Buttercup3 wrote:

MathRevolution wrote:

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 independent equations ensures a solution.

Is 1+x+x^2+x^3+x^4<1/(1-x)?

1) x>0

2) x<1

The question is (1-x^5)/(1-x)<1/(1-x)? and –x^5/(1-x)<0?, x^5/(x-1)<0? is derived when delete 1/(1-x) from the both equations of 1/(1-x)-x^5/(1-x)<1/(1-x)?. When you multiply (x-1)^2 to the both equations, x^5(x-1)<0? is derived. Divide it with x^4 and it becomes x(x-1)<0?. That is, 0<x<1? is derived, which makes C the answer.

Thanks for such clear and precise explanation. But can this question be done by taking numbers and plugging in?

Here algebra is slightly better but yes, it can be done by plugging in numbers. Let me try:

(1) x > 0

Now RHS of the question stem is 1/(1-x). Since its 1-x in the denominator, and this statement says x > 0, we should immediately look for two cases: one where 0<x<1 and one where x>1 (x cannot be 1 as for that RHS would be undefined).

So lets take a value of x between 0 and 1, say 0.5. RHS becomes 1/(1-0.5) = 2 LHS becomes 1+0.5+0.25+0.125+0.0625=1.9375. LHS is less than RHS

If we try any other value in the same range say 0.2 or 0.9, still RHS would be greater.

Now lets take another case where x > 1, in this case clearly we can see that 1-x would be negative, so RHS would be negative while LHS positive. So LHS would become greater.

So not sufficient.

(2) x < 1

We have got to take cases where 0<x<1, where -1<x<0, and where x<-1

If 0<x<1, RHS would be greater as discussed in first statement.

If -1<x<0, lets say x=-0.5. RHS = 1/(1+0.5) = 0.666.. and LHS = 1-0.5+0.25-0.125+0.0625 = 0.6875. LHS is greater

We dont even need to look at third case because we can see that this statement is not sufficient.

Combining the two statements: 0<x<1. In that case RHS would be greater as discussed. Sufficient.

Hence C answer