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

*w**x**y*'=12,13 1100,1101 here you will see it is also one bit different

*x**y*'*z**=5,13 0101,1101 *here you will see it is one bit different

*xy**z*=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

*x**y*'*z**=5,13 0101,1101 *here you will see it is one bit different

*xy**z*=7,15 0111, 1111 here you will see it is one bit different

*w**x**y*'=12,13 1100,1101 here you will see it is also one bit different

So option d satisfy whole f so it is Answer .