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2016 Paper 1 Q1
D: 1500.0 B: 1516.0

  1. For \(n=1\), \(2\), \(3\) and \(4\), the functions \(\p_n\) and \(\q_n\) are defined by \[ \p_n(x) = (x+1)^{2n} - (2n+1)x (x^2+x+1)^{n-1} \] and \[ \q_n(x) = \frac{x^{2n+1}+1}{x+1} \ \ \ \ \ \ \ \ \ \ \ \ (x\ne -1) \,. \ \ \ \ \ \ \ \ \ \ \] Show that \(\p_n(x)\equiv \q_n(x)\) (for \(x\ne-1\)) in the cases \(n=1\), \(n=2\) and \(n=3\). Show also that this does not hold in the case \(n=4\).
  2. Using results from part (i):
    • [\bf (a)] express \( \ \dfrac {300^3 +1}{301}\,\) as the product of two factors (neither of which is 1);
    • [\bf (b)] express \( \ \dfrac {7^{49}+1}{7^7+1}\,\) as the product of two factors (neither of which is 1), each written in terms of various powers of 7 which you should not attempt to calculate explicitly.


Solution:

  1. \(n=1\): \begin{align*} && p_1(x) &= (x+1)^2 - 3x(x^2+x+1)^0 \\ &&&= x^2+2x+1-3x \\ &&&= x^2-x+1\\ && q_1(x) &= \frac{x^3+1}{x+1} \\ &&&= x^2-x+1 = p_1(x) \\ \\ && p_2(x) &= (x+1)^4-5x(x^2+x+1)^1 \\ &&&= x^4+4x^3+6x^2+4x+1 - 5x^3-5x^2-5x \\ &&&= x^4-x^3+x^2-x+1 \\ &&q_2(x) &= \frac{x^5+1}{x+1} \\ &&&= x^4-x^3+x^2-x+1 = p_2(x) \\ \\ && p_3(x) &= (x+1)^6-7x(x^2+x+1)^2 \\ &&&= x^6+6x^5+15x^4+20x^3+15x^2+6x+1 - 7x(x^4+2x^3+3x^2+2x+1) \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 \\ && q_3(x) &= \frac{x^7+1}{x+1} \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 = p_3(x) \\ \\ && p_4(1) &= 2^8 - 9 \cdot 1 \cdot 3^3 \\ &&&= 256 - 243 = 13 \\ && q_4(1) &= \frac{2}{2} = 1 \neq 13 \end{align*}
    • [\bf (a)] \(\,\) \begin{align*} && \frac{300^3+1}{300+1} &= (300+1)^2 - 3 \cdot 300 \\ &&&= 301^2 - 30^2 \\ &&&= 271 \cdot 331 \end{align*}
    • [\bf (b)] \(\,\) \begin{align*} && \dfrac {7^{49}+1}{7^7+1} &= (7^7+1)^6 - 7 \cdot 7^7 \cdot (7^2+7+1)^2 \\ &&&= (7^7+1)^6 - 7^8 \cdot (7^2+7+1)^2 \\ &&&= ((7^7+1)^3 - 7^4(7^2+7+1)) \cdot ((7^7+1)^3 + 7^4(7^2+7+1)) \end{align*}

2016 Paper 1 Q2
D: 1516.0 B: 1516.0

Differentiate, with respect to \(x\), \[ (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \,, \] where \(a\), \(b\), \(c\), \(d\) and \(e\) are constants. You should simplify your answer as far as possible. Hence integrate:

  1. \( \ln \big( x+\sqrt{1+x^2}\,\big) \,;\)
  2. \(\sqrt{1+x^2} \,; \)
  3. \( x\ln \big( x+\sqrt{1+x^2}\,\big) \,.\)


Solution: \begin{align*} && y &= (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \\ && y' &= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + (ax^2+bx+c) \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}} \right) + d\sqrt{1+x^2} + \frac{x(dx+e)}{\sqrt{1+x^2}} \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (ax^2+bx+c) + d(1+x^2) + x(dx+e) \right) \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (a+2d)x^2+(b+e)x+(d+c) \right) \\ \end{align*}

  1. We want \(a = 0, b = 1, d = 0, e = -1, c =0\), so \begin{align*} I &= \int \ln \big( x+\sqrt{1+x^2}\,\big) \,\d x \\ &= x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C \end{align*}
  2. We want \(a = b =0, e = 0, d = \frac12, c = \frac12\), so \begin{align*} I &= \int \sqrt{1+x^2}\, \d x \\ &= \frac12\ln(x+\sqrt{1+x^2}) + \frac12x\sqrt{1+x^2}+C \end{align*}
  3. We want \(a = \frac12, b = 0, d = -\frac14, e = 0, c = \frac14\), so \begin{align*} I &= \int x \ln (x+\sqrt{1+x^2}) \, \d x \\ &= \left (\frac12 x^2+\frac14 \right)\ln(x+\sqrt{1+x^2}) -\frac14x\sqrt{1+x^2}+C \end{align*}

2016 Paper 1 Q3
D: 1500.0 B: 1487.6

In this question, \(\lfloor x \rfloor\) denotes the greatest integer that is less than or equal to \(x\), so that (for example) \(\lfloor 2.9 \rfloor = 2\), \(\lfloor 2\rfloor = 2\) and \(\lfloor -1.5 \rfloor = -2\). On separate diagrams draw the graphs, for \(-\pi \le x \le \pi\), of:

(i) \(y = \lfloor x \rfloor\); (ii) \(y=\sin\lfloor x \rfloor\); (iii) \(y = \lfloor \sin x\rfloor\); (iv) \(y= \lfloor 2\sin x\rfloor\).
In each case, you should indicate clearly the value of \(y\) at points where the graph is discontinuous.


Solution:

  1. TikZ diagram
  2. TikZ diagram
  3. TikZ diagram
  4. TikZ diagram

2016 Paper 1 Q4
D: 1516.0 B: 1502.9

  1. Differentiate $\displaystyle \; \frac z {(1+z^2)^{\frac12}} \;$ with respect to \(z\).
  2. The {\em signed curvature} \(\kappa\) of the curve \(y=\f(x)\) is defined by \[ \kappa = \frac {\f''(x)}{\big({1+ (\f'(x))^2\big)^{\frac32}}} \,.\] Use this definition to determine all curves for which the signed curvature is a non-zero constant. For these curves, what is the geometrical significance of \(\kappa\)?


Solution:

  1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
  2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.

2016 Paper 1 Q5
D: 1484.0 B: 1516.0

  1. \noindent\vspace{-4cm} %%%%%%%The diagram requires scale of 1 unit = 15cm and 11 pt font.
    \psset{xunit=15.0cm,yunit=15.0cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.8pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(0.579583766333,2.71182136869)(1.83739646118,3.59292873869) \psline(0.630644718793,3.13692729767)(1.31015089163,3.14219478738) \pscircle(0.785157750343,3.25807956104){1.75692695495} \pscircle(1.00990397805,3.24754458162){1.58572438898} \pscircle(0.900067062948,3.16853223594){0.425605184678} \rput[tl](0.771128846767,3.26889831096){\(A\)} \rput[tl](0.890844522038,3.17472197974){\(B\)} \rput[tl](1.00098294329,3.26570589295){\(C\)} \rput[tl](0.77272505577,3.12683570963){\(P\)} \rput[tl](0.892440731042,3.12683570963){\(Q\)} \rput[tl](1.01215640631,3.12523950063){\(R\)} \end{pspicture*}
    \vspace{-6cm} The diagram shows three touching circles \(A\), \(B\) and \(C\), with a common tangent \(PQR\). The radii of the circles are \(a\), \(b\) and \(c\), respectively. Show that \[ \frac 1 {\sqrt b} = \frac 1 {\sqrt{a}} + \frac1{\sqrt{c}} \tag{\(*\)} \] and deduce that \[ 2\left(\frac1{a^2} + \frac1 {b^2} + \frac1 {c^2} \right) = \left(\frac1 a + \frac1 {b} + \frac1 {c} \right)^{\!2} . \tag{\(**\)} \]
  2. Instead, let \(a\), \(b\) and \(c\) be positive numbers, with \(b

2016 Paper 1 Q6
D: 1500.0 B: 1484.7

The sides \(OA\) and \(CB\) of the quadrilateral \(OABC\) are parallel. The point \(X\) lies on \(OA\), between \(O\) and \(A\). The position vectors of \(A\), \(B\), \(C\) and \(X\) relative to the origin \(O\) are \(\bf a\), \(\bf b\), \(\bf c\) and \(\bf x\), respectively. Explain why \(\bf c\) and \(\bf x\) can be written in the form \[ {\bf c} = k {\bf a} + {\bf b} \text{ \ \ \ \ and \ \ \ \ } {\bf x} = m {\bf a}\,, \] where \(k\) and \(m\) are scalars, and state the range of values that each of \(k\) and \(m\) can take. %

  1. The lines \(OB\) and \(AC\) intersect at \(D\), the lines \(XD\) and \(BC\) intersect at \(Y\) and the lines \(OY\) and \(AB\) intersect at \(Z\). Show that the position vector of \(Z\) relative to \(O\) can be written as \[ \frac{ {\bf b} + mk {\bf a}}{mk+1}\,. \] %
The lines \(DZ\) and \(OA\) intersect at \(T\). Show that \[ OT \times OA = OX\times TA \text{ \ \ \ \ and \ \ \ \ } \frac 1 {OT} = \frac 1 {OX} + \frac 1 {OA} \,, \] where, for example, \(OT\) denotes the length of the line joining \(O\) and \(T\).

2016 Paper 1 Q7
D: 1500.0 B: 1500.0

The set \(S\) % = \{1, 5, 9, 13, \,\ldots \}$ consists of all the positive integers that leave a remainder of 1 upon division by 4. The set \(T\) % = \{1, 5, 9, 13, \,\ldots \}$ consists of all the positive integers that leave a remainder of 3 upon division by 4.

  1. Describe in words the sets \(S \cup T\) and \(S \cap T\).
  2. Prove that the product of any two integers in \(S\) is also in \(S\). Determine whether the product of any two integers in \(T\) is also in \(T\).
  3. Given an integer in \(T\) that is not a prime number, prove that at least one of its prime factors is in \(T\).
  4. For any set \(X\) of positive integers, an integer in \(X\) (other than 1) is said to be {\em \(X\)-prime} if it cannot be expressed as the product of two or more integers {\em all in \(X\)} (and all different from 1).
    • [\bf (a)] Show that every integer in \(T\) is either \(T\)-prime or is the product of an odd number of \(T\)-prime integers.
    • [\bf (b)] Find an example of an integer in \(S\) that can be expressed as the product of \hbox{\(S\)-prime} integers in two distinct ways. [Note: \(s_1s_2\) and \(s_2s_1\) are not counted as distinct ways of expressing the product of \(s_1\) and \(s_2\).]

2016 Paper 1 Q8
D: 1500.0 B: 1530.6

Given an infinite sequence of numbers \(u_0\), \(u_1\), \(u_2\), \(\ldots\,\), we define the {\em generating function}, \(\f\), for the sequence by \[ \f(x) = u_0 + u_1x +u_2 x^2 +u_3 x^3 + \cdots \,. \] Issues of convergence can be ignored in this question.

  1. Using the binomial series, show that the sequence given by \(u_n=n\,\) has generating function \(x(1-x)^{-2}\), and find the sequence that has generating function \(x(1-x)^{-3}\). Hence, or otherwise, find the generating function for the sequence \(u_n =n^2\). You should simplify your answer.
    • [\bf (a)] The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\,\) is determined by \(u_{n} = ku_{n-1}\) (\(n\ge1\)), where \(k\) is independent of \(n\), and \(u_0=a\). By summing the identity \(u_{n}x^n \equiv ku_{n-1}x^n\), or otherwise, show that the generating function, f, satisfies \[ \f(x) = a + kx \f(x) \,. \] Write down an expression for \(\f(x)\). \vspace{3mm}
    • [\bf (b)] The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\,\) is determined by \(u_{n} = u_{n-1}+ u_{n-2}\) (\(n\ge2\)) and \(u_0=0\), \(u_1=1\). Obtain the generating function.

2016 Paper 1 Q9
D: 1516.0 B: 1469.4

A horizontal rail is fixed parallel to a vertical wall and at a distance \(d\) from the wall. A~uniform rod \(AB\) of length \(2a\) rests in equilibrium on the rail with the end \(A\) in contact with the wall. The rod lies in a vertical plane perpendicular to the wall. It is inclined at an angle~\(\theta\) to the vertical (where \(0<\theta<\frac12\pi\)) and \(a\sin\theta < d\), as shown in the diagram.

\newrgbcolor{wwwwww}{0.4 0.4 0.4} \psset{xunit=1.0cm,yunit=1.0cm} \begin{pspicture*}(-2.02,-0.1)(6.98,4.98) \psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.8pt,arrowsize=3pt 2,arrowinset=0.25} \psline[linewidth=6.2pt,linestyle=dashed,dash=11pt 1pt,linecolor=wwwwww](-0.12,5)(-0.12,0) \psline[linewidth=2pt](0,1.86)(5.86,3.34) %\psline[linestyle=dotted](0,1.82)(4.02,1.82) \psline[linestyle=dotted](4,4.48)(4,0.58) \rput[tl](0.15,2.4){\( \theta \)} \psline[linestyle=dotted](0,4 )(3.98,4 ) \rput[tl](1.86,4.40){\( d \)} \rput[tl](0.12,1.78){\( A \)} \rput[tl](5.66,3.14){\( B \)} \parametricplot{0.2473863631292545}{1.5708}{1*0.80*cos(t)+0*0.80*sin(t)+0|0*0.80*cos(t)+1*0.80*sin(t)+1.86} \begin{scriptsize} \psdots[dotsize=5pt 0,dotstyle=*](4,2.78) \end{scriptsize} \end{pspicture*}
The coefficient of friction between the rod and the wall is \(\mu\), and the coefficient of friction between the rod and the rail is \(\lambda\). Show that in limiting equilibrium, with the rod on the point of slipping at both the wall and the rail, the angle \(\theta\) satisfies \[ d\cosec^2\theta = a\big( (\lambda+\mu)\cos\theta + (1-\lambda \mu)\sin\theta \big) \,. \] Derive the corresponding result if, instead, \( a\sin\theta > d \).

2016 Paper 1 Q10
D: 1484.0 B: 1500.0

Four particles \(A\), \(B\), \(C\) and \(D\) are initially at rest on a smooth horizontal table. They lie equally spaced a small distance apart, in the order \(ABCD\), in a straight line. Their masses are \(\lambda m\), \(m\), \(m\) and \(m\), respectively, where \(\lambda>1\). Particles \(A\) and \(D\) are simultaneously projected, both at speed \(u\), so that they collide with \(B\) and \(C\) (respectively). In the following collision between \(B\) and \(C\), particle \(B\) is brought to rest. The coefficient of restitution in each collision is \(e\).

  1. Show that \(e = \dfrac {\lambda-1}{3\lambda+1}\) and deduce that \(e < \frac 13\,\).
  2. Given also that \(C\) and \(D\) move towards each other with the same speed, find the value of \(\lambda\) and of \(e\).


Solution:

TikZ diagram
Collision between A & B. Since the speed of approach is \(u\) and the coefficient of restitution is \(e\) we must have \(v_B = v_A + eu\). \begin{align*} \text{COM}: && \lambda m u &= \lambda m (v_B - eu) + m v_B \\ \Rightarrow && v_B(\lambda + 1) &=\lambda (1+ e) u \\ \Rightarrow && v_B &= \frac{\lambda(1+ e)}{1+\lambda} u \end{align*}
TikZ diagram
Collision between A & B. Since the speed of approach is \(u\) and the coefficient of restitution is \(e\) we must have \(v_D = v_C + eu\). \begin{align*} \text{COM}: && m(-u) &= mv_C + m(v_C + eu) \\ \Rightarrow && 2v_C &= -(1+e)u \\ \Rightarrow && v_C &= -\frac{1+e}{2} u \end{align*}
  1. TikZ diagram
    \begin{align*} \text{NEL}: && w_C &= e(v_B - v_C) \\ \text{COM}: && mv_B+ mv_C &= m w_C \\ \Rightarrow && w_C &= v_B + v_C\\ \Rightarrow && e(v_B - v_C) &= (v_B + v_C) \\ \Rightarrow && (1-e)v_B &= -(1+e)v_C \\ \Rightarrow && (1-e) \frac{\lambda(1+ e)}{1+\lambda} &= (1+e) \frac{1+e}{2} \\ \Rightarrow && 2\lambda - 2\lambda e &= 1+\lambda + e + \lambda e \\ \Rightarrow && (3\lambda +1)e &= \lambda - 1 \\ \Rightarrow && e &= \frac{\lambda -1}{3\lambda + 1} \\ &&&< \frac{\lambda - 1 + \frac{4}{3}}{3\lambda + 1} \\ &&& = \frac13 \end{align*}
  2. Since they move towards each other at the same speed \(w_C = - v_D\) \begin{align*} && w_C &= - v_D \\ \Rightarrow && v_B + v_C &= -(v_C+eu) \\ \Rightarrow && -eu &= v_B +2v_C \\ &&&= \frac{\lambda(1+ e)}{1+\lambda} u -(1+e)u \\ \Rightarrow && 1 &= \frac{\lambda(1+e)}{1+\lambda} \\ \Rightarrow && 1+\lambda &= \lambda \left ( 1 + \frac{\lambda -1}{3\lambda+1} \right) \\ &&&= \lambda \frac{4\lambda}{3\lambda +1} \\ \Rightarrow && 1+4\lambda + 3\lambda^2 &= 4\lambda^2 \\ \Rightarrow && 0 &= \lambda^2 - 4\lambda - 1 \\ \Rightarrow && \lambda &= \frac{4 \pm \sqrt{20}}{2} \\ &&&= 2\pm \sqrt{5} \\ \Rightarrow && \lambda &= 2 + \sqrt{5} \\ && e &= \frac{1+\sqrt{5}}{7+3\sqrt{5}} \\ &&&=\sqrt{5}-2 \end{align*}