Answer: A
s^2 + t^2 < 1 - 2st
A: 1-s
B: t

s^2 + t^2 + 2st < 1
(s+t)^2 < 1
(s+t) < 1
t < 1-s
So A is bigger than B.

\(s^2\) + \(t^2\) \(< 1 - 2st\)

Take 2st to left hand side we get
\(s^2\) + \(t^2\) + 2st < 1
=> \(s^2\) + 2st + \(t^2\) < 1
This is of the form \((a+b)^2\) = \(a^2\) + 2ab + \(b^2\)
=> \((s+t)^2\) < 1
Using \(a^2\) < 1 => -1 < a < 1, we get
- 1 < s + t < 1

Take s+t < 1
Take s to right hand side we get
t < 1 - s
=> 1 - s > t

=> Quantity A (1-s) > Quantity B (t)

So, Answer will be A
Hope it helps!

To learn more about inequalities watch this video

from what is given; we know that s^2+t^2+2st<1
therefore, t and s must of course be decimals.

Option (A) 1-s Option (B) t by adding both side S
(A) 1
(B) t+s
if t+s can equal 1, then the equation above is wrong. You can try yourself
then (a) must be true

s^2 + t^2 + 2st < 1
(s + t)^2 - 1 < 0

(s + t + 1)(s + t - 1) < 0

its in form of (a + 1)(a - 1) < 0. Let a = s + t

-1 < a < 1

-1 < s + t < 1

-1 - s < t < 1 - s

t < 1 - s

Note: (s + t)^2 < 1
(s + t) < 1. I think this is wrong.
Anyone clarify last part in note
Carcass

No sir

it is correct

\(s^2+t^2+2st<1\)

It is a double product

\(s^2+t^2<1\)

square both sides

\(\sqrt{s^2+t^2} < \sqrt{1}\)

\(s+t<1\)

\(s-1<-t\)

or

\(-s+1>t\)

or

\(t<-s+1\)

which perfectly matches the two quantities

So A is > B

hard question but in the end pretty silly

The right form is this

\((a+b)^2\) = \(a^2\) + 2ab + \(b^2\)

not this

\((a + 1)(a - 1)\)

see more here https://gre.myprepclub.com/forum/gre-ma ... tml#p81624

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