ISI 2017

Question

I find alpha <x+y which gives me x+y<2. But the answer is A. Can someone please help.

 

Consider the statement$:$

$x(\alpha-x)<y(\alpha-y)$ for all $x,y$ with $0<x<y<1.$

The statement is true

• A. if and only if $\alpha\geq 2$
• B. if and only if $\alpha >2$
• C. if and only if $\alpha <-1$
• D. for no values of $\alpha$

Answer 1

$x(\alpha-x) < y(\alpha-y)$

$\alpha x-x^2 -\alpha y+y^2<0 $

$\alpha(x-y)-(x-y)(x+y)<0 $

$(x-y)(\alpha-(x+y))<0 $

$x<y \implies \alpha>(x+y)$

$x+y<2 \implies \alpha\geq 2$

Comments

thanks got it. I overlooked the x<y condition.

Answer 2

I took x as 1/4 and y as 1/2 and got α>5/4,so option a is correct.

Comments

Can you please explain how do you get a(5/4)>x+y(1/2 + 1/4)