I found a better way to answer this question .
Go according to the Question . i1 and i2 differs in exactly one bit position without becoming false momentarily.
Go to option (a)
Draw K map and find the binary covered by implicant
w'xz =5,7 0101 , 0111 here you will see it is one bit different
wxy'=12,13 1100,1101 here you will see it is also one bit different
xy'z=5,13 0101,1101 here you will see it is one bit different
xyz=7,15 0111, 1111 here you will see it is one bit different
wyz=11,15 1011,1111 here you will see it is one bit different
Option (a) misses wxz possible one bit change.So it is false.
(b)
wxy = 15,14 but 14 is not covered by K map so b is wrong option
(c)
wxy'z' = 12 =1100 here there is not bit change so it is wrong option
(d)
wyz=11,15 1011,1111 here you will see it is one bit different
wxz=13,15 1101,1111 here you will see it is one bit different
w'xz =5,7 0101 , 0111 here you will see it is one bit different
xy'z=5,13 0101,1101 here you will see it is one bit different
xyz=7,15 0111, 1111 here you will see it is one bit different
wxy'=12,13 1100,1101 here you will see it is also one bit different
So option d satisfy whole f so it is Answer .