For any fraction \(x\), non integer decimal part = \(x - 1\) (example non-integer decimal part of 1.2 = 1.2 - 1 = 0.2)

The reciprocal part of \( x\)'s non-integer decimal part equals \(x + 1\)

=> Reciprocal of \(( x - 1 )\) = \(x + 1\)

=> \(\frac{1}{(x - 1)}\) = x + 1

cross multiply, we get,

=> 1 = ( \(x\) - 1 ) ( \(x\) + 1 )

Using ( a - b ) * ( a + b ) = \(a^2\) - \(b^2\)

=> 1 = \(x^2\) - \(1^2\) = \(x^2\) - 1

=> \( x^2 \)= 1 + 1 = 2

=> \(x\) = +√2 or \(x\) = -√2

But \(x\) > 0

=> \(x\) = √2

Ans: C

the explanation won't hold for x>2

Did you want to say x > 2 or x > 0? and give us the reason why it not hold (whatever u said)..............?

In such difficult question, we should ask ourselves WHY a numeric value is given other sector.
Lets play with calculator,
√2=1.4142.......
Then 1.4142.......-1=0.4142.......
then M+
Then 1/MR=2.4142...... which is Exactly √2+1.
Everything matches. So, X=√2.
c

You are purporting that x can have only 1 possible value, which is true for 1<x<2 (I'm going to assume the phrasing excludes 0<x<1) but as soon as you go above that range, x can have multiple values, which are the positive solutions for:

\(\frac{1}{(x - n)}\) = x + 1

where n={2,3,4,5...}

So, the answer should be D as there is no information allowing us to tell whether x is √2 or some other value.

The given answer is wrong.

Assume \(x = N+f\), where \(N\) is a non-negative integer (because \(x\) is positive) and \(f\) is the fractional part ranging between \(0\) and \(1\). According to the question, reciprocal of non-integral part of \(x\) is \(x+1\), which according to how we have defined \(x\) can be written as \(1/f = x+1 = (N+1) + f\)

Multiply both sides by \(f\) to get \(1=(N+1)f+f^2\).

This is a simple quadratic equation whose roots can be found by using the quadratic formula.

The positive root will be \((-(N+1)+\sqrt{(N+1)^2+4})/2\) and this will always lie between \(0\) and \(1\). Thus, we have a value of \(f\) independent of \(N\), which implies value of \(x\) is not fixed.

Hence, D is the correct option. eg - consider the numbers 2.30277563773199464655961063373525 or 0.61803398874989484820458683436564 (this is the golden ratio, btw).

PS:

An easier method is to realize that the function \(1/t\) when \(t\) is between \(0\) and \(1\) has the range from \(1\) to infinity. Thus, we can always create the desired number \(x\) because the number \(x+1\) will itself be in the range of the function \(1/t\)

Thank you. Fixed.

Regards

If the condition given was 1<x<2, then what would have been the answer? Was it going to be C?

the fourth root of 40=2.51..
the calc. shows 2.445..

Exactly. So how can this be solved algebraically?

In the previous replies it is solved algebraically such as this one https://gre.myprepclub.com/forum/the-re ... tml#p46710

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.