2019 Paper 1

•¹ evidence of product rule with one term correct
$\displaystyle 6x^{5}\cot 5x \pm x^{6}(...)$

•² complete differentiation
$\displaystyle 6x^{5}\cot 5x - 5x^{6}\csc^{2} 5x$ OR $\displaystyle -5x^{6}\csc^{2} 5x + (...)\cot 5x$

•³ evidence use of quotient rule with denominator and one term of numerator correct
$\displaystyle \frac{6x^2(x^3-4)-...}{(x^3-4)^2}$ OR $\displaystyle \frac{...-(2x^3+1)(3x^2)}{(x^3-4)^2}$

•⁴ complete differentiation
$\displaystyle \frac{6x^2(x^3-4)-(2x^3+1)(3x^2)}{(x^3-4)^2}$

•⁵ simplify
$\displaystyle \frac{-27x^2}{(x^3-4)^2}$

•⁶ start differentiation
$\displaystyle \frac{-1}{\sqrt{1-(2x)^2}}$

•⁷ complete differentiation
$\displaystyle \frac{-1}{\sqrt{1-(2x)^2}} \times 2$

•⁸ evaluate
$\displaystyle -4$

•¹ begin process
eg $\displaystyle 2\begin{vmatrix} p & 2 \\ -2 & 5 \end{vmatrix} - 1\begin{vmatrix} -3 & 2 \\ -1 & 5 \end{vmatrix} + 4\begin{vmatrix} -3 & p \\ -1 & -2 \end{vmatrix}$

•² find determinant
$\displaystyle 14p + 45$

•³ equate to 3 and find p
$\displaystyle -3$

•⁴ any two simplified entries

•⁵ complete multiplication
$\displaystyle \begin{pmatrix} q+16 & 5 \\ -3q+8 & -12 \\ -2q+20 & -7 \end{pmatrix}$

•⁶ explain
AB is not a square matrix AND a general statement about square matrices (e.g. "Only square matrices have an inverse").

•¹ state why function is even
graph is symmetrical about the y-axis... even OR $\displaystyle f(-x) = (-x)^2 - a^2 = x^2 - a^2 = f(x)$ $\displaystyle \therefore$ even

•² sketch graph
Valid sketch showing parabola symmetrical in the y-axis with x-intercepts at $\displaystyle -a$ and $\displaystyle a$, and a maximum turning point on the y-axis.

•¹ complete algebraic division and express in required form
$\displaystyle 3 + \frac{4x+19}{x^2-x-12}$

•² state expression
$\displaystyle \frac{A}{x+3} + \frac{B}{x-4}$

•³ form linear equation and obtain one constant
$\displaystyle 4x+19 = B(x+3) + A(x-4)$
$\displaystyle B=5$ or $\displaystyle A=-1$

•⁴ obtain final constant and state full expression
$\displaystyle 3 - \frac{1}{x+3} + \frac{5}{x-4}$

•¹ find $\displaystyle \frac{dx}{dt}$
$\displaystyle \frac{2}{2t+7}$

•² find $\displaystyle \frac{dy}{dx}$
$\displaystyle \frac{t}{\frac{2}{2t+7}}$ leading to $\displaystyle \frac{1}{2}(2t^2+7t)$

•³ differentiate $\displaystyle \frac{dy}{dx}$ w.r.t. t and evidence of strategy
$\displaystyle \frac{1}{2}(4t+7) \times \dots$

•⁴ find $\displaystyle \frac{d^2y}{dx^2}$
$\displaystyle \frac{1}{2}(2t+7)(4t+7)$

•¹ evidence of relationship
$\displaystyle \frac{dV}{dr} = 4\pi r^2$ AND $\displaystyle \frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$ OR $\displaystyle \frac{dr}{dt} = \frac{dV}{dt} \times \frac{dr}{dV}$

•² substitute
$\displaystyle -60 = 4\pi(3)^2 \frac{dr}{dt}$ OR $\displaystyle \frac{dr}{dt} = \frac{-60}{4\pi(3)^2}$

•³ evaluate
$\displaystyle -\frac{5}{3\pi} \text{ cm s}^{-1}$

•¹ find expression
$\displaystyle 3n^2 + 16n$

•² substitute 20 and evidence of subtraction from this term
$\displaystyle (3 \times 20^2 + 16 \times 20) - \dots$

•³ substitute for p and find expression
$\displaystyle 1520 - 3p^2 - 16p$

•¹ solve auxiliary equation
$\displaystyle m = -4, -7$

•² state general solution
$\displaystyle y = Ae^{-4x} + Be^{-7x}$

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

•⁴ form equations and solve for a constant
$\displaystyle A = 3$ or $\displaystyle B = -3$

•⁵ find second constant and state particular solution
$\displaystyle y = 3e^{-4x} - 3e^{-7x}$

•¹ state general term
$\displaystyle \binom{7}{r} (2x^2)^{7-r} \left(\frac{-d}{x^3}\right)^r$

•² simplify powers of x or coefficients
$\displaystyle x^{14-5r}$ or $\displaystyle 2^{7-r}(-d)^r$

•³ state simplified general term
$\displaystyle \binom{7}{r} 2^{7-r} (-d)^r x^{14-5r}$

•⁴ obtain value of r
$\displaystyle r = 3$

•⁵ find value of d
$\displaystyle d = 5$

•¹ apply chain or product rule
$\displaystyle 2y \frac{dy}{dx}$ or $\displaystyle y + x \frac{dy}{dx}$

•² complete differentiation
$\displaystyle 2x + 2y \frac{dy}{dx} = y + x \frac{dy}{dx}$

•³ express $\displaystyle \frac{dy}{dx}$ in terms of x and y
$\displaystyle \frac{dy}{dx} = \frac{y-2x}{2y-x}$

•⁴ equate denominator of $\displaystyle \frac{dy}{dx}$ to zero
$\displaystyle 2y - x = 0$

•⁵ calculate values of k
$\displaystyle k = \pm 4$

•¹ state counterexample
eg when $\displaystyle n = 4$, $\displaystyle n^2+n+1 = 21$ which is not prime.

•² write down contrapositive statement
If $\displaystyle n$ is even then $\displaystyle n^2-2n+7$ is odd.

•³ write down appropriate form for n AND substitute
$\displaystyle n = 2k$, $\displaystyle k \in \mathbb{N}$ and $\displaystyle (2k)^2 - 2(2k) + 7$

•⁴ show $\displaystyle n^2-2n+7$ is odd
eg $\displaystyle 2(2k^2-2k+3) + 1$ which is odd since $\displaystyle 2k^2-2k+3 \in \mathbb{N}$

•⁵ communicate
contrapositive statement is true AND therefore original statement is true.

•¹ convert to base 10
$\displaystyle 276$

•² method leading to a quotient of 0 or equivalent
$\displaystyle 276 = 7 \times 39 + 3$ $\displaystyle \dots$

•³ express in base 7
$\displaystyle 543_7$

•¹ separate variables and write integral equation
$\displaystyle \int \frac{1}{12-V} \, dV = \int k \, dt$

•² integrate LHS
$\displaystyle -\ln(12-V)$

•³ integrate RHS
$\displaystyle kt + c$

•⁴ evaluate constant of integration
$\displaystyle -\ln 10$

•⁵ express V in terms of k and t
$\displaystyle V = 12 - 10e^{-kt}$

•¹ show true when $\displaystyle n=1$
LHS $\displaystyle = 1! \times 1 = 1$, RHS $\displaystyle = (1+1)! - 1 = 1$

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

•³ state sum to $\displaystyle (k+1)$ terms using inductive hypothesis
$\displaystyle (k+1)! - 1 + (k+1)!(k+1)$

•⁴ extract $\displaystyle (k+1)!$ as common factor
$\displaystyle (k+1)!(1 + k + 1) - 1$

•⁵ express sum explicitly in terms of $\displaystyle (k+1)$ or achieve stated aim/goal AND communicate suitable statement
$\displaystyle ((k+1)+1)! - 1$ AND "If true for $\displaystyle n=k$ then true for $\displaystyle n=k+1.$ Also shown true for $\displaystyle n=1$ therefore, by induction, true for all positive integers $\displaystyle n.$"

•¹ verify that the line lies on one plane
eg $\displaystyle 2(2\lambda+3) - 3(\lambda-1) - \lambda = 9$

•² verify for other plane and state conclusion
eg $\displaystyle 2\lambda+3 + \lambda-1 - 3\lambda = 2$; therefore the line lies on both planes

•³ identify vectors
$\displaystyle \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$, $\displaystyle \begin{pmatrix} -2 \\ 4 \\ 3 \end{pmatrix}$

•⁴ start to calculate angle
$\displaystyle \cos \theta = \left(\frac{3}{\sqrt{6}\sqrt{29}}\right)$

•⁵ calculate complement
any answer which rounds to $\displaystyle 0.229$ or $\displaystyle 13^\circ$

•⁶ parametric equations for $\displaystyle L_2$
$\displaystyle x = -2\mu+1$; $\displaystyle y = 4\mu+3$; $\displaystyle z = 3\mu-2$

•⁷ two equations for two parameters
eg $\displaystyle 2\lambda+3 = -2\mu+1$; $\displaystyle \lambda-1 = 4\mu+3$

•⁸ solve for two possible parameters
eg $\displaystyle \mu = -1$; $\displaystyle \lambda = 0$

•⁹ substitute into remaining equation and state conclusion
eg LHS = $\displaystyle 0$, RHS = $\displaystyle -5$ so lines do not intersect.

•¹ evidence of integration by parts
$\displaystyle \frac{e^{4x}}{4}(x^2-2x+1) - \dots$

•² complete first application
$\displaystyle \dots - \int (2x-2)\frac{e^{4x}}{4} \, dx$

•³ second application of integration by parts
$\displaystyle \dots - \left[ \frac{e^{4x}}{16}(2x-2) - \frac{1}{8} \int e^{4x} \, dx \right]$

•⁴ complete integration and include limits
$\displaystyle \left[ \frac{e^{4x}}{4}(x^2-2x+1) \right]_0^1 - \left[ \frac{1}{16}(2x-2)e^{4x} - \frac{1}{32}e^{4x} \right]_0^1$

•⁵ evaluate
$\displaystyle \frac{1}{32}(e^4 - 13)$

•⁶ correct form of integral
$\displaystyle \pi \int_0^1 y^2 \, dx$

•⁷ find expression to integrate
$\displaystyle 16\pi \int_0^1 (x^2-2x+1)e^{4x} \, dx$

•⁸ integrate and evaluate
$\displaystyle \frac{\pi}{2}(e^4 - 13)$

•¹ substitute and calculate one ratio
$\displaystyle \frac{-21}{63} = -\frac{1}{3}$ or $\displaystyle \frac{7}{-21} = -\frac{1}{3}$

•² calculate second ratio and state common ratio
$\displaystyle \frac{7}{-21} = -\frac{1}{3}$ or $\displaystyle \frac{-21}{63} = -\frac{1}{3}$ so $\displaystyle r = -\frac{1}{3}$

•³ state condition
$\displaystyle |-\frac{1}{3}| < 1$

•⁴ begin to substitute
$\displaystyle \frac{63}{1 - (-\frac{1}{3})}$

•⁵ calculate sum
$\displaystyle \frac{189}{4}$ or $\displaystyle 47.25$

•⁶ equate ratios
$\displaystyle \frac{-2x+1}{5x+8} = \frac{x-4}{-2x+1}$

•⁷ perform algebraic manipulation leading to formation of quadratic equation
$\displaystyle x^2 - 8x - 33 = 0$

•⁸ calculate second value of x
$\displaystyle x = -3$

•⁹ find first three terms
$\displaystyle -7, 7, -7$

•¹⁰ state $\displaystyle S_{2n}$ and justify
$\displaystyle 0$ since eg $\displaystyle 2n$ is even and so pairs of terms cancel each other out.

•¹ write in Cartesian form
$\displaystyle a - a\sqrt{3}i$

•² calculate modulus
$\displaystyle 2a$

•³ calculate argument
$\displaystyle -\frac{\pi}{3}$

•⁴ write in polar form
$\displaystyle 2a\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$

•⁵ begin process
$\displaystyle z_1 = 8^{\frac{1}{3}}\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)^{\frac{1}{3}}$

•⁶ complete process
$\displaystyle z_1 = 2\left(\cos\left(-\frac{\pi}{9}\right) + i\sin\left(-\frac{\pi}{9}\right)\right)$

•⁷ state value of k
$\displaystyle k = 2$

•⁸ state value of m
$\displaystyle m = -9$

•⁹ begin to add or subtract to or from argument of $\displaystyle z_1$
$\displaystyle \dots \pm \frac{2\pi}{3}$

•¹⁰ state roots
$\displaystyle z_2 = 2\left(\cos\frac{5\pi}{9} + i\sin\frac{5\pi}{9}\right)$
$\displaystyle z_3 = 2\left(\cos\left(-\frac{7\pi}{9}\right) + i\sin\left(-\frac{7\pi}{9}\right)\right)$