4037/22

Additional Mathematics 4037/22October/November 2011

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

11
questions
80
marks
120
minutes

Topics Trigonometry · Quadratic functions · Calculus · [Legacy] Set language and notation · Factors of polynomials · [Legacy] Matrices · +5 more

Q1[Legacy] Set language and notationFree sample
(a)

The universal set E\mathscr{E} and the sets AA and BB shown in the Venn diagram below are such that

n(A)=15,n(B)=20,n(AB)=6andn(E)=30.n(A) = 15, \quad n(B) = 20, \quad n(A' \cap B) = 6 \quad \text{and} \quad n(\mathscr{E}) = 30.

In the Venn diagram below insert the number of elements in the set represented by each of the four regions.

4M
DifficultyMedium-Easy
Worked solution

Approach

Work from the given counts inward: n(AB)n(A' \cap B) is the part of BB outside AA, so subtracting it from n(B)n(B) gives the intersection, then subtracting that from n(A)n(A) gives the part of AA alone, and finally the universal-set total gives the region outside both circles.

Working

The region in BB but not in AA has n(AB)=6n(A' \cap B) = 6 elements.

n(AB)=n(B)n(AB)=206=14n(A \cap B) = n(B) - n(A' \cap B) = 20 - 6 = 14 Region in A only=n(A)n(AB)=1514=1\text{Region in } A \text{ only} = n(A) - n(A \cap B) = 15 - 14 = 1 Outside both circles=n(E)(1+14+6)=3021=9\text{Outside both circles} = n(\mathscr{E}) - (1 + 14 + 6) = 30 - 21 = 9

Answer

In the two-circle Venn diagram: AA only =1= 1, intersection =14= 14, BB only =6= 6, outside both circles =9= 9.

Final answer

A only: 1, intersection: 14, B only: 6, outside both circles: 9

Detailed explanation

Walkthrough

The four regions of a two-circle Venn diagram are: inside AA only, inside both (the intersection), inside BB only, and outside both circles.

First, n(AB)=6n(A' \cap B) = 6 means exactly 6 elements lie in BB but not in AA. Since all of BB contains 20 elements, the remaining elements of BB must be in the overlap:

n(AB)=206=14n(A \cap B) = 20 - 6 = 14

Next, AA as a whole contains 15 elements, of which 14 are already accounted for in the overlap, so the part of AA outside BB holds just 1514=115 - 14 = 1 element.

Finally, the universal set holds 30 elements in total. Adding the three regions inside the circles gives 1+14+6=211 + 14 + 6 = 21, so the region outside both circles contains 3021=930 - 21 = 9 elements.

Key Takeaways

  • Each number given about a set pins down one or more regions by subtraction.
  • ABA' \cap B is read directly as "in BB but not in AA".
  • The universal-set total lets you find the outside region once every other region is known.
  • Always check that the four region values sum to n(E)n(\mathscr{E}).

Common Mistakes

  • Writing 6 in the intersection instead of the "BB only" region — ABA' \cap B excludes AA.
  • Putting n(A)=15n(A) = 15 straight into the "AA only" region without subtracting the overlap.
  • Forgetting to find the outside region at all, so losing the final B1FT mark.
  • Mark scheme notes: each value must be correctly positioned — a correct number in the wrong region scores nothing.

Things to Be Careful About

  • The last value (99) is marked B1FT: it follows from your own three earlier values, so an arithmetic slip earlier still earns this mark if your subtraction is consistent.
  • Check the sum: 1+14+6+9=30=n(E)1 + 14 + 6 + 9 = 30 = n(\mathscr{E}) confirms the diagram.
Techniques used
interpret the complement and intersection of setssubtract region counts to fill the Venn diagramuse the universal set total to find the outside region
(b)

In the Venn diagram below shade the region that represents (PQ)R(P \cup Q) \cap R'.

1M
DifficultyEasy
Worked solution

Approach

(PQ)(P \cup Q) is everything in PP or QQ (or both); RR' is everything outside RR. Intersecting them shades the parts of PP and QQ that lie outside RR.

Working

Shade all of circles PP and QQ except any part lying inside circle RR: that is, the whole of PP outside RR together with the whole of QQ outside RR, including the PQP \cap Q overlap where it does not meet RR.

Answer

The shaded region consists of the portions of PP and QQ lying outside circle RR.

Final answer

The parts of P and Q outside R are shaded (P union Q excluding R)

Detailed explanation

Walkthrough

Build the expression in two stages. First, PQP \cup Q covers both upper circles completely, including their overlap. Second, RR' removes everything belonging to RR, so intersecting with RR' deletes from that union every piece that falls inside the lower circle RR. What remains — and what should be shaded — is the whole of PP and the whole of QQ apart from their portions inside RR.

Key Takeaways

  • Union means "or": shade everything in either set.
  • Complement means "not": intersecting with RR' erases anything inside RR.
  • Overlap regions between PP and QQ stay shaded provided they do not touch RR.

Common Mistakes

  • Shading only PP and forgetting QQ (or vice versa).
  • Leaving unshaded the PQP \cap Q overlap because it looks like a separate region — it belongs to PQP \cup Q and stays shaded unless it meets RR.
  • Shading inside RR by misreading the complement.

Things to Be Careful About

  • This is a single B1 mark: the entire region must be shaded correctly; partial shading scores zero.
Techniques used
interpret union and complement of setsshade the required region on a three-circle Venn diagram

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