Correct Answer - B
If we assume that `clta` and `cltb` then we observe that options (a) and ( c ) do not hold.
In order to check option (b), let us consider the following cases:
CASE I When `altb`
In this case, we have
`min(a,b)=b" and "|a-b|=a-b`
`:." "(1)/(2){a+b-|a-b|}=(1)/(2){a+b-(a-b)}=b`
Hence, `min(a,b)=(1)/(2){a+b-|a-b|}`
Thus, in both the cases, we have
`min(a,b)=(1)/(2){a+b-|a-b|}`
Hence, option (b) is true.