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

I have a query - If I take (C)
--> it means x = 0.5, 0.3, 0.2
With x = 0.5
1+ x + x sq + x^3 + x^4 < 1/1-x
1+ 0.5+ 0.25 + 0.125 + 0.0625 = 1/1-0.5
1.93< 2 SUFF
if x= 0.3
1+0.3+0.09 + 0.027 + 0.0081 = < 1/1 - 0.3
1.42 no less than 0.7
if x = 0.2
1+0.2+0.04 + 0.008 + 0.0016 <1/ 1-0.2
1.24 not less than 0.8
So I chose E.
Can anyone explain the issue in my reasoning here?

It's \(\frac{1}{1 - 0.3}\) NOT \(\frac{1}{1}- 0.3\).

\((1 + 0.3 +0.3^2 + 0.3^3 + 0.3^4 = 1.4251)< (\approx 1.428=\frac{1}{1 - 0.3})\)

\((1 + 0.2 +0.2^2 + 0.2^3 + 0.2^4 = 1.2496)< (1.25=\frac{1}{1 - 0.3})\)

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

Good Q MathRevolution

\(1+x+x^2+x^3+x^4<\frac{1}{(1-x)}\)..
\(1+x+x^2+x^3+x^4=\frac{1(1-x^5)}{1-x}\).... sum of a Geometric Progression

so \(\frac{1-x^5}{1-x}<\frac{1}{1-x}..........\frac{1-x^5}{1-x}-\frac{1}{1-x}<0.........\frac{1-x^5-1}{1-x}<0.............-\frac{x^5}{1-x}<0\)
so both numerator and denominator are of opposite sign..
two cases
(A) 1-x<0 or x>1,
so\(-x^5 >0........... x^5<0........x<0\)
x cannot be <0 and >1 at same time
Not possible
(B) 1-x>0 or x<1,
so\(-x^5 <0........... x^5>0........x>0\)
so x lies between 0 and 1

lets see the statements..
(1) \(x>0\)
(2) \(x<1\)

combined both tells us that x lies between 0 and 1
suff

C

In case one doesn't know about or remember the GP series as explained by chetan2u, then you can use below approach -

\(1+x+x^2+x^3+x^4<\frac{1}{(1-x)}\)---------------(1)

Case 1: \(x>1\), for eg. \(x=2\), then LHS of the above inequality will be greater than \(2\)

but RHS of inequality (1) will be negative. So we have a NO for this case

Case 2: \(0<x<1\), for eg. \(x=\frac{1}{2}\), then LHS of inequality will be \(1+0.5+0.25+0.125+0.0625<2\)

but RHS of the inequality will be \(\frac{1}{(1-0.5)}=2\). So we have a YES for this case

Case 3: \(x<0\), for eg. \(x=-1\), then LHS of the inequality will be \(1-1+1-1+1=1\)

but RHS of the inequality will be \(\frac{1}{(1+1)}=0.5\). So we have a NO for this case

With these understandings, we can now check the statements

Statement 1: from this statement, both Case 1 & Case 2 are possible. Hence we have a Yes & a No. Insufficient

Statement 2: from this statement, both Case 2 & Case 3 are possible. Hence we have a Yes & a No. Insufficient

Combining 1 & 2, we have \(0<x<1\), so only Case 2 possible, hence we have a Yes. Sufficient

Option C

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.

Modifying the question:
\(1+x+x^2+x^3+x^4<\frac{1}{(1-x)}\)
\(⇔ (1+x+x^2+x^3+x^4)(1-x)^2<(1-x)\)
\(⇔ (1-x^5)(1-x)<(1-x)\)
\(⇔ 1-x^5-x+x^6<1-x\)
\(⇔ x^6-x^5<0\)
\(⇔ x^5(x-1)<0\)
\(⇔ x(x-1)<0\)
\(⇔ 0<x<1\)

The question asks if \(0<x<1\).
This question has the unique answer, ‘yes’, if we require both \(x>0\) and \(x<1\).
Thus, both conditions together are sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Answer: C

Hi chetan2u

Sum of an GP is a\((r^n - 1)/r - 1\) for r > 0 and for r < 0 Sum of an GP is a \((1-r^n)/1-r\)

Now, from Stmnt 1: As x > 0, \(x^5 - 1/x-1 > 1 / 1- x\)
\(x^5 - 2/ x - 1 > 0\)

As x > 0, Can we make the Statemnt 1 insufficient based on x =1.. Please advise. Other than 1 on which values of x, statement 1 would be insufficient. I can only think 1.

Please let me know if there is anything wrong with my approach.

approach is correct but you have made two errors..
1) while solving \(\frac{x^5 - 1}{x-1} > \frac{1}{1- x}\), you have not made th edenominator equal you have taken x-1 same as 1-x and got answer \(\frac{x^5 - 2}{x - 1} > 0\)..
\(\frac{x^5 - 1}{x-1} > \frac{1}{1- x}........\frac{x^5 - 1}{x-1} - \frac{1}{1- x}>0.............\frac{x^5 - 1}{x-1} -(- \frac{1}{x-1})>0.........\frac{x^5 - 1}{x-1} + \frac{1}{x-1}>0.............\frac{x^5}{x-1}>0\)

2) even if you have an equation \(\frac{x^5 - 2}{x - 1}> 0\), the values will be x^5>2, so 1 or 1.05 etc will not hold true as x need not be integer

I would be endlessly grateful, if any expert could help me see a mistake in my method. For me the answer is A.

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

1. x>0

Additionally, x cannot be 0 because in case we plug in 0 we get 1<1; Therefore, we can devide both sides of the inequality by x.

1+x+x^2+x^3+x^4<1/(1-x)
x+x^2+x^3+x^4<1/(1-x)-1
x+x^2+x^3+x^4<x/(1-x) (devide both sides by x)
1+x+x^2+x^3<1/(1-x)
x+x^2+x^3<1/(1-x)-1
x+x^2+x^3<x/(1-x) (devide both sides by x)
1+x+x^2<1/(1-x)
x+x^2<1/(1-x)-1
x+x^2<x/(1-x) (devide both sides by x)
1+x<1/(1-x)
x<1/(1-x)-1
x<x/(1-x) (devide both sides by x)
1<1/(1-x)
0<x/(1-x)

th values x can accept here 0<x/(1-x) are within 0<x<1 A. sufficient

Hi anud33p,

Actually we can while analyzing St1, as I did in my previous post. St 1 says that x>0, so we are sure that x doesn’t affect the inequality. I’m dividing both sides by x, considering St 1, so that’s fine. I keep dividing both sides until I get much much simple inequality x/(x-1)<0.

What I failed to do in my rather old post is to infer the correct conclusion. Since x/(x-1)<0 and x*(x-1)<0 have the same set of solutions, the answer is 0<x<1. We can see that for the original inequality to hold true x needs to be both greater than 0 and less than 1. We get that constraint only when we combine both statements, so the correct answer is C.

To my mind that’s a much easier way to come to the correct answer. We just need to keep dividing both sides by x until we get a simple inequality to manipulate.

Similar questions to practice:
https://gmatclub.com/forum/if-x-0-is-1- ... 56227.html
https://gmatclub.com/forum/is-1-x-x-2-x ... 56226.html

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