Is 1 + x + x^2 + x^3 + x^4 + x^5 0 (2) x

Question

Is  1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}  ?

(1)  x > 0

(2)  x < 1

M36-01

Bunuel · Answer

Similar questions to practice: https://gmatclub.com/forum/is-1-x-x-2-x ... 15368.html https://gmatclub.com/forum/is-1-x-x-2-x ... 56226.html

chetan2u · Answer

1+x+x^2+x^3+x^4+x^5<\frac{1}{(1-x)} .. 1+x+x^2+x^3+x^4+x^5-\frac{1}{(1-x)}<0 .. \frac{{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}}{(1-x)}<0 \frac{{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}}{(1-x)}<0 \frac{{(1+x+x^2+x^3+x^4+x^5-x-x^2-x^3-x^4-x^5-x^6)-1}}{(1-x)}<0 \frac{-x^6}{1-x}<0 \frac{x^6}{x-1}<0 x^6 is always non negative, so the question becomes : Is x-1<0? OR Is x<1? But when x=0, the equation becomes 1+0+0+0+0+0<1/1-0.....1<1. So x cannot be 0 lets check the statements 1) x>0 if x<1.. ans is YES if x>1.. ans is NO Insuff 2) x<1 Almost what we were looking for But if x=0, answer is No Insuff Combined x<1 and x is not 0 Sufficient C

WannaManaConsult · Answer

Multiply both sides by (1-x) (1+x+x^2+x^3+x^4+x^5)*(1-x)=1-x^6 Case 1: if x<1 1-x^6<1 -x^6<0 This means if x<1, then any value of x, except 0, will be sufficient to establish the inequality in the prompt is true. Case 2: if x>1 -x^6>0 Notice this is not possible so if we establish x>1, then we establish the inequality in the prompt is false. 1) x>0 means f can be in either of the first two cases. (x can be > or < than 1) 2) x<1 This confines us to case 1, where any value of x except 0 works. However, x can still be zero so 2) by itself is insufficient. 1)+2) Sufficient, because 1) tells us x can't be zero and 2) tells us x<1. Answer: C

Bunuel · Answer

Official Solution:

Is  1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}?   
 
Let's simplify the question:
 
Is  1+x+x^2+x^3+x^4+x^5-\frac{1}{1-x} < 0? 
 
Is  \frac{(1+x+x^2+x^3+x^4+x^5)(1-x)-1}{1-x} < 0? 
 
Is  \frac{(1+x+x^2+x^3+x^4+x^5-x-x^2-x^3-x^4-x^5-x^6)-1}{1-x} < 0? 
 
Is  \frac{-x^6}{1-x} < 0? 
 
Is  \frac{x^6}{x-1} < 0? 
 
For  \frac{x^6}{x-1}  to be negative  x  must not be 0 (because if  x=0 , then  \frac{x^6}{x-1}  would be 0) AND  x-1  must be negative. In this case we'd have  \frac{x^6}{x-1}=\frac{positive}{negative}=negative . So, the question basically asks whether  x < 1   ( x -1 < 0  means  x < 1  ) ,  bearing in mind that  x  cannot be 0 .
 
(1)  x > 0 . Not sufficient. 
 
(2)  x < 1 . Not sufficient because if  x = 0 , then  \frac{x^6}{x-1}=0 , not   < 0  . 
 
(1)+(2) When combining the statements we get  0 < x < 1 , so  \frac{x^6}{x-1} < 0  . Sufficient.

Answer: C

GMATGuruNY · Answer

Critical points occur when the two sides of an inequality are EQUAL or when the inequality is UNDEFINED.
Here:
The two sides of the inequality are equal when x=0.
The right side is undefined when x=1.

To determine valid ranges for x, test one value to the left of and one value to the right of each critical point.
If we test x=-1, x=1/2, and x=2,  1 + x + x^2 + x^3 + x^4 + x^5 < \frac{1}{1 - x}  when x=-1 (implying that x<0 is a valid range) and when x=1/2 (implying that 0<x<1 is a valid range).
Since the inequality is valid only if x<0 or if 0<x<1, the question stem can be rephrased as follows:
Is x a nonzero value less than 1?

Only when the two statements are combined is it guaranteed that x will be a nonzero value less than 1.

C

Stormik1 · Answer

1. Simplify the expressionThe left-hand side is a geometric series with 6 terms, a first term of \(1\), and a common ratio of \(x\). For \(x \neq 1\), it can be rewritten using the geometric sum formula:\(1+x+x^{2}+x^{3}+x^{4}+x^{5}=\frac{1-x^{6}}{1-x}\)

The question asks whether:\(\frac{1-x^{6}}{1-x}<\frac{1}{1-x}\)

2.Analyze Statement 1Statement (1) tells us that \(x > 0\).If \(x = 0.5\), then \(\frac{1 - 0.5^6}{1 - 0.5} < \frac{1}{1 - 0.5}\) is True.If \(x = 2\), then \(\frac{1 - 64}{1 - 2} = 63\), and \(\frac{1}{1 - 2} = -1\). Since \(63 < -1\) is False, this statement does not yield a consistent yes/no answer.Statement (1) alone is not sufficient.

3. Analyze Statement 2Statement (2) tells us that \(x < 1\), which means \((1 - x) > 0\). We can multiply both sides of the inequality by the positive value \((1 - x)\) without flipping the sign:\(1-x^{6}<1\implies -x^{6}<0\implies x^{6}>0\)For any non-zero value of \(x\), \(x^6 > 0\) is True.If \(x = 0\), then \(0^6 > 0\) is False (the two sides become equal: \(1 = 1\)).Because \(x = 0\) is included in \(x < 1\), Statement (2) alone is not sufficient.

4. Combine both statementsWhen we combine both statements, we get \(0 < x < 1\).Since \(x > 0\), \(x\) cannot be \(0\), ensuring that \(x^6 > 0\) is always strictly true.Since \(x < 1\), \((1 - x)\) is always strictly positive.Thus, the inequality always holds true, providing a definite "Yes".

Is 1 + x + x^2 + x^3 + x^4 + x^5 < 1/(1 - x) ? (1) x > 0 (2) x < 1

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