The question is a bit tricky. However, it could be solved without picking numbers but just with logic

Now, x is positive and squared is a fraction between 0 and 1

But if a fraction squared is between 0 and 1 that means the original fraction - notice how is mentioned that x>0 at the origin - is still between zero and one but is bigger than the squared one

\(x^2<x\)

For example \(x=\frac{1}{2}\)

\(x^2=\frac{1}{4}
\)

\(\frac{1}{4}<\frac{1}{2}\)

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

I fixed the OA because the book reported D as OA but indeed is A

x must be a positive fraction!

Col. A: \(1-x^2\)
Col. B: \(1-x\)

Adding \(x^2\) and \(x\) to both sides;

Col. A: \(1 + x\)
Col. B: \(1 + x^2\)

Subtracting \(1\) from both sides;

Col. A: \(x\)
Col. B: \(x^2\)

Dividing both sides by \(x\);

Col. A: \(1\)
Col. B: \(x\)

Hence, option A

Hello - Couldnt we also just factor A to the following:

(1+x)(1-x)

Then divide each A & B by (1-x) leaving us:

A: 1+x
B: 0

We know X is positive, so A is greater. Is this allowed?

yes this is another and faster approach

Hi nurakib!

Apologies! Missed this one! Updated the explanation!


Kind regards

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