Is x^n/x^(n+1) > x^(n+1)/x^n?

can you please explain it in detail?

x^(n+1) = x^n.x

Now, while reducing the given inequality we need to take care of the inequality sign. ONLY if a -ve value is multiplied/divided on BOTH sides of the equation we need to reverse the inequality.

But, in this case we are not multiplying or dividing each side. We are just canceling out the same factor(-ve or +ve) from each side. So the inequality will remain the same.

In other words, x^n/x^n = 1, regardless of the sign of x^n. So, basically we are just multiplying each side by 1

Is it only for x= -1 or for the range -1<x<0?

+1 E

When we simplify the original inequality, using exponents theory, we get 1/x > x.

So both statements are insufficient.

Why have we flipped the inequality sign here? \(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)? --> is \(x^{n-n-1}<x^{n+1-n}\)?

It was a typo edited. Thank you. +1.

The correct answer is E

From question stem we can simplify and get is 1/x > x?

1/x - x > 0 --> Using key points is x<-1 or 0<x<1?

Statement 1

x<0 not sufficient

Statement 2

I don't care about 'n' at this point

Statements 1 and 2 together

I still don't have enough info

Hence E

Hope it helps
Cheers!

J

Sol: Given question can be rephrased as is 1/x>x ------> (1-x^2)/x>0

Given x<0, let x=-2 then the above equation holds true but if x= -1/2 then the equation doesn't hold true
(1-1/4)/-1/2 ------>3/4/-1/2---->-6/4 or -3/2 which is less than 0
Ans Statement A is not sufficient

A and D ruled out

St2: given n<0 -----> There is no use for it. Hence B ruled out
With both statements we still don't have anything new.

Hence Ans E

Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.



Is x^n/x^(n+1)>x^(n+1)/x^n?

(1) x < 0

(2) n < 0

==>transforming the original condition and the question by variable approach method, we have x^n/x^(n+1)>x^(n+1)/x^n? --> 1/x>x?. multiplying x^2 to both sides (since multiplying square values maintain the direction of the inequality sign), x>x^3? , x^3-x>0?, x(x-1)(x+1)>0? gives us -1<x<0 or 1<x?. Using both 1) and 2), the range of que doesn't include the range of con. Therefore E is the answer.

Is
/frac{x^n}{x^n+1} > /frac{x^n+1}{x^n}

1) x<0
2) n<0

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Is x^n/x^(n+1) > x^(n+1)/x^n?

(1) x < 0

(2) n < 0

x^n/x^(n+1) > x^(n+1)/x^n ==> x^2n > x^2(n+1)
ie X^2 < 1 , therefor the range of X is(-1, 1)

So statement 1 & 2 are not enough to get the sollution
so Ans> E

DS:
\(\frac{x^n}{x^{n+1}}>\frac{x^{n+1}}{x^n}\)
\(\frac{1}{x}>\frac{x[}{fraction]\)

Statement 1 : x<0
-1<x<0, \([fraction]1/x}>\frac{x[}{fraction]\)
x = -1, \([fraction]1/x}=\frac{x[}{fraction]\)
x<-1, \([fraction]1/x}<[fraction]x[/fraction]\)
NOT SUFFICIENT

Statement 2: n<0 has no significance
NOT SUFFICIENT

Answer E

First simplify the question stem: x^n/x^n+1 > x^n+1/x^n --> x^-1>x --> 1/x >x ----> is x negative or a fraction or a negative fraction?

(1) x<0

Let x=-1 1/-1 >-1 No
let x = -3 1/-3 >-3 Yes
NS

(2) n<0 Question does not depend on n, NS

(1) and (2) Since (1) NS and question does not depend on (2) NS OA is E

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