Explanation
propositional logic formulas is TRUE when exactly two of p,q and r are TRUE means possibilities are pq or qr or pr to be TRUE
option A)
((p↔q)∧r)∨(p∧q∧∼r)
=((T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
=((T)∧r)∨(T∧∼r)
=(r)∨(∼r)
=T
It doesn't depend upon the value of r, so, if the value of r is T then it also to be T so the condition exactly 2 are T is false. Hence option A is FALSE.
option B)
(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
=(∼(T)∧r)∨(T∧∼r)
=(F∧r)∨(T∧∼r)
=(F)∨(∼r)
=∼r
so if r=F then this is T so condition exactly 2 are T is true
(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔q)∧T)∨(T∧q∧∼T) ( take p=T and r=T)
=(∼(q)∧T)∨(F)
=∼(q)∨(F)
=∼q
so if q=F then this is T so condition exactly 2 are T is true
(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(p↔T)∧T)∨(p∧T∧∼T) ( take q=T and r=T)
=(∼(p)∧T)∨(F)
=∼(p)∨(F)
=∼p
so if p=F then this is T so condition exactly 2 are T is true
Hence option B is CORRECT.
option C)
((p→q)∧r)∨(p∧q∧∼r)
=((T→T)∧r)∨(T∧T∧∼r) ( take p=T and q=T)
=((T)∧r)∨(T∧∼r)
=(r)∨(∼r)
=T
It doesnt depend upon value of r , so , if value of r is T then it also to be T so the condition exactly 2 are T is false . Hence option C is FALSE.
option D)
(∼(p↔q)∧r)∧(p∧q∧∼r)
=(∼(T↔T)∧r)∧(T∧T∧∼r) ( take q=T and r=T)
=(∼(T)∧r)∧(T∧∼r)
=(F∧r)∧(T∧∼r)
=F∧(∼r)
=F
It doesnt depend upon value of r , so , if value of r is F then it also to be F so the condition exactly 2 are T is false . Hence option D is FALSE.