Is xy < 4 ? (1) x^2 + y^2 < 8 (2) x y < 1

The question asks is xy < 4?



Statement 1: \(x^2 + y^2 <8 \)

Let us first solve this using a logic that is likely to be helpful in the exam. And then another approach that is a proper mathematical way in the end.

We know if the sum of two numbers is given then the product becomes maximum when they are equal.

For example if we know a + b = 10 then ab will be maximum when a = b = 5.

(This is an important concept to know. In case anyone is not aware of this, you can ask here.)

So let us start with \(x^2 + y^2 = 8 \)

Now this tells me that the product of\(x^2 y^2\) will be maximum when \(x^2 = y^2 = 4\)
So we can say that the maximum value of \(x^2 y^2\) can be 4 × 4 = 16
So the maximum value of xy will be 4

Now all that we have done assuming \(x^2 + y^2 = 8 \) but since \(x^2 + y^2 <8 \) so definitely for this question \(x^2 = y^2 < 4\)

Thus, \(x^2 y^2 < 16\)
And \(xy < 4\)

Statement 2: x - y < 1

Statement 2 analysis should be easier for most.

The easiest way is to assume values logically.

How should we assume values? We should try to take two scenarios where we get one YES and one NO to our answer.

One easy way to get yes as the answer is when we take them very small values. In fact we can just assume x = 0 and y = some number like 1, 2, 3 etc.
To get NO, we need to take big values for both x and y and thus we can take x = 20, y = 21, or something like this

Thus we get A as the answer.


Now another mathematical way for statement 1 could be as below.

We are looking for xy > 4 or not or this can be written as xy - 4 > 0 or not.

\(x^2 + y^2 <8 \)
OR, \(x^2 + y^2 - 2xy <8 - 2xy \)(I take - 2xy as we need (xy - 4) as my unit and -2xy does that
Or, \(2xy <8 - {(x - y)}^2 \)


Thus we can say \( xy < 4 - \frac{{(x - y)}^2}{2} \)
Since, \(\frac{{(x - y)}^2}{2} \) must be 0 or more thus 4 - \( \frac{{(x - y)}^2}{2} \) must be less than 4.
So yes, xy < 4.

It's good to come out with such a way but may not always happen in the exam.