Past Exam Questions

Answer: (a) \(y^4-2y^3+4y^2+2y+11=0\)

(b) \(S_2=-4\)

(c) \(S_4=-76\)

(a)

Let \(y=2x+1\). Then \(x=\frac12(y-1)\). Substitute into the original equation:

\(\left(\frac{y-1}{2}\right)^4+\left(\frac{y-1}{2}\right)^3+\left(\frac{y-1}{2}\right)^2+\frac{y-1}{2}+1=0.\)

That is

\(\frac1{16}(y-1)^4+\frac18(y-1)^3+\frac14(y-1)^2+\frac12(y-1)+1=0.\)

Expanding and simplifying gives

\(\frac1{16}y^4-\frac18y^3+\frac14y^2+\frac18y+\frac{11}{16}=0.\)

Multiplying by \(16\),

\(y^4-2y^3+4y^2+2y+11=0.\)

(b)

For the transformed quartic \(y^4-2y^3+4y^2+2y+11=0\), the sum of roots is \(S_1=2\), and the sum of products in pairs is \(4\). Hence

\(S_2=(\text{sum of roots})^2-2(\text{sum of products in pairs})=2^2-2(4)=-4.\)

(c)

Each transformed root \(y\) satisfies

\(y^4-2y^3+4y^2+2y+11=0.\)

Sum this equation over the four roots:

\(S_4-2S_3+4S_2+2S_1+44=0.\)

Using \(S_1=2\), \(S_2=-4\), and \(S_3=-22\),

\(S_4-2(-22)+4(-4)+2(2)+44=0,\)

so \(S_4+76=0\), and therefore

\(S_4=-76.\)

Mark scheme

Answer: \(p^2+4r=0\), and \(q=0\).

Let the two pairs of equal roots be \(\alpha,\alpha,\beta,\beta\).

For \(x^4-px^2+qx-r=0\), Vieta's formulae give

\(2\alpha+2\beta=0\), so \(\alpha+\beta=0\), hence \(\beta=-\alpha\).

The sum of the pairwise products is

\(\alpha^2+4\alpha\beta+\beta^2=-p\).

Using \(\beta=-\alpha\),

\(-p=\alpha^2-4\alpha^2+\alpha^2=-2\alpha^2\), so \(p=2\alpha^2\).

The product of the roots is

\(\alpha^2\beta^2=-r\).

Again using \(\beta=-\alpha\), this gives \(\alpha^4=-r\).

Therefore

\(p^2=(2\alpha^2)^2=4\alpha^4=-4r\),

so \(p^2+4r=0\).

For \(q\), the sum of the triple products is

\(2\alpha^2\beta+2\alpha\beta^2=-q\).

This is \(2\alpha\beta(\alpha+\beta)\), which is zero because \(\alpha+\beta=0\). Hence \(q=0\).

Answer: \(n(A)=6\), \(n(A\cap B)=1\); \((S\cup T)'\); \(X\cap(Y\cup Z)\).

The integers in \(\mathcal E\) are \(2,3,\ldots,19\).

The multiples of 3 in this range are

\(3,\ 6,\ 9,\ 12,\ 15,\ 18,\)

so

\(n(A)=6.\)

The multiples of 4 in the same range are \(4,8,12,16\). The only number common to both sets is \(12\), hence

\(n(A\cap B)=1.\)

For \(P\subset Q\), draw \(P\) completely inside \(Q\). Since \(Q\cap R=\varnothing\), draw \(R\) separately from \(Q\).

In the first shaded diagram, the shaded region is outside both \(S\) and \(T\), so it is

\((S\cup T)'.\)

In the second shaded diagram, the shaded region is the part of \(X\) that is also in \(Y\) or \(Z\). Therefore it is

\((X\cap Y)\cup(X\cap Z)=X\cap(Y\cup Z).\)

Answer: Shade the required Venn regions; for part (b), draw \(P\) and \(Q\) as disjoint subsets inside \(R\).

For \((A\cap B)\cup C\), shade the whole of \(C\), and also shade the overlap of \(A\) and \(B\).

For \((A'\cup B)\cap C\), first restrict attention to \(C\). Within \(C\), shade the part that is outside \(A\), together with any part that is in \(B\).

For part (b), \(P\subset R\) and \(Q\subset R\) mean that both \(P\) and \(Q\) must lie inside \(R\). The condition \(P\cap Q=\varnothing\) means they must not overlap.

Answer: (a)(i) outside all three sets; (a)(ii) all of \(A\) and the overlap \(B\cap C\). (b) \(P\cap Q=\{30^\circ,150^\circ\}\).

For \(A'\cap B'\cap C'\), shade the region outside all three circles \(A\), \(B\) and \(C\).

For \(A\cup(B\cap C)\), shade the whole of \(A\), and also shade the overlap of \(B\) and \(C\).

For \(P\), solve

\(\cos 2x=\frac12.\)

For \(0^\circ\leq x\leq360^\circ\), this gives

\(P=\{30^\circ,150^\circ,210^\circ,330^\circ\}.\)

For \(Q\), solve

\(\sin x=\frac12,\)

so

\(Q=\{30^\circ,150^\circ\}.\)

Therefore

\(P\cap Q=\boxed{\{30^\circ,150^\circ\}}.\)

Answer: \(A\cap B=\varnothing\), and \(Z\subset (X\cap Y)\).

In the first diagram, the circles \(A\) and \(B\) do not overlap. This means that they have no common elements:

\(A\cap B=\varnothing.\)

In the second diagram, \(Z\) lies completely inside the overlap of \(X\) and \(Y\). Therefore every element of \(Z\) is also in \(X\cap Y\):

\(Z\subset (X\cap Y).\)

Answer: \(A'\cap B\); \((X\cap Y)\cup(X\cap Z)\).

For the first diagram, the shaded region is inside \(B\) but outside \(A\). This is

\(A'\cap B.\)

For the second diagram, the shaded region is the part of \(X\) that overlaps with \(Y\) or with \(Z\). This is

\((X\cap Y)\cup(X\cap Z).\)

Equivalently, this can be written as \(X\cap(Y\cup Z)\).

Answer: \(x=15\); mathematics only \(=43\).

The pairwise totals include the students who study all three subjects. Therefore the regions for exactly two subjects are

\(M\cap C\text{ only}=24-x,\)

\(P\cap C\text{ only}=23-x,\)

\(M\cap P\text{ only}=28-x.\)

The mathematics-only region is

\(80-(24-x)-(28-x)-x=28+x.\)

The physics-only region is

\(75-(28-x)-(23-x)-x=24+x.\)

The chemistry-only region is

\(50-(24-x)-(23-x)-x=3+x.\)

Since all \(145\) students study at least one subject, add all seven regions:

\((28+x)+(24+x)+(3+x)+(24-x)+(23-x)+(28-x)+x=145.\)

This simplifies to

\(130+x=145,\)

so

\(x=15.\)

The number who study mathematics only is

\(28+x=28+15=43.\)

Hence

\(\boxed{x=15,\qquad \text{mathematics only}=43}.\)

Answer: (i) shade the two non-overlapping parts of \(A\) and \(B\); (ii) shade all of \(C\) and the overlap \(A\cap B\); (iii) shade the part common to \(A\) and \(B\) but outside \(C\).

(i) The expression \((A'\cap B)\cup(A\cap B')\) means elements in \(B\) but not \(A\), together with elements in \(A\) but not \(B\). So shade the two outer crescent regions of \(A\) and \(B\), leaving the overlap unshaded.

(ii) The expression \((A\cap B)\cup C\) means everything in \(C\), together with the overlap of \(A\) and \(B\). So shade the whole of \(C\), and also shade any part of \(A\cap B\) which lies outside \(C\).

(iii) The expression \(A\cap B\cap C'\) means elements that are in both \(A\) and \(B\), but not in \(C\). So shade only the part of the overlap \(A\cap B\) which lies outside \(C\).

Answer: (a) \((P\cup Q)\cap R'\). (b) Outside all three sets is \(9\), so \(n((A\cup B\cup C)')=1\).

The shaded region in part (a) is inside \(P\) or \(Q\), but outside \(R\). Therefore it is

\((P\cup Q)\cap R'.\)

For part (b), first list the sets:

\(A=\{2,3,5,7\}\),

\(B=\{2,4,6,8,10\}\),

\(C=\{1,2,3,4,8\}\).

The element in all three sets is \(2\).

The elements in \(A\cap C\), but not in \(B\), are \(3\). The elements in \(B\cap C\), but not in \(A\), are \(4\) and \(8\).

The elements only in \(A\) are \(5\) and \(7\). The elements only in \(B\) are \(6\) and \(10\). The element only in \(C\) is \(1\).

The only element outside all three sets is \(9\), so

\(\boxed{n((A\cup B\cup C)')=1}.\)

Answer: (i)(a) \(A\) is not a subset of \(B\). (i)(b) \(A\) and \(C\) are mutually exclusive. (ii)(a) \(n(A\cup B)=3\). (ii)(b) \(x\in A\cap C'\).

(i)(a) The notation \(A\not\subseteq B\) means that \(A\) is not a subset of \(B\).

(i)(b) The notation \(A\cap C=\varnothing\) means that \(A\) and \(C\) have no elements in common. In words, \(A\) and \(C\) are mutually exclusive.

(ii)(a) Elements in \(A\) or \(B\) or both are elements of \(A\cup B\). Therefore

\(n(A\cup B)=3.\)

(ii)(b) If \(x\) is in \(A\) but not in \(C\), then \(x\) is in \(A\cap C'\). Thus

\(x\in A\cap C'.\)

Answer: (a) Shade the part of \(A\) outside \(B\). (b) The regions are \(11,2,5,0,8,0,4,0\), so \(n(R)=12\).

(a) The set \(A\cap B'\) means elements that are in \(A\) and not in \(B\). So shade only the left-only part of circle \(A\).

(b) Since \((P\cup Q\cup R)'=\varnothing\), the region outside all three sets is \(0\).

The condition \(P'\cap(Q\cap R)=\varnothing\) means the part in \(Q\cap R\) but not in \(P\) is \(0\).

Also

\(n(Q\cap R)=8,\qquad n(P\cap R)=8,\)

so the central triple intersection is \(8\), and the \(P\cap R\) only and \(Q\cap R\) only regions are both \(0\).

Now \(n(P\cap Q)=10\), so the \(P\cap Q\) only region is

\(10-8=2.\)

Using \(n(P)=21\), the \(P\)-only region is

\(21-2-8-0=11.\)

Using \(n(Q)=15\), the \(Q\)-only region is

\(15-2-8-0=5.\)

Finally, the total number of elements is \(30\), so the \(R\)-only region is

\(30-(11+2+5+8)=4.\)

Therefore

\(n(R)=0+8+0+4=12.\)

Answer: (i)(a) \(A\) is not a subset of \(B\). (i)(b) \(A\) and \(C\) are mutually exclusive. (ii)(a) \(n(A\cup B)=3\). (ii)(b) \(x\in A\cap C'\).

(i)(a) The notation \(A\not\subseteq B\) means that \(A\) is not a subset of \(B\).

(i)(b) The notation \(A\cap C=\varnothing\) means that \(A\) and \(C\) have no elements in common. In words, \(A\) and \(C\) are mutually exclusive.

(ii)(a) Elements in \(A\) or \(B\) or both are elements of \(A\cup B\). Therefore

\(n(A\cup B)=3.\)

(ii)(b) If \(x\) is in \(A\) but not in \(C\), then \(x\) is in \(A\cap C'\). Thus

\(x\in A\cap C'.\)

Answer: \(x=8\), and \(13\) girls play netball only.

From the diagram, the region \(R\cap H\cap N\) is \(9\).

The region \(R\cap H\cap N'\) is \(14-x\).

The region \(R\cap N\cap H'\) is \(5\).

The region \(N\cap(R\cup H)'\), which is netball only, is \(21-x\).

The total of all regions in the universal set is \(70\). Using the diagram,

\(x+9+x+15+(14-x)+5+(21-x)+(x-2)=70.\)

Simplify:

\(x+62=70.\)

Therefore

\(x=8.\)

The number of girls who play netball only is

\(21-x=21-8=13.\)

Answer: \(x=15\). The regions are football only \(50\), both \(15\), cricket only \(25\), neither \(15\).

The number who play cricket is \(40\). Since \(x\) boys play both sports, the number who play cricket only is

\(40-x.\)

The number who play football only is twice this, so it is

\(2(40-x)=80-2x.\)

The number who play neither game is equal to the number who play both, so it is also \(x\).

Now add all four regions of the Venn diagram:

\((80-2x)+x+(40-x)+x=105.\)

Simplify:

\(120-x=105.\)

Hence

\(x=15.\)

Therefore the regions are:

\(\text{football only}=50,\quad \text{both}=15,\quad \text{cricket only}=25,\quad \text{neither}=15.\)

Answer: Shade outside all three sets; shade the part common to \(A\) and \(B\) but outside \(C\); shade all of \(C'\) together with \(A\cap B\).

For \((A\cup B\cup C)'\), shade the region outside every one of the three circles.

For \(A\cap B\cap C'\), shade only the part that is inside both \(A\) and \(B\), but outside \(C\).

For \((A\cap B)\cup C'\), shade the overlap \(A\cap B\), and also shade every region outside \(C\).

Answer: Shade the part of \(C\) that lies in \(A\) or \(B\); shade all of \(C\) together with the overlap \(A\cap B\); shade the part of \(C\) outside both \(A\) and \(B\).

For \((A\cup B)\cap C\), first take everything in \(A\) or \(B\), then keep only the part that is also in \(C\). So shade the two lens-shaped parts of \(C\) inside \(A\) or \(B\), including the central triple intersection.

For \((A\cap B)\cup C\), first take the overlap of \(A\) and \(B\), then combine it with all of \(C\). So shade the whole circle \(C\), and also shade the part where \(A\) and \(B\) overlap outside \(C\).

For \((A'\cap B')\cap C\), take the elements outside both \(A\) and \(B\), then keep only the part in \(C\). So shade only the part of \(C\) that is not inside \(A\) and not inside \(B\).

Answer: (a) Shade the parts of \(A\) and \(C\) that lie outside \(B\). (b) Draw a larger set \(Y\) around \(X\), and a smaller set \(Z\) entirely inside \(X\).

(a) The set \(A\cap B'\) means the part of \(A\) which is not in \(B\). The set \(C\cap B'\) means the part of \(C\) which is not in \(B\). Taking the union means shading both of these regions.

So shade all points that are in \(A\) but not \(B\), together with all points that are in \(C\) but not \(B\).

(b) The notation \(Z\subset X\subset Y\) means every element of \(Z\) lies in \(X\), and every element of \(X\) lies in \(Y\).

Therefore \(Y\) must be a larger set containing the whole of \(X\), and \(Z\) must be a smaller set drawn entirely inside \(X\).

Answer: (b)(i) \(M\cap D=\varnothing\) (or \(n(M\cap D)=0\)). (b)(ii) Art only \(=2\), music only \(=33\), design only \(=11\), and \(17\) students study none.

(a) For \(X\cap Y'\), shade the part of \(X\) which lies outside \(Y\). Since \(Y\) is inside \(X\), this is the ring-shaped region inside \(X\) but not inside \(Y\).

For \(P'\cap Q\cap R\), shade the part common to \(Q\) and \(R\) that is not in \(P\).

(b)(i) “No student studies both music and design” means the intersection is empty, so

\(M\cap D=\varnothing.\)

(b)(ii) Because no student studies both music and design, there is no overlap between \(M\) and \(D\), and also no triple intersection.

From the art total:

\(\text{art only}=39-12-25=2.\)

From the music total:

\(\text{music only}=45-12=33.\)

From the design total:

\(\text{design only}=36-25=11.\)

The total number studying at least one subject is

\(2+12+25+33+11=83.\)

So the number studying none is

\(100-83=17.\)

Hence the completed diagram has \(2\) in art only, \(12\) in \(A\cap M\), \(25\) in \(A\cap D\), \(33\) in music only, \(11\) in design only, and

\(\boxed{17}\)

outside the circles.

Answer: (i) \(A'\cap B\). (ii) \(A\cap B\cap C\). (iii) \(A\cup B\).

In diagram (i), the shaded region is inside \(B\), but outside \(A\). Therefore it is

\(A'\cap B.\)

In diagram (ii), only the central overlap of all three sets is shaded. Therefore it is

\(A\cap B\cap C.\)

In diagram (iii), everything inside \(A\) or \(B\) is shaded. Therefore it is

\(A\cup B.\)