A and B

Algebra Level pending

There exists non-negative integers $a$ and $b$ such that for any $x$ satisfying $a<x<b$ , it is true that the square root of $x$ is greater than $x$ . Find the ordered pair ( $a$ , $b$ ).

(0, π/6) (-1,1) (2,4) (0,1)

2 solutions

Viki Zeta
Mar 19, 2017

Since a and b are integers, options (-1,1) and $(0,\dfrac{\pi}{6})$ are excluded. Now for (2,4). Take 3, 3<3<4, $3^3=9>3$ , So the only option left is (0,1)

Clarify that a and b are specifically non-negative integers. Also fix 3^3=9. Otherwise, great use of elimination!

Amogha Pokkulandra - 4 years, 2 months ago

Amogha Pokkulandra
Mar 18, 2017

We can rewrite $√x$ > $x$ as $x>(x^2$ ). The only number that satisfy this are positive decimals that lie in between 0 and 1. Therefore $a$ is equal to 0 and $b$ is equal to 1. Therefore the ordered pair ( $a$ , $b$ ) is equal to $\boxed{(0,1)}$ .

A and B

2 solutions

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