2016 Paper 1

•¹ evidence of use of product rule
$\displaystyle (...)tan^{-1}2x+x(...)$

•² one resultant term of the product correct
$\displaystyle 1.tan^{-1}2x$ or $\displaystyle x.\frac{1}{1+(2x)^2}.2$

•³ complete differentiation
$\displaystyle tan^{-1}2x+\frac{2x}{1+4x^2}$

•⁴ evidence of use of quotient or product rule and one term of numerator correct
$\displaystyle (-2x)(1+4x^2)- \dots$

•⁵ complete differentiation correctly
$\displaystyle \frac{...(1-x^2).8x}{(1+4x^2)^2}$ or $\displaystyle \frac{-10x}{(1+4x^2)^2}$

•⁶ simplify answer
$\displaystyle -\frac{10x}{(1+4x^2)^2}$

•⁷ correct derivatives
$\displaystyle 76$ and $\displaystyle \sin t$

•⁸ find $\displaystyle \frac{dy}{dx}$
$\displaystyle \frac{1}{6}\sin t$

•¹ interpret geometric series
$\displaystyle ar=108$ and $\displaystyle ar^4=4$

•² evidence of strategy
$\displaystyle \frac{ar^4}{ar}$ OR $\displaystyle r^3=\frac{1}{27}$

•³ value
$\displaystyle r=\frac{1}{3}$

•⁴ know condition
$\displaystyle -1<\frac{1}{3}<1$

•⁵ calculate the first term
$\displaystyle a=324$

•⁶ value
$\displaystyle \frac{324}{1-\frac{1}{3}}$ leading to $\displaystyle 486$

•¹ state general term
$\displaystyle ^{13}C_r(\frac{3}{x})^{13-r}(-2x)^r$

•² simplify powers of x OR coefficients
$\displaystyle (3)^{13-r}(-2)^r$ or $\displaystyle x^{2r-13}$

•³ state simplified general term
$\displaystyle ^{13}C_r(3)^{13-r}(-2)^rx^{2r-13}$

•⁴ determine value of r
$\displaystyle 2r-13=9 \Rightarrow r=11$

•⁵ evaluate term
$\displaystyle -1437696x^9$

•¹ Construct augmented matrix
$\displaystyle \begin{pmatrix}1&2&3&13\\ 2&-1&4&15\\ 1&-3&2\lambda&12\end{pmatrix}$

•² Use row operations to establish first two zero elements
$\displaystyle \begin{pmatrix}1&2&3&13\\ 0&5&2&1\\ 0&-5&2\lambda-3&-1\end{pmatrix}$

•³ Establish third zero element OR recognise linear relationship between two rows
$\displaystyle \begin{pmatrix}1&2&3&13\\ 0&5&2&1\\ 0&0&2\lambda-1&0\end{pmatrix}$ OR $\displaystyle 2\lambda-3=-2$

•⁴ State value of $\displaystyle \lambda$
$\displaystyle \lambda=\frac{1}{2}$

•¹ show true for $\displaystyle n=1$
LHS: $\displaystyle 1(3-1)=2$ RHS: $\displaystyle 1^2(1+1)=2 .$ So true for $\displaystyle n=1$

•² assume true for $\displaystyle n=k$ AND consider $\displaystyle n=k+1$
$\displaystyle \sum_{r=1}^{k}r(3r-1)=k^2(k+1)$ and $\displaystyle \sum_{r=1}^{k+1}r(3r-1)= \dots$

•³ correct statement of sum to $\displaystyle (k+1)$ terms using inductive hypothesis
$\displaystyle \dots=\sum_{r=1}^{k}r(3r-1)+(k+1)(3(k+1)-1) = k^2(k+1)+(k+1)(3k+2)$

•⁴ express explicitly in terms of $\displaystyle (k+1)$ or achieve stated aim/goal AND communicate
$\displaystyle =(k+1)^2((k+1)+1)$, thus if true for $\displaystyle n=k$ then true for $\displaystyle n=k+1$ but since true for $\displaystyle n=1$, then by induction true for all $\displaystyle n \in \mathbb{N}$

•¹ for either function: first derivative and two evaluations OR all three derivatives OR all four evaluations
$\displaystyle f(x)=\sin 3x$, $\displaystyle f(0)=0$
$\displaystyle f'(x)=3\cos 3x$, $\displaystyle f'(0)=3$
$\displaystyle f''(x)=-9\sin 3x$, $\displaystyle f''(0)=0$
$\displaystyle f'''(x)=-27\cos 3x$, $\displaystyle f'''(0)=-27$

•² complete derivatives and evaluations AND substitute
$\displaystyle f(x)=3x-\frac{27}{3!}x^3 = 3x-\frac{9}{2}x^3$

•³ for second function: first derivative and two evaluations OR all three derivatives OR all four evaluations
$\displaystyle f(x)=e^{4x}$, $\displaystyle f(0)=1$
$\displaystyle f'(x)=4e^{4x}$, $\displaystyle f'(0)=4$
$\displaystyle f''(x)=16e^{4x}$, $\displaystyle f''(0)=16$
$\displaystyle f'''(x)=64e^{4x}$, $\displaystyle f'''(0)=64$

•⁴ complete derivatives and evaluations AND substitute
$\displaystyle f(x)=1+4x+\frac{16x^2}{2}+\frac{64x^3}{6} = 1+4x+8x^2+\frac{32}{3}x^3$

•⁵ multiply expressions
$\displaystyle e^{4x}\sin 3x=(3x-\frac{9}{2}x^3...)(1+4x+8x^2+\frac{32}{3}x^3...)$

•⁶ multiply out and simplify
$\displaystyle =3x+12x^2+\frac{39}{2}x^3...$

•¹ calculate determinant
$\displaystyle -2$

•² find $\displaystyle A^2$
$\displaystyle A^2=\begin{pmatrix}4&0\\\lambda&1\end{pmatrix}$

•³ use an appropriate method
$\displaystyle \begin{pmatrix}4&0\\\lambda&1\end{pmatrix}=\begin{pmatrix}2&0\\\lambda&-1\end{pmatrix}+2\begin{pmatrix}1&0\\0&1\end{pmatrix}$

•⁴ write in required form and explicitly state values of p and q
$\displaystyle A^2=A+2I \Rightarrow p=1$ and $\displaystyle q=2$

•⁵ square expression found in (b)
$\displaystyle A^4=(A+2I)^2 = A^2+4A+4I$

•⁶ substitute for $\displaystyle A^2$ and complete process
$\displaystyle =5A+6I$

•¹ correctly plot z on Argand diagram
Point in quadrant 4 with coordinates/labels for $\displaystyle \sqrt{3}$ and $\displaystyle -1$

•² find modulus or argument
$\displaystyle w = 2a$ or $\displaystyle \arg(w)=-\frac{\pi}{6}$

•³ complete and express in polar form
$\displaystyle w=2a(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6}))$

•⁴ process modulus
$\displaystyle 256a^8$

•⁵ process argument
$\displaystyle \dots(\cos(-\frac{8\pi}{6})+i\sin(-\frac{8\pi}{6}))$

•⁶ evaluate and express in form $\displaystyle ka^n(x+i\sqrt{y})$
$\displaystyle w^8=128a^8(-1+i\sqrt{3})$

•¹ know to use integration by parts and start process
$\displaystyle \frac{1}{8}x^8(\ln x)^2 - \dots$

•² correct choice of functions to differentiate and integrate AND application thereof
$\displaystyle \dots - \frac{1}{8}\int x^8 \times \frac{d}{dx}((\ln x)^2)dx$

•³ differentiate $\displaystyle (\ln x)^2$
$\displaystyle \frac{1}{8}x^8(\ln x)^2 - \frac{1}{4}\int x^7(\ln x)dx$

•⁴ know to use second application and begin process
$\displaystyle \dots - [\frac{1}{32}x^8(\ln x) - \frac{1}{32}\int x^8(\frac{1}{x})dx]$

•⁵ complete second application
$\displaystyle \dots - [\frac{1}{32}x^8(\ln x) - \frac{1}{256}x^8]$

•⁶ simplify
$\displaystyle \frac{1}{8}x^8(\ln x)^2 - \frac{1}{32}x^8(\ln x) + \frac{1}{256}x^8 + c$

•¹ give counterexample
eg. choose $\displaystyle p=7$, $\displaystyle 2(7)+1=15$ and since $\displaystyle 15=5\times3$, hence not prime, statement is false

•² set up n
$\displaystyle n=3a+1$, $\displaystyle a\in\mathbb{N}_0$

•³ consider expansion of $\displaystyle n^3$
$\displaystyle n^3=27a^3+27a^2+9a+1$

•⁴ complete proof with conclusion
$\displaystyle =3(9a^3+9a^2+3a)+1$ and statement such as "so $\displaystyle n^3$ has remainder 1 when divided by 3... statement is true"

•¹ state differential equation
$\displaystyle \frac{dh}{dt}=5$

•² state relationship or apply chain rule
$\displaystyle \frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}$ OR $\displaystyle V=h^3$

•³ find the rate of change of volume with respect to height
$\displaystyle \frac{dV}{dh}=3h^2$

•⁴ evaluate
$\displaystyle \frac{dV}{dt}=3h^2\times 5=3(3)^2\times 5=135 \text{ cm}^3\text{s}^{-1}$

•¹ correct shape
V shape reflecting in x-axis

•² graph passes through 2c on the positive y-axis
Passes through (0, 2c) and meets x axis at c

•³ graph of $\displaystyle y=|2f(x)|$ passing through 2c on the positive y-axis
Passes through $\displaystyle 2c$ on y-axis

•⁴ correct shape (symmetrical V) meeting positive x-axis at c
V shape meeting x-axis at $\displaystyle c$

•¹ correct application of partial fractions
$\displaystyle \frac{3x+32}{(x+4)(6-x)}=\frac{A}{x+4}+\frac{B}{6-x}$

•² starts process
$\displaystyle 3x+32=A(6-x)+B(x+4)$

•³ calculate one value
$\displaystyle A=2$

•⁴ calculate second value
$\displaystyle B=5$

•⁵ re-state integral in partial fractions
$\displaystyle \int_{3}^{4}(\frac{2}{(x+4)}+\frac{5}{(6-x)})dx$

•⁶ one term correctly integrated
$\displaystyle [2\ln|x+4|\dots$

•⁷ Integrate second term correctly
$\displaystyle \dots-5\ln|6-x|]_3^4$

•⁸ substitute limits
$\displaystyle (2\ln|4+4|-5\ln|6-4|) - (2\ln|3+4|-5\ln|6-3|)$

•⁹ evaluate to expected form
$\displaystyle =\ln\frac{486}{49}$

•¹ convert any two components of $\displaystyle L_2$ to parametric form
two from $\displaystyle x=3-2\mu, y=8+\mu, z=-1+3\mu$

•² two linear equations involving two distinct parameters
two from $\displaystyle 4+3\lambda=3-2\mu, 2+4\lambda=8+\mu, -7\lambda=-1+3\mu$

•³ find parameter values
$\displaystyle \lambda=1, \mu=-2$

•⁴ verify third component in both equations or equivalent
eg $\displaystyle z_1=-7\times1$ and $\displaystyle z_2=3(-2)-1$ therefore the lines intersect

•⁵ find point of intersection
$\displaystyle (7,6,-7)$

•⁶ identify first direction vector
$\displaystyle d_1=3i+4j-7k$

•⁷ identify second direction vector
$\displaystyle d_2=-2i+j+3k$

•⁸ calculate magnitudes and scalar product
$\displaystyle |d_1|=\sqrt{74}, |d_2|=\sqrt{14}$ and $\displaystyle d_1 \cdot d_2=-6+4-21=-23$

•⁹ calculate obtuse angle
$\displaystyle \cos^{-1}(\frac{-23}{\sqrt{74}\sqrt{14}})\approx 135.6^\circ$

•¹ state auxiliary equation
$\displaystyle m^2+5m+6=0 \Rightarrow m=-3, m=-2$

•² solve auxiliary equation and state complementary function
$\displaystyle y=Ae^{-3x}+Be^{-2x}$

•³ construct particular integral
$\displaystyle y=Cx^2+Dx+E$

•⁴ differentiate particular integral
$\displaystyle \frac{dy}{dx}=2Cx+D$ and $\displaystyle \frac{d^2y}{dx^2}=2C$

•⁵ calculate one coefficient of the particular integral
$\displaystyle C=2$

•⁶ calculate remaining coefficients
$\displaystyle D=-3, E=1 \Rightarrow y=Ae^{-3x}+Be^{-2x}+2x^2-3x+1$

•⁷ differentiate general solution
$\displaystyle \frac{dy}{dx}=-3Ae^{-3x}-2Be^{-2x}+4x-3$

•⁸ construct equations using given conditions
$\displaystyle A+B=-7$ and $\displaystyle 3A+2B=-6$ or equivalent

•⁹ Find one coefficient
$\displaystyle A=8$ or $\displaystyle B=-15$

•¹⁰ Find other coefficient and state particular solution
$\displaystyle y=8e^{-3x}-15e^{-2x}+2x^2-3x+1$

•¹ construct integral equation
$\displaystyle \int\frac{1}{T-T_F}dT=\int-k dt$

•² integrate
$\displaystyle \ln(T-T_F)=-kt+c$

•³ find constant, c
$\displaystyle \ln(9.8-4)=-k(0)+c \Rightarrow c=\ln 5.8$

•⁴ substitute using given information
$\displaystyle \ln(6.5-4)=-15k+\ln 5.8$

•⁵ find constant, k
$\displaystyle k=\frac{\ln 2.5-\ln 5.8}{-15}=0.05610\dots$

•⁶ substitute given condition
$\displaystyle \ln(25-4)=-0.05610\dots t+\ln 5.8$

•⁷ know how to find time
$\displaystyle t=\frac{\ln 21-\ln 5.8}{-0.05610\dots}$

•⁸ calculate time
$\displaystyle t=-22.93\dots$

•⁹ state the time to the nearest minute
The liquid was placed in the fridge at 11:37 (am)