gmatrant, request u to post one Qs per post! However answer for the first Qs -
IMO E. Because in the Qs, it has not been mentioned that a not = 0 or b not = 0.
For Qs 2, IMO B. Because -x|x| > 0 means x is -ve number. So it suits the condition mentioned in the Qs - sqrt(3 - x)^2 = 3 -x - This is only possible when the x is -ve.
Are u getting me, gmatrant?

camitava wrote:gmatrant, request u to post one Qs per post! However answer for the first Qs -
IMO E. Because in the Qs, it has not been mentioned that a not = 0 or b not = 0.
For Qs 2, IMO B. Because -x|x| > 0 means x is -ve number. So it suits the condition mentioned in the Qs - sqrt(3 - x)^2 = 3 -x - This is only possible when the x is -ve.
Are u getting me, gmatrant?

Thanks a lot camitava..

The first problem is clear.

Second Problem
Case 2:
where we find out x is negative. Now hereafter I assume x as -5
so when I substitute they don't match
sqrt[(x-3)^2] = 3-x
sqrt[(-5-3)^2] = 3+5
sqrt[64] =8
LHS can be + or - 8 ... while RHS is 8 ..so condition does not match..
hence E.
Well the OA is B,but I fail to understand.

Pls explain.
Thanks
gmatrant

gmatrant, dnt take the second Qs as a equ rather take the RHS as value of LHS. Now with ur example, as x has been decided that it is -ve, so the expression - sqrt((x - 3)^2) becomes sqrt((-x -3)^2). Now if x is +ve, sqrt((x -3) ^ 2) will come to sqrt((x - 3) ^ 2). With x = 5, the exp becomes sqrt(4) whereas sqrt((-5 - 3)^2) becomes sqrt(64) which somehow comes equal to RHS. Are u getting me, gmatrant?

camitava wrote:gmatrant, dnt take the second Qs as a equ rather take the RHS as value of LHS. Now with ur example, as x has been decided that it is -ve, so the expression - sqrt((x - 3)^2) becomes sqrt((-x -3)^2). Now if x is +ve, sqrt((x -3) ^ 2) will come to sqrt((x - 3) ^ 2). With x = 5, the exp becomes sqrt(4) whereas sqrt((-5 - 3)^2) becomes sqrt(64) which somehow comes equal to RHS. Are u getting me, gmatrant?

camitava, sorry for the trouble, but I did not get what you have written.
Is it possible for you to give this is a detailed fashion with some comments by the side for me to follow, will be of great help.