Correct Ans: D

(a)= sqrt of a^2+b^2
(b)= sqrt of a^2 = a & sqrt of b^2 = b
therefore, (b) = a+b

Lets take an example: a = 2 & b=3
(a)=sqrt of {2^2 + 3^2}
= sqrt of {4+9}
= sqrt of 13
= 3.6

(b)= sqrt of {2^2} + sqrt of {3^2}
= sqrt of {4} + sqrt of {9}
= 2+3
= 5

Here, B is greater than A

Lets take another example, a =-2 & b=3

(a)= sqrt of {(-2)^2 + 3^2}
= sqrt of {4+9}
= sqrt of 13
= 3.6

(b)= sqrt of (-2)^2 + sqrt of (3)^2
= -2+3
= 1

Here, A is greater than B

Therefore, D