Written Instructions:
For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..
For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.
Leave your answers in the simplest form or correct to two decimal places.
| 1) Given that U = {r,p,q,d,x,a,e,p}, A = {r,p,a,e}, and B = {r,p,q}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer:_______________ |
| 2) Given that U = {u,c,o,p,f,c,g,e}, A = {u,c,c,g}, and B = {u,c,o}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer:_______________ |
| 3) Given that U = {h,s,n,v,r,x,g,j}, A = {h,s,x,g}, and B = {h,s,n}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer:_______________ |
| 4) Given that U = {x,q,v,d,p,q,c,w}, A = {x,q,q,c}, and B = {x,q,v}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer:_______________ |
| 5) Given that U = {a,a,p,o,l,z,d,x}, A = {a,a,z,d}, and B = {a,a,p}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer:_______________ |
| 6) Given that U = {k,d,o,u,m,a,h,f}, A = {k,d,a,h}, and B = {k,d,o}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer:_______________ |
| 7) Given that U = {k,g,k,u,q,d,c,f}, A = {k,g,d,c}, and B = {k,g,k}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer:_______________ |
| 8) Given that U = {m,r,c,i,g,u,o,m}, A = {m,r,u,o}, and B = {m,r,c}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer:_______________ |
| 9) Given that U = {j,f,i,o,w,j,l,v}, A = {j,f,j,l}, and B = {j,f,i}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer:_______________ |
| 10) Given that U = {l,u,w,r,q,y,m,u}, A = {l,u,y,m}, and B = {l,u,w}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer:_______________ |
| 1) Given that U = {r,v,s,p,z,b,f,t}, A = {r,v,b,f}, and B = {r,v,s}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer: s,p,z,b,f,t SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∩B)‘ = A‘∪B‘ A∩B = {r,v,b,f}∪{r,v,s} = {r,v,s,b,f} (A∩B)‘ = U∖(A∩B) = {r,v,s,p,z,b,f,t}∖{r,v} (A∩B)‘ = {s,p,z,b,f,t} ...... (i) A‘ = U∖A = {r,v,s,p,z,b,f,t}∖{r,v,b,f} A‘ = {s,p,z,t} B‘ = U∖B = {r,v,s,p,z,b,f,t}∖{r,v,s} B‘ = {p,z,b,f,t} A‘∪B‘ = {s,p,z,t}∖ {p,z,b,f,t} A‘∪B‘ = {s,p,z,b,f,t} ....... (2) From (i) and (2), we have (A∩B)‘ = A‘∪B‘ |
| 2) Given that U = {n,t,j,r,q,b,u,k}, A = {n,t,b,u}, and B = {n,t,j}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer: r,q,k SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∪B)‘ = A‘∩B‘ (ii) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∪B)‘ = A‘∩B‘ A∪B = {n,t,b,u}∪{n,t,j} = {n,t,j,b,u} (A∪B)‘ = U∖(A∪B) = {n,t,j,r,q,b,u,k}∖{n,t,j,b,u} (A∪B)‘ = {r,q,k} ...... (i) A‘ = U∖A = {n,t,j,r,q,b,u,k}∖{n,t,b,u} A‘ = {j,r,q,k} B‘ = U∖B = {n,t,j,r,q,b,u,k}∖{n,t,j} B‘ = {r,q,b,u,k} A‘∩B‘ = {j,r,q,k}∖ {r,q,b,u,k} A‘∩B‘ = {r,q,k} ....... (2) From (i) and (2), we have (A∪B)‘ = A‘∩B‘ |
| 3) Given that U = {d,n,l,i,x,x,i,f}, A = {d,n,x,i}, and B = {d,n,l}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer: l,i,x,x,i,f SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∩B)‘ = A‘∪B‘ A∩B = {d,n,x,i}∪{d,n,l} = {d,n,l,x,i} (A∩B)‘ = U∖(A∩B) = {d,n,l,i,x,x,i,f}∖{d,n} (A∩B)‘ = {l,i,x,x,i,f} ...... (i) A‘ = U∖A = {d,n,l,i,x,x,i,f}∖{d,n,x,i} A‘ = {l,i,x,f} B‘ = U∖B = {d,n,l,i,x,x,i,f}∖{d,n,l} B‘ = {i,x,x,i,f} A‘∪B‘ = {l,i,x,f}∖ {i,x,x,i,f} A‘∪B‘ = {l,i,x,x,i,f} ....... (2) From (i) and (2), we have (A∩B)‘ = A‘∪B‘ |
| 4) Given that U = {d,l,d,w,z,n,c,x}, A = {d,l,n,c}, and B = {d,l,d}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer: w,z,x SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∪B)‘ = A‘∩B‘ (ii) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∪B)‘ = A‘∩B‘ A∪B = {d,l,n,c}∪{d,l,d} = {d,l,d,n,c} (A∪B)‘ = U∖(A∪B) = {d,l,d,w,z,n,c,x}∖{d,l,d,n,c} (A∪B)‘ = {w,z,x} ...... (i) A‘ = U∖A = {d,l,d,w,z,n,c,x}∖{d,l,n,c} A‘ = {d,w,z,x} B‘ = U∖B = {d,l,d,w,z,n,c,x}∖{d,l,d} B‘ = {w,z,n,c,x} A‘∩B‘ = {d,w,z,x}∖ {w,z,n,c,x} A‘∩B‘ = {w,z,x} ....... (2) From (i) and (2), we have (A∪B)‘ = A‘∩B‘ |
| 5) Given that U = {k,a,y,c,j,q,v,p}, A = {k,a,q,v}, and B = {k,a,y}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer: y,c,j,q,v,p SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∩B)‘ = A‘∪B‘ A∩B = {k,a,q,v}∪{k,a,y} = {k,a,y,q,v} (A∩B)‘ = U∖(A∩B) = {k,a,y,c,j,q,v,p}∖{k,a} (A∩B)‘ = {y,c,j,q,v,p} ...... (i) A‘ = U∖A = {k,a,y,c,j,q,v,p}∖{k,a,q,v} A‘ = {y,c,j,p} B‘ = U∖B = {k,a,y,c,j,q,v,p}∖{k,a,y} B‘ = {c,j,q,v,p} A‘∪B‘ = {y,c,j,p}∖ {c,j,q,v,p} A‘∪B‘ = {y,c,j,q,v,p} ....... (2) From (i) and (2), we have (A∩B)‘ = A‘∪B‘ |
| 6) Given that U = {e,a,h,d,f,i,y,z}, A = {e,a,i,y}, and B = {e,a,h}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer: d,f,z SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∪B)‘ = A‘∩B‘ (ii) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∪B)‘ = A‘∩B‘ A∪B = {e,a,i,y}∪{e,a,h} = {e,a,h,i,y} (A∪B)‘ = U∖(A∪B) = {e,a,h,d,f,i,y,z}∖{e,a,h,i,y} (A∪B)‘ = {d,f,z} ...... (i) A‘ = U∖A = {e,a,h,d,f,i,y,z}∖{e,a,i,y} A‘ = {h,d,f,z} B‘ = U∖B = {e,a,h,d,f,i,y,z}∖{e,a,h} B‘ = {d,f,i,y,z} A‘∩B‘ = {h,d,f,z}∖ {d,f,i,y,z} A‘∩B‘ = {d,f,z} ....... (2) From (i) and (2), we have (A∪B)‘ = A‘∩B‘ |
| 7) Given that U = {k,v,y,o,b,c,f,n}, A = {k,v,c,f}, and B = {k,v,y}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer: y,o,b,c,f,n SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∩B)‘ = A‘∪B‘ A∩B = {k,v,c,f}∪{k,v,y} = {k,v,y,c,f} (A∩B)‘ = U∖(A∩B) = {k,v,y,o,b,c,f,n}∖{k,v} (A∩B)‘ = {y,o,b,c,f,n} ...... (i) A‘ = U∖A = {k,v,y,o,b,c,f,n}∖{k,v,c,f} A‘ = {y,o,b,n} B‘ = U∖B = {k,v,y,o,b,c,f,n}∖{k,v,y} B‘ = {o,b,c,f,n} A‘∪B‘ = {y,o,b,n}∖ {o,b,c,f,n} A‘∪B‘ = {y,o,b,c,f,n} ....... (2) From (i) and (2), we have (A∩B)‘ = A‘∪B‘ |
| 8) Given that U = {g,o,f,a,m,q,l,r}, A = {g,o,q,l}, and B = {g,o,f}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer: a,m,r SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∪B)‘ = A‘∩B‘ (ii) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∪B)‘ = A‘∩B‘ A∪B = {g,o,q,l}∪{g,o,f} = {g,o,f,q,l} (A∪B)‘ = U∖(A∪B) = {g,o,f,a,m,q,l,r}∖{g,o,f,q,l} (A∪B)‘ = {a,m,r} ...... (i) A‘ = U∖A = {g,o,f,a,m,q,l,r}∖{g,o,q,l} A‘ = {f,a,m,r} B‘ = U∖B = {g,o,f,a,m,q,l,r}∖{g,o,f} B‘ = {a,m,q,l,r} A‘∩B‘ = {f,a,m,r}∖ {a,m,q,l,r} A‘∩B‘ = {a,m,r} ....... (2) From (i) and (2), we have (A∪B)‘ = A‘∩B‘ |
| 9) Given that U = {e,p,o,v,l,t,s,x}, A = {e,p,t,s}, and B = {e,p,o}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘ (must use , ) A‘∪B‘ = Answer: o,v,l,t,s,x SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∩B)‘ = A‘∪B‘ A∩B = {e,p,t,s}∪{e,p,o} = {e,p,o,t,s} (A∩B)‘ = U∖(A∩B) = {e,p,o,v,l,t,s,x}∖{e,p} (A∩B)‘ = {o,v,l,t,s,x} ...... (i) A‘ = U∖A = {e,p,o,v,l,t,s,x}∖{e,p,t,s} A‘ = {o,v,l,x} B‘ = U∖B = {e,p,o,v,l,t,s,x}∖{e,p,o} B‘ = {v,l,t,s,x} A‘∪B‘ = {o,v,l,x}∖ {v,l,t,s,x} A‘∪B‘ = {o,v,l,t,s,x} ....... (2) From (i) and (2), we have (A∩B)‘ = A‘∪B‘ |
| 10) Given that U = {q,g,u,i,e,b,e,i}, A = {q,g,b,e}, and B = {q,g,u}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘ (must use , ) A‘∩B‘ = Answer: i,e,i SOLUTION 1 : verify De Morgan‘s laws of complementation. (i) (A∪B)‘ = A‘∩B‘ (ii) (A∩B)‘ = A‘∪B‘ (1) to verify : (A∪B)‘ = A‘∩B‘ A∪B = {q,g,b,e}∪{q,g,u} = {q,g,u,b,e} (A∪B)‘ = U∖(A∪B) = {q,g,u,i,e,b,e,i}∖{q,g,u,b,e} (A∪B)‘ = {i,e,i} ...... (i) A‘ = U∖A = {q,g,u,i,e,b,e,i}∖{q,g,b,e} A‘ = {u,i,e,i} B‘ = U∖B = {q,g,u,i,e,b,e,i}∖{q,g,u} B‘ = {i,e,b,e,i} A‘∩B‘ = {u,i,e,i}∖ {i,e,b,e,i} A‘∩B‘ = {i,e,i} ....... (2) From (i) and (2), we have (A∪B)‘ = A‘∩B‘ |