Sets Past Year

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Contents

Preparation

  • Make sure you have have mastered all the materials in the previous parts.
  • If you are using the material here to learn this topic (as opposed to just a revision to supplement your school lessons), its' best you take a day (or more) off after learning the previous parts.
  • In fact, try take a week off, and then revise back all the materials before trying the following questions. This will be a good practice for how you are going to revise for the actual STPM.
  • Find a comfortable place & comfortable time.
  • DO NOT do this while you are half-awake, or directly after a long day in school. Else, you will frustrate yourself. Trust me.
  • Saving trees is a good thing, but DO NOT do this (in fact, any of the exercises) on rough paper/recycled paper.
  • "ROUGH PAPER = ROUGH WORK = CARELESS MISTAKES = LOSS OF MARKS"
  • Know that you will face questions that you have NEVER SEEN which will require you to adapt on the spot.
  • Keep a clock or watch handy and give yourself the suggested time to complete it. Check your answers too during that time limit.
  • After finishing, check the answers given. If you made mistakes or couldn't find the solution, you can refer to the answers/answers with guidance.

Questions

Estimated time : 19.8 minutes

1) Using the laws of algebra of sets, show that \left(A \cap B \right)'-\left(A' \cap B \right) = B' [4]

2) Using the laws of algebra of sets, show that \left(A \cup B \right)-\left(A \cap B \right) = \left(A - B \right) \cup \left( B-A \right) [7]

Answers

1)\left(A \cap B \right)'-\left(A' \cap B \right) = B'

  • \begin{align} \mathrm{L.H.S} & = \left(A \cap B \right)'-\left(A' \cap B \right) \\ & = \left(A \cap B \right)' \cap \left(A' \cap B \right)' \\ & = \left(A' \cup B' \right) \cap \left(A \cup B' \right) \\ & = \left(B' \cup A' \right) \cap \left(B' \cup A \right) \\ & = B' \cup \left(A' \cap A \right) \\ & = B' \cup \phi \\ & = B' \\ & = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}

2)\left( A \cup B \right)-\left( A \cap B \right)=\left( A - B \right) \cup \left( B-A \right)

  • \begin{align} \mathrm{L.H.S} & = \left( A \cup B \right)-\left( A \cap B \right) \\ & = \left( A \cup B \right) \cap \left( A \cap B \right)' \\ & = \left( A \cup B \right) \cap \left( A' \cup B' \right) \\ & = \left[ \left( A \cup B \right) \cap A' \right] \cup \left[ \left( A \cup B \right) \cap B' \right] \\ & = \left[ A' \cap \left( A \cup B \right) \right] \cup \left[ B' \cap \left( A \cup B \right) \right] \\ & = \left[ \left( A' \cap A \right) \cup \left( A' \cap B \right) \right] \cup \left[ \left( B' \cap A \right) \cup \left( B' \cap B \right)\right] \\ & = \left[ \phi \cup \left( A' \cap B \right) \right] \cup \left[ \left( B' \cap A \right) \cup \phi \right] \\ & = \left( A' \cap B \right) \cup \left( B' \cap A \right) \\ & = \left( B \cap A' \right) \cup \left( A \cap B' \right) \\ & = \left( B-A \right) \cup \left( A-B \right) \\ & = \left( A-B \right) \cup \left( B-A \right) \\ & = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}

Answers(With guidance)

1)\left(A \cap B \right)'-\left(A' \cap B \right) = B'

  • The 4 marks given signals a easy/medium difficult question
    • obviously we will start from LHS
    • \mathrm{L.H.S}= \left(A \cap B \right)'-\left(A' \cap B \right) ... deal with the difference first
    • =\left(A \cap B \right)' \cap \left(A' \cap B \right)' ... no identity, so apply De Morgan's
    • =\left(A' \cup B' \right) \cap \left(A \cup B' \right) ... this is the tricky part. What should we do?
    • We "factor"
    • = \left(B' \cup A' \right) \cap \left(B' \cup A \right) ... be careful of the operations after "factoring"
    • = B' \cup \left(A' \cap A \right)
    • = B' \cup \phi
    • = B'\,
    • = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

2)\left( A \cup B \right)-\left( A \cap B \right)=\left( A - B \right) \cup \left( B-A \right)

  • The 7 marks given is a signal that this is a difficult question.
    • Do we start from the right or left?
    • It's hard to see. Both sides are similar. So instead of wasting time analyzing it further, just choose any side and start first
    • \mathrm{L.H.S} = \left( A \cup B \right)-\left( A \cap B \right) ... as always, deal with the difference first
    • = \left( A \cup B \right) \cap \left( A \cap B \right)' ... no identity, so apply De Morgan's
    • = \left( A \cup B \right) \cap \left( A' \cup B' \right) ... What now?
    • Hmm....
    • Hmm....
    • Factor?.... Can't
    • Rearrange?.... Can't
    • I can't see what to do!
    • Was there any careless mistakes?.... No
    • Hmm....
    • So we are stuck. What do we do when we are stuck?
    • LEAVE this part and try from the other side ... Who knows? It might work from there, or at least let us know what step to do.
    • \mathrm{R.H.S}=\left(A-B \right) \cup \left( B-A \right) ... as always, deal with the difference first
    • =\left(A \cap B' \right) \cup \left( B \cap A' \right) ... hmmm....
    • Stuck again....
    • Lets see what we have so far
    • \begin{align} \mathrm{L.H.S} & = \left(A \cup B \right)-\left( A \cap B \right) \\ & = \left(A \cup B \right) \cap \left( A \cap B \right)' \\ & = \left(A \cup B \right) \cap \left( A' \cup B' \right) \end{align} \begin{align} \mathrm{R.H.S} & =\left(A-B \right) \cup \left( B-A \right) \\ & =\left(A \cap B' \right) \cup \left( B \cap A' \right) \end{align}
    • There must be something we can do to change from one to the other.
    • The positions of the sets are changed, but we know we can't just simply rearrange
    • We can't deal with all four sets at once, so we...
    • Draw an apple around one of the brackets, lets say {\color{Red}\left(A \cup B \right)} \cap \left( A' \cup B' \right), and see that we can ...
    • Expand
    • So, continuing from our right hand side
    • = \left(A \cup B \right) \cap \left( A' \cup B' \right) ... expand carefully
    • = \left[ \left( A \cup B \right) \cap A' \right] \cup \left[ \left( A \cup B \right) \cap B' \right] ... seems like nothing we can do but expand again.
    • = \left[ A' \cap \left( A \cup B \right) \right] \cup \left[ B' \cap \left( A \cup B \right) \right] ... Its going to be long... but you don't get 7 marks for nothing
    • = \left[ \left( A' \cap A \right) \cup \left( A' \cap B \right) \right] \cup \left[ \left( B' \cap A \right) \cup \left( B' \cap B \right)\right]
    • = \left[ \phi \cup \left( A' \cap B \right) \right] \cup \left[ \left( B' \cap A \right) \cup \phi \right]
    • = \left( A' \cap B \right) \cup \left( B' \cap A \right)
    • Okay, finally we see that this will equal to RHS.
    • We can either just rearrange it and then say LHS=RHS.
    • But since its just going to be a few more lines, we might as well continue further till RHS. (just cut away the RHS working with a single straight line)
    • = \left( B \cap A' \right) \cup \left( A \cap B' \right)
    • = \left( B-A \right) \cup \left( A-B \right)
    • = \left( A-B \right) \cup \left( B-A \right)
    • = \mathrm{R.H.S} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}

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