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1969 Paper 1 Q1
D: 1500.0 B: 1500.0

A party of seven people arrives at a tavern which has six vacant rooms. In how many ways can they be accommodated if each room can take only two persons?

1969 Paper 1 Q2
D: 1500.0 B: 1500.0

  1. Show that \(8(p^4 + q^4) > (p + q)^4\).
  2. If \(a > b > c\) and \(c > 0\) show that \(\left(\frac{a+c}{a-c}\right)^a < \left(\frac{b+c}{b-c}\right)^b.\)
Does this inequality remain true if the condition \(c > 0\) is not given?

1969 Paper 1 Q3
D: 1500.0 B: 1500.0

Obtain the condition for the equation \(ax^2 + 2bx + c = 0\) to have real roots, where \(a\), \(b\) and \(c\) are real numbers. The real numbers \(p\), \(q\) and \(r\) are such that none has unit modulus, and \(p^2 + q^2 + r^2 + 2pqr = 1.\) Prove that \(p\), \(q\), and \(r\) either all lie between \(+1\) and \(-1\), or all lie outside this range.

1969 Paper 1 Q4
D: 1500.0 B: 1500.0

Prove that \(\sum_{r=1}^n r(r+1)(r+2)\ldots(r+s-1) = n(n+1)\ldots(n+s)/(s+1).\) Evaluate \(\sum_{r=1}^n r^4\).

1969 Paper 1 Q5
D: 1500.0 B: 1500.0

A zero of the polynomial \(f(x) = a_0 x^n + a_1 x^{n-1} + \ldots + a_n\) is \(p/q\), where \(p/q\) is a fraction in its lowest terms, and each \(a_r\) is an integer. Show that

  1. \(q\) divides \(a_0\) and \(p\) divides \(a_n\),
  2. if \(m\) is any integer \(p - mq\) divides \(f(m)\).

1969 Paper 1 Q6
D: 1500.0 B: 1500.0

In a group with identity \(e\), an element \(g\) is said to have order \(n\) if \(n\) is the least positive integer such that \(g^n = e\).

  1. Prove that a group in which every element other than the identity has order 2 is Abelian (i.e. the group operation is commutative).
  2. Show that no group can have precisely two elements of order 2.


Solution:

  1. Suppose every element satisfies \(x^2 = e\) (since \(e^2 = e\) and every other element has order \(2\)) and in particular \(x^{-1} = x\). Now \begin{align*} && e &= (xy)^2 \tag{order 2} \\ && e &= xyxy \tag{expand} \\ \Rightarrow && x &= yxy \tag{multiply on the left by \(x\)} \\ \Rightarrow && xy &= yx \tag{multiply on the right by \(y\)} \\ \end{align*} Therefore the group is commutative.
  2. Suppose \(x\), \(y\) are the only two elements of order \(2\). Then consider \(yxy\): \begin{align*} && (yxy)^2 &= yxyyxy \\ &&&= yxxy \\ &&&= yy \\ &&&= e \end{align*} Therefore either \(yxy = e\) or \(yxy = x\) or \(yxy = y\). Case 1: \(yxy = e\), then multiplying on the left and right by \(y\) we find \(x = e\), contradiction. Case 2: \(yxy = y\), so \(xy = e\) so \(x = y\), contradiction. Case 3: \(yxy = x\) so \(yx = xy\). Then consider the element \(xy\). \((xy)^2 = xyxy = xxyy = e\) so \(xy\) has order at most \(2\). Therefore \(xy\) is \(e\) or \(x\) or \(y\), all of which quickly lead to contradictions

1969 Paper 1 Q7
D: 1500.0 B: 1500.0

Let \(N\) denote the non-negative integers. A subset \(S \subseteq N\) is called convex if \(x \in S\), \(y \in N\), \(x < y < z\), implies that \(y \in S\). Let \(*\) be a composition on \(N\) defined by \(x * y = \max(x, y)\). Prove that if an equivalence relation \(R\) on \(N\) has convex equivalence classes then \((x * y) R (x' * y')\) whenever \(x R x'\) and \(y R y'\). Is the converse true?

1969 Paper 1 Q8
D: 1500.0 B: 1500.0

\(ABC\) is a non-isosceles triangle, with \(M\) the mid-point of \(BC\). A line passes through \(A\), \(B\) in \(P\), \(Q\) respectively, and \(AP = AQ\). Prove that \(BP = CQ\). If \(ABC\) is a non-isosceles triangle with points \(B\), \(C\), \(P\), \(Q\) cannot be concyclic, and that the triangles \(BMP\), \(CMQ\) cannot have equal areas.

1969 Paper 1 Q9
D: 1500.0 B: 1500.0

A straight line meets the sides \(BC\), \(CA\), \(AB\) of a triangle \(ABC\) in \(L\), \(M\), \(N\) respectively. Prove that \(\frac{BL \cdot CM \cdot AN}{LG \cdot MA \cdot NB} = -1,\) due regard being paid to sign. The mid-points of the sides \(PQ\), \(RS\) of a parallelogram \(PQRS\) are \(X\), \(Y\) respectively. \(H\) is a point on the diagonal \(PR\) and \(HX\), \(HY\) meet \(QR\), \(PS\) respectively in \(U\), \(V\). Prove that \(UV\) is parallel to \(PQ\). If \(UV\) cuts \(PR\) in \(W\) prove that \(\frac{2}{HW} = \frac{1}{HP} + \frac{1}{HR},\) due regard being paid to sign.

1969 Paper 1 Q10
D: 1500.0 B: 1500.0

A fixed point \(K\) lies inside a triangle \(ABC\) and a circle through \(A\) and \(K\) meets \(AB\), \(AC\) again in \(R\), \(Q\) respectively. Prove that the circles \(BRK\) and \(CQK\) meet again in a point \(P\) of \(BC\). Show further that all triangles \(PQR\) obtained in this way are similar, and that the one of smallest area has its vertices at the feet of the perpendiculars from \(K\) to the sides of \(ABC\).