1.If the truth value of the Boolean expression((p∨q)∧(q⇒r)∧(∼r))⇒(p∧q)((p \vee q) \wedge (q \Rightarrow r) \wedge (\sim r)) \Rightarrow (p \wedge q)((p∨q)∧(q⇒r)∧(∼r))⇒(p∧q)is false, then the truth values of the statements p,q,rp, q, rp,q,r respectively can be:a.F T Fb.T F Fc.T F Td.F F TLogin to continueOnly logged in users canattempt or see the solution.