(a) Factorise : \(px - 2px - 4qy + 2py\)
(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \((P \cup Q)' = P' \cap Q'\)
(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \((P \cup Q)' = P' \cap Q'\)
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Correct Answer: Option n
Explanation:
(a) \(px - 2qx - 4qy + 2py\)
= \(x(p - 2q) + 2y(p - 2q)\)
= \((x + 2y)(p - 2q)\)
(b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\)
\(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\)
\((P \cup Q) = {1, 2, 3, 4, 6, 9, 10}\)
\((P \cup Q)' = {5, 7, 8}\)
\(P' \cap Q' = {3, 5, 7, 8, 9} \cap {1, 4, 5, 7, 8, 10}\)
= \({5, 7, 8}\)
Hence, \((P \cup Q)' = P' \cap Q'\).
(a) \(px - 2qx - 4qy + 2py\)
= \(x(p - 2q) + 2y(p - 2q)\)
= \((x + 2y)(p - 2q)\)
(b) \(U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}\)
\(P = {1, 2, 4, 6, 10} ; Q = {2, 3, 6, 9}\)
\((P \cup Q) = {1, 2, 3, 4, 6, 9, 10}\)
\((P \cup Q)' = {5, 7, 8}\)
\(P' \cap Q' = {3, 5, 7, 8, 9} \cap {1, 4, 5, 7, 8, 10}\)
= \({5, 7, 8}\)
Hence, \((P \cup Q)' = P' \cap Q'\).