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Best answer
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7 votes
$p\vee (p\wedge q)`$ = p+p`+q` = 1+q`= 1 so tautology.

C and b cant be tautology so no option is matching.
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–1 votes

p∨~(p∧q)

p∨~(p∧q) =p ∨ ~p v~q

p ∨ ~p is always true

a is a tautology

 (p∧~q)∨~(p∧q)

(p∧~q)∨~(p∧q) = (p∧~q)∨(~pv~q) is also tautology

 p∧(q∨r)

 p∧(q∨r) =  (p^q) v (p^r) is not a tautology

Answer 

C  a and b

Answer:

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