number, then the quantity (1 − |x|)(1 + x) is positive

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If x is a real number, then the quantity (1 − |x|)(1 + x) is positive

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