given d>0 and 0<1-c/d<1>0 C is also > 0

So I must be true

c^2 + d^2 is some thing like 1.01d^2 to 1.81d^2 If d=1/10 we will have a value less than 1. If d is 1 or more we will have a value greater than 1. So it need not be true.

I & II must be true and hence C

Chidec is correct.

c has to be less than d in order for c/d to be less than 1. However, if d is a decimal (or some fraction), then c can still be a decimal less than d.

When you square both of them, they get smaller, and do not need to add to 1.

For example
d = 0.5 = 1/2
c = 0.25 = 1/4

c / d = . 5

c^2 = 1/16 d^2 = 1/4 c^2+d^2 = 5/16 < 1

I admit, I got this wrong, so it was a good reminder problem. You just have to always look at different types of answers, namely negative, positive, zero, and fractions when solving questions

In this case, 0 and negative is already ruled out. but I just imagined positives greater than 1, rather than including the fractions.

I will go with option B as the answer..only 2 is correct

0 < 1 - c/d <1> -1 < -c/d <1> -1 < c/d <1>0

=> -d < c < d

Thus c can be +ve or -ve

and we clearly see that c/d < 1

nothing can be said abt c^2+ d^2

so B will be the answer

Neo

I can't tell what you did, but 'I' is pretty obvious if you think about it this way:
if 1-(c/d)<1, then c obviously can't be 0 or negative.

algebraically:
1-(c/d)<1
1<1+(c/d)
0<(c/d)
0<c

getneonow

Here is by your approach I think when u subtracted 1 then you didnt apply it to the 0

0 < 1 -c/d < 1
-1 < -c/d <0> c/d > 0

As d > 0 , for c/d to be betweeen 0 and 1
1. c < d (Fraom c/d <1> 0 (From c/d > 0)

Also that doesnt imply anything about values of C nad D

So we can have c = 2 and d = 4 where c^2+ d^2 > 1
or c = 0.4 and d= 0.5 where c^2+ d^2 < 1

So C is the answer