Is x > 0? (1) 1/x < 1 (2) 1/x > x

Hello,

Multiplying or dividing by a variable is not a valid operation in inequations; because if the variable is negative, it may alter the inequality.

Example:

3 > 2

but it doesn't mean that 3x > 2x always holds

For negative values of x,

3x < 2x

This is one of the earliest concepts explained in our inequations videos; please revise the concept videos.

All the best!
Experts' Global

Oh. Okay I got my mistake. I only considered x as positive, x can be negative as well. I subconsciously took x as positive after seeing the question.
My bad. Apologies.

Now it all makes sense.

Thank you

st 1 --> 1/x < 1

=> 1 < x OR x < 0

st. 2 --> 1/x > x
=> 1 > x > 0 OR x < -1

Merge the two statements and the only common point is that x can be negative.

First of all, testing numbers is not the only way to solve this question.

The reason we wouldn’t multiply each side of 1/x < 1 by x is that we have no information on the sign of x. If x were positive, then we would obtain x > 1, as you noticed, but if x were negative, we would obtain x < 1 since we have to change the direction of an inequality whenever we multiply each side by a negative number.

To solve 1/x < 1, move the 1 on the right hand side to the left hand side:

1/x - 1 < 0

Now get the common denominator x and then combine the two fractions:

1/x - x/x < 0

(1 - x)/x < 0

Now, there are two ways this expression can be negative: Case 1: 1 - x is negative and x is positive, or Case 2: 1 - x is positive and x is negative.

For Case 1, we get 1 - x < 0 (which is equivalent to x > 1) and x > 0. The set of numbers that satisfy both of the inequalities x > 0 and x > 1 are x > 1. So, one way to satisfy the inequality 1/x < 1 is to pick values for x which are greater than 1.

For Case 2, we get 1 - x > 0 (which is equivalent to x < 1) and x < 0. The set of numbers that satisfy both inequalities are the set of x such that x < 0. Thus, the inequality 1/x < 1 can also be satisfied by picking a negative value for x.

As we can see, x could be either greater than 1 or less than 0. If we were to multiply each side of 1/x < 1 by x, we would completely miss the fact that negative values of x also satisfy the inequality 1/x < 1.

The explanation of why it’s wrong to rewrite 1/x > x as x^2 < 1 is similar to the explanation just presented.

Statement#1: Not Sufficient

1/x < 1
1/x - 1 < 0
1-x/x < 0
(x-1)/x > 0 (x can be positive or negative)

Statement#2: Not Sufficient

1/x > x
1/x - x > 0
1 - x^2/x > 0
x^2 - 1/x < 0
(x+1)(x-1)/x < 0 (x can be positive or negative)

Combined 1&2:

Statement#1 tells us that (x-1)/x > 0
If we put this in statement#2 i.e (x+1)(x-1)/x < 0, then it implies that x+1<0 because (x-1)/x > 0.
Hence, x<-1.

If x < -1, then it is definitely not greater than 0. Hence, we have a definite NO.

Sufficient. Answer (C)

For statement 2, if we choose a positive integer, say, 3, wouldn't that violate the inequality? i.e. isn't the LHS supposed to be negative?

Is x > 0 ?

Or

Is X = (+)Positive Value?


(1) 1/X < 1


Case 1: X = Pos.

X >1 ------ YES


Case 2: X = Neg.

X < 1 ----- and Since X < 0 b/c we are assuming it is Negative?

the Limiting Inequality stands as: X < 0 ------NO



S1 NOT SUFFICIENT


(2) 1/x > x


Testing Values in the 4 Ranges:

X > +1

0 < X < +1

-1 < X < 0

X < -1

the Ranges of X that Satisfy the Inequality are:

X < -1 ----- NO

or

0 < X < 1 ------YES


S2 NOT SUFFICIENT ALONE



TOGETHER:


If X is Positive, it must Satisfy BOTH the Inequalities in Statement 1 and Statement 2:

S1: if X is Positive: X > 1

S2: 0 < X < 1


since it is impossible for a Positive Value to Satisfy BOTH Ranges at the same time, X can NOT be a Positive Value


S1: if X is Negative: X < 1

thus, S1 tells us that any Value in the Range: X < 0 satisfies S1's Inequality


S2: X < -1


therefore, any Value in the Range of: X < -1

will Satisfy Both Statements at the Same Time


X must be negative. Definite NO.

(C)Together Sufficient

Sure, 3, a positive number, may violate the inequality, but what if x= 1/2, which is also a positive number? In that case (x=1/2), x can assume a positive value without violating the inequality. Ultimately, even if one and only one value satisfies an inequality, the value stands valid. This means that statement 2 is still insufficient. Hope it helps.

i tried in this way-

State 1: 1/X < 1 = (1-x)/x < 0;
x-x^2 < 0 or x < X^2

this implies that X could either positive or negative, so Not sufficient.

State II:

1/X > X;
1/X-X>0;
1-X^2/X > 0;
(X-X^3) > 0;
X > X^3;
inference; either X is negative or X is a fraction. so so Statement II also both Yes and no therefore is sufficient.

However combining Statement I and II, we can say that X is negative. (statement i says X either positive or negative) (statement 2 says, either negative or fraction), hence X must be negative.

So answer C

what i did:

to check for C,

1) 1/x < 1
2) x < 1/x

Adding both, 1/x + x < 1 + 1/x

=> subtracting 1/x from both sides, x < 1.

So, answer should be E, but i know this is wrong because values like x = 0.5 doesnt satisfy St.1.

Can anyone explain what am i doing wrong when i am adding and trying to solve them algebraically?

Thanks in advance.

Bunuel

Solved it by trying cases and drawing inferences from the statements

St 1:
X can be negative - Not satisfied
X can be positive - Satisfied
2 possibilities- one satisfies the condition, other doesnt, thus not sufficient

St 2:
X can be negative - Not satisfied
X can be a value between 0 and 1 (This is a catch)- Satisfied
2 possibilities- one satisfies the condition, other doesnt, thus not sufficient

St 1+ St 2:
Common and the only possibility. X is negative - Not satisfied
Only one solution with a single result- Sufficient