ps17

Imo C

whats the OA?

i more for C

if v take C=1/2
D= 2/3

raunekk wrote:i more for C

if v take C=1/2
D= 2/3

Hi raunekk,


if we consider c=1/2 and d = 2/3
then c/d = (1/2 * 3/2) = 3/4

now c= 3 and d = 4

so c^2 + d^2 > 1


please correct me if I am wrong.

thanks,
Ket

.
0 < 1 - c/d < 1
subtract 1 from all sides

-1 < -c/d < 0

multiply all sides by d............as we know d>0, it wont affect the signs or inequality

-d < -c < 0

multiply all sides by -1...........as you do that the signs would change

d > c > 0

this means c > 0 and c< d

you can see I and II will always be true..a direct deduction from ineuality

for III you can prove it wrong by using any two +ve fractions, whose sum is less than or equal to 1

ket_gmat wrote:

raunekk wrote:i more for C

if v take C=1/2
D= 2/3

Hi raunekk,


if we consider c=1/2 and d = 2/3
then c/d = (1/2 * 3/2) = 3/4

now c= 3 and d = 4

so c^2 + d^2 > 1


please correct me if I am wrong.

thanks,
Ket

hi Ket..

if we once take C=1/2 and D=2/3

then we cant change d values of C and D again as 3 and 4 respectively(ie if we simplify)

the values will remain the same,,,

thus 1/4 + 4/9 = 25/36 <1

correct me if i m wrong,,,

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


I) c > 0 - This (I) must be true because if either c is o and c is negative then the inequality would be false.

Eg: c=0

0 < 1 - 0 / d < 1 i.e 0<1<1 - Not possible

c= -ve

0<1+somevalue<1 - Not possible

2) 0 < 1 - c / d < 1

Add c/d to all sides of the inequality

c/d<1<1+c/d

II must be true (proved above)

3)Consider fractional values for c and d

Let c=1/4 d=1/2

0<1- (1/4) / (1/2) < 1
0< 1/2<1

c^2+d^2 < 1 and not > 1

Therefore C) Hope this helps!


OA?