S1: assuming x to be +ve, no way a +ve number can be the median along with 2 other -ve numbers. Thus, x has to be < 0 . Suff.

S2:
take x as 2, we have -1/4 is the median of -1/4, -2, 32 - true.
take x as -2, we have -1/4 is the median of -1/4, 2, -32 - also true.
hence S2 is insuff.

Option A

I tried to explain everything in detail in the attached file.
My answe is A

_________________

Asked: If x ≠ 0, is x < 1 ?

(1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).
1/x lies between -x^2 and - x^3 .
If x > 0
-x^2 < 0
-x^3 < 0
1/x can not lie between -x^2 and -x^3.
Then x<=0 < 1
SUFFICIENT

(2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).
-1/x^2 lies between - x and x^5
If x>0
-1/x^2 < 0
-x < 0
x^5 >0
-x < -1/x^2 < 0 < x^5
x - 1/x^2 >0
(x^3 - 1) / x^2 > 0
(x-1)(x^2 +1 +x) / x^2 >0
x > 1

But if x<0
-1/x^2 < 0
-x >0
x^5 < 0
x^5 < -1/x^2 < 0 < -x
x^5 + 1/x^2 < 0
(x^7 + 1)/x^2 < 0
-1 < x < 0
NOT SUFFICIENT

IMO A
_________________

I always get confused if -x^2 = (-x)^2 or -(x^2), but I am going to answer as if it is -(x^2)

1) If x is the median, ie between 2 negative numbers, we know it is negative and therefore less than 1. Sufficient
2) 2 and -2 satisfy this constraint. Insuff

A is my answer.

IMO OA should be A

is x < 1

Statement 1

Let's take x = -2

\(\frac{1}{x} = -\frac{1}{2}\)
\(-x^2 = -4\)
\(-x^3 = 8\)

\(\frac{1}{x}\) is median. x < 1 satisfies

Let's take x = 2
\(\frac{1}{x} = \frac{1}{2}\)
\(-x^2 = -4\)
\(-x^3 = -8 \)

Median is not \(\frac{1}{x }\)

So, if \(\frac{1}{x }\)is the median of \(\frac{1}{x}, -x^2\), and \(-x^3\) then x has to be less than 1

Statement 1 is sufficient

Statement 2

Let's take x = 2
\(-\frac{1}{x^2} = -\frac{1}{4}\)
-x = -2
\(x^5 = 32\)

Median is \(\frac{-1}{x^2}\)

Let's take x = -2

\(-\frac{1}{x^2} = -\frac{1}{4}\)
-x = 2
\(x^5 = -32\)

Median is \(\frac{-1}{x^2}\)

So, as per statement 2, x can be less than 1 and greater than 1

Insufficient

My answer is A).

Easiest to consider 4 intervals broken up by 3 key nodes: -1, 0, 1.

Statement 1: notice that if x>=1, 1/x > 0 > -x^2 or -x^3. So for 1/x to be between -x^2 and -x^3, x must be <1, so already sufficient.

Statement 2: say x=2 (in the x>=1 interval), -1/x^2 = -1/4, -x = -2, x^5 = 32. So -1/x^2 is the median. But when x = -2 (in the x<=-1 interval), -1/x^2 = -1/4, -x = 2, and x^5 = -32. So -1/x^2 is again the median. Insufficient.

So A).

If x ≠ 0, is x < 1 ?

(1) 1/x is the median of 1/x, −x2 and −x3

(2) −1/x2 is the median of −1/x2, −x and x5

1-

If x > 1, then 1/x cannot be the median of 1/x, -x^2 and -x^3 since the other two are negative

Sufficient

2.

If x = 2, -2, -1/4, 32
If x = -2, -32,-1/4, 2

Not sufficient

A

The answer is A.

Statement 1 : This condition is satisfied only when x<0. For all nos. > 0 this condition does not satisfy. If x<0 then x<1.
Sufficient.

Statement 2 : This condition is satisfied both when x<0 and when x>1.
Insufficient.

Hence, A is the correct choice.

My answer is (A).

Evaluate (1) independently.
Can x be a positive number (i.e. x > 0)?
If x is a positive number, 1/x is positive, but -x^2 and -x^3 are both negative. Therefore, x must be a negative number. We can answer "YES" to the question.
Sufficient.

Evaluate (2) independently.
Let x = 2, we have -1/4, -2 and 32. So x can be larger than 1.
(The next step is actually unnecessary. Even though we are evaluating 2 regardless of 1, there is no inherent contradiction between 1 and 2. )
Let x = -1. we have -1, 1, and -1. X can also be less than 1.
Insufficient.

target is x<1

#1
\(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).

x=-2 we get
1/x = -1/2
-x^2 = -4
-x^3 = +8
sufficient to say that -x^2,1/x ,-x^3 where 1/x is median and x<1

#2
\(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\)
x=-2
-1/x^2 = -1/4
-x=+2
x^5 = -32
sufficient that x<1 and \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\)
x= 2
-1/x^2 = -1/4
-x=-2
x^5 = 32
x>1 and \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\)
insufficient as x>1


OPTION A is correct

As shown in the image, try various cases as shown below for both statements and in each of the statements the answer we get si sometimes yes and sometimes no.
So option E is correct.
Cases:
x = 1/2, -1/2,-1, -2.

_________________

Cases:
1) 0<x<1 = 1/2
2) -1<x<0 = -1/2
3) x>1 = 2
4) x<-1 = -2

Statement 1:
1) 2, -1/4, 1/8 Median is 1/8 => not valid
2) -2, -1/4, 1/8. Median is -1/4 => invalid
3) 1/2, -4, -8. Median is -4. => invalid
4) -1/2, -4, 8. Median is -1/2 => valid.
Only one option is valid hence sufficient

Statement 2:
1) -4, -1/2,1/32 => invalid
2) -4,1/2,-1/32 => invalid
3) -1/4,-2,32 => valid
4) -1/4,-2,-32 => valid
Insufficient since x>1 and x<-1 both give us answers.

Hence the correct answer is A.

From Statement 1

there exists 2 possibilities

-x^2<1/x<-x^3 this is trues if x<-1
-x^3<1/x<-x^2 this is not possible for any values of x

Hence A is sufficient to ans

From Statement 2

there exists 2 possibilities

x^5<-1/x^2<-x this is trues if x<-1
-x<-1/x^2<x^5 this is true if x>1

Hence statement 2 is insufficient to answer

thus A ans

(1)
Case 1: -x^2 < 1/x < -x^3
=> 1/x + x^3 < 0
=> (1 + x^4)/x < 0
=> x<0
Case 2: -x^2 > 1/x > -x^3
=> -x^2 + x^3 > 0
=> x^2(x - 1) > 0
=> x>1

=> x<0 or x>1 Insufficient

(2)
Case 1: -x < -1/x^2 < x^5
=> x - 1/x^2 > 0
=> (x^3 - 1)/x^2 > 0
=> x^3 - 1 > 0
=> (x-1)(x^2+x+1) > 0
We have x^2+x+1 = (x+1/2)^2 + 3/4 > 0
=> x-1 > 0 or x>1
Case 2: -x > -1/x^2 > x^5
=> x^5 + x < 0
=> x*(x^4 + 1) < 0
=> x<0

=> x>1 or x<0 Insufficient

Combine (1) and (2) we would still have x>1 or x<0 Insufficient

The answer is E

(A) is the answer IMO


If x ≠ 0, is x < 1 ?

Statement (1) \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).

Take values to check median in the following range
For \(x>1 \) Median is\( -x^2\)
For \(0<x<1 \) Median is \(-x^3\)
For \(-1<x<0 \) Median is \(-x^2\)
For \(x<-1 \) Median is \(1/x\)

\(1/x\) is median only for values \(x<-1 \)

Hence, we can conclude using statement 1 that x < 1

Sufficient


Statement (2) \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).

Take values to check median in the following range
For \(x>1 \) Median is\( \frac{-1}{x^2}\)
For \(0<x<1 \) Median is \(-x\)
For \(-1<x<0 \) Median is \(x^5\)
For \(x<-1 \) Median is \(\frac{-1}{x^2}\)

x can be in the range x>1 or x<-1

Not sufficient

(A) IMO

1) 1/x is the median of 1/x, -x^2, -x^3

It means 1/x should be the middle value.
For x>0, 1/x will be the greatest value as -x^2, -x^3 both are negative.

This condition is holding true only for x<= -1
as -x^2 will be always be negative and will be the lowest value, and -x^3 will always be positive.
1/x will be the mid value.

Hence, 1) is sufficient.

2) -1/x^2 is the median of -1/x^2, -x and x^5

This statement is valid for x = 2 as well as x = -2
Hence, not sufficient.

(A) is our answer.

(1) If x is positive:
- \(\frac{1}{x}\) is positive
- \(-x^2\) and \(-x^3\) are negative

So we have two negative numbers and one positive, so the median must be one of the negative numbers, so \(\frac{1}{x}\) can not be the median, so x MUST BE negative, so \(x<1\)

SUFFICIENT

(2) If \(x=-2<1\) we have \(-32, -\frac{1}{4}, 2\)
If \(x=2>1\) we have \(-2, -\frac{1}{4}, 32\)

INSUFFICIENT

IMO A

x can be of 4 groups

Case 1: x is greater than 1, example:2
Case 2: x is greater than 0 but less than 1, example: 1/2
Case 3: x is greater than -1 but less than 0, example: -1/2
Case 4: x is less than -1, example: -2

1.

Case 1: 1/x is not median
Case 2: 1/x is not median
Case 3: 1/x is not median
Case 4: 1/x is median

So 1 is enough

2.

Case 1: -1/x^2 is median
Case 2: -1/x^2 is not median
Case 3: -1/x^2 is not median
Case 4: -1/x^2 is median

So 2 alone is not enough.

So answer is A

If x ≠ 0, is x < 1 ?

Stat1: \(\frac{1}{x}\) is the median of \(\frac{1}{x}\), \(-x^2\) and \(-x^3\).

if x >1 then, sequence will be \(-x^3\), \(-x^2\) and \(\frac{1}{x}\).
if 0<x<1 then, sequence will be \(-x^2\), \(-x^3\) and \(\frac{1}{x}\).
if x<0 then, sequence will be \(-x^2\), \(\frac{1}{x}\) and \(-x^3\), which satisfies the condition. Sufficient

Stat2: \(-\frac{1}{x^2}\) is the median of \(-\frac{1}{x^2}\), \(-x\) and \(x^5\).
Similarly, for if x<0 then, sequence will be \(x^5\), \(-\frac{1}{x^2}\) and \(-x\) , which satisfies the condition.

And, if x >0 still, sequence will be \(x^5\), \(-\frac{1}{x^2}\) and \(-x\) , which satisfies the condition. So, Not Sufficient

So, I think A.