shashank.ism wrote:If x, xy, xy2 are the sides of a triangle, where x and y are real numbers and y ≥ 1, then which of the following is the value that y cannot take?

a)1
b)2
c)3/2
d)1.2
e)None of these

xy2 > xy + x
=>xy2 > x(y+1)
=>y2-y-1 >0
Discriminant = 1^2 -4x1x-1 = 5
roots are (1+sqrt5)/2 and (1-sqrt5)/2
y>(1+sqrt5)/2 or y<(1-sqrt5)/2

Hence,1 A

SIDES ARE X, XY , XY^2

A) IF Y=1--->SIDES ARE X,X,X POSSIBLE

B) IF Y=2 SIDES ARE X,2X,4X NOT POSSIBLE AS LNTH OF LONGEST SIDE< SUM OF OTHER TWO SIDE BUT
4X>3X(SUM OF OTHER TWO SIDE ; HENCE NOT POSSIBLE

CORRECT

C)Y=3/2
POSSIBLE
D)Y=1.2
LNTH OF LONGEST SIDE>DIFF OF OTHER TWO SIDE--->1.44X>0.2 X TRUE
AND LNTH LONGEST SIDE < SUM OF OTHER TWO SIDE---->1.44X<2.2X TRUE

HENCE b