well the function y+|2-x|=2
can be broken into two parts
for x>2 x+y=4
and y=x for x<2
x+y=4 intersects at x=3
y=x intersects at sqroot(3)
so required area is
[sqroot(3) ]
[2] (x- 3/x) .dx +
[2 ]
[3] (4-x- 3/x) .dx
=4-3log3/2
hence (a) option is correct
bye!!