If x is a prime number, what is the value of x + c?
(1) \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = - |c|\)
The RHS - |c| can only be possible as 0, because the LHS has SUM of a modulus and a square root, both of which are positive. so it will be positive and -|c| cannot be positiveNow, the LHS, \(|a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} =0\) , but since both terms are positive, they should be 0 themselves too..
so a + b + c - 2x=0......a+b+c=2x...(i)
Also \(\sqrt{(x^2 - a - b - c)^2} =0......=>x^2 - a - b - c=0.........x^2=a+b+c\)...(ii)
From (i) and (ii), we get \(a+b+c=2x=x^2........x^2-x=0....x(x-2)=0\), so x is 0 or 2
But x is a prime number, so x+c=2+0=2
sufficient
(2) \(\sqrt{(a + b + c - 1 + x)^2} + |\frac{1}{x-1}- a - b - c| + \sqrt{c^2}=0\)
All terms on LHS are positive, so their SUM can only be 0 when all terms are 0..\(\sqrt{c^2}=0, so c=0.\)
\(|\frac{1}{x-1}- a - b - c|=0..............|\frac{1}{x-1}= a + b|\)...(i)
\(\sqrt{(a + b + c - x+1)^2}=0.........a + b + c - x+1 =0..........a+b=x-1.\)...(ii)
from (i) and (ii).. \(\frac{1}{x-1}= a + b=x-1........x-1=\frac{1}{x-1}......(x-1)^2=1\), so x can be 0 or 2
But x is prime number so 2
hence x+c=2+0=2
suff
D
Note:-
Bunuel, there seems to be a typo in statement II.
Either it is 1/(1-x) and not 1/(x-1) or it is a+b+c-x+1 and not a+b+c-1+x...
Pl relook
_________________