2017 Paper 1

•¹ write down binomial expansion
$\displaystyle \binom{3}{0}(\frac{2}{y^2})^3 + \binom{3}{1}(\frac{2}{y^2})^2(-5y) + \binom{3}{2}(\frac{2}{y^2})(-5y)^2 + \binom{3}{3}(-5y)^3$

•² resolve signs, •³ simplify coefficients or powers of y, •⁴ complete simplification and obtain expression
$\displaystyle \frac{8}{y^6} - \frac{60}{y^3} + 150 - 125y^3$

•¹ state expression
$\displaystyle \frac{x^2-6x+20}{(x+1)(x-2)^2} = \frac{A}{(x+1)} + \frac{B}{(x-2)} + \frac{C}{(x-2)^2}$

•² form equation
$\displaystyle x^2-6x+20 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)$

•³ obtain two of A, B and C
$\displaystyle A=3$, $\displaystyle B=-2$, $\displaystyle C=4$

•⁴ obtain final constant and state expression
$\displaystyle \frac{3}{(x+1)} - \frac{2}{(x-2)} + \frac{4}{(x-2)^2}$

•¹ evidence use of quotient rule with one term of numerator correct
$\displaystyle \frac{2xe^{x^2-1}(x^2-1) - \dots}{(x^2-1)^2}$

•² complete differentiation
$\displaystyle \dots - e^{x^2-1}(2x)$

•³ simplify
$\displaystyle \frac{2x^3e^{x^2-1}-4xe^{x^2-1}}{(x^2-1)^2}$

•¹ evidence use of valid strategy
eg $\displaystyle a+4d=-6$ and $\displaystyle a+11d=-34$

•² obtain values of a and d
$\displaystyle a=10$, $\displaystyle d=-4$

•³ set up equation
$\displaystyle \frac{n}{2}[20-4(n-1)] = -144$

•⁴ rearrange to standard form
$\displaystyle 2n^2-12n-144=0$

•⁵ determine the value of n
$\displaystyle n>0 \dots n=12$

•¹ set up augmented matrix
$\displaystyle \begin{pmatrix} 1 & 2 & -1 & | & -3 \\ 4 & -2 & 3 & | & 11 \\ 3 & 1 & 2\lambda & | & 8 \end{pmatrix}$

•² obtain two zeros
$\displaystyle \begin{pmatrix} 1 & 2 & -1 & | & -3 \\ 0 & -10 & 7 & | & 23 \\ 0 & -5 & 2\lambda+3 & | & 17 \end{pmatrix}$

•³ complete row operations
$\displaystyle \begin{pmatrix} 1 & 2 & -1 & | & -3 \\ 0 & -10 & 7 & | & 23 \\ 0 & 0 & 4\lambda-1 & | & 11 \end{pmatrix}$

•⁴ obtain expression for z
$\displaystyle z = \frac{11}{4\lambda-1}$

•⁵ state value of $\displaystyle \lambda$
$\displaystyle \lambda = \frac{1}{4}$

•⁶ find solution
$\displaystyle z=-1$, $\displaystyle y=-3$, $\displaystyle x=2$

•¹ differentiate $\displaystyle 5x^2$
$\displaystyle \frac{du}{dx} = 10x$ or $\displaystyle du = 10xdx$

•² find limits for u
$\displaystyle u=0$, $\displaystyle u=\frac{1}{2}$

•³ replace $\displaystyle x dx$
$\displaystyle \frac{1}{10}\int \dots du$

•⁴ obtain integrand
$\displaystyle \frac{1}{10}\int_0^{1/2} \frac{1}{\sqrt{1-u^2}} du$

•⁵ integrate
$\displaystyle \frac{1}{10}[\sin^{-1}u]_0^{1/2}$

•⁶ evaluate
$\displaystyle \frac{\pi}{60}$

•¹ determine value of x
$\displaystyle x=8$

•² find inverse
$\displaystyle P^{-1} = \frac{1}{2}\begin{pmatrix} -1 & -2 \\ 5 & x \end{pmatrix} = \frac{1}{2}\begin{pmatrix} -1 & -2 \\ 5 & 8 \end{pmatrix}$

•³ state transpose
$\displaystyle Q' = \begin{pmatrix} 2 & 4 \\ -3 & y \end{pmatrix}$

•⁴ obtain product
$\displaystyle P^{-1}Q' = \begin{pmatrix} 2 & -2-y \\ -7 & 10+4y \end{pmatrix}$ OR $\displaystyle \frac{1}{2}\begin{pmatrix} 4 & -4-2y \\ -14 & 20+8y \end{pmatrix}$

•⁵ state condition for singularity
$\displaystyle \det R = 0$ or one row is a multiple of the other

•⁶ obtain value for z
$\displaystyle z=15$

•¹ start process
$\displaystyle 1595 = 1 \times 1218 + 377$ and $\displaystyle 1218 = 3 \times 377 + 87$

•² obtain remainder of 29
$\displaystyle 377 = 4 \times 87 + 29$ and $\displaystyle 87 = 3 \times 29 + 0$

•³ express gcd in terms of 377 and 1218
$\displaystyle 29 = 377 - 4(1218 - 3 \times 377)$

•⁴ state values of a and b
$\displaystyle a=13$, $\displaystyle b=-17$

•¹ separate variables and write down integral equation
$\displaystyle \int \frac{dy}{1+y^2} = \int e^{2x} dx$

•² integrate LHS
$\displaystyle \tan^{-1} y$

•³ integrate RHS
$\displaystyle \frac{1}{2}e^{2x} + c$

•⁴ evaluate constant of integration
$\displaystyle c = \frac{\pi}{4} - \frac{1}{2}$

•⁵ express y in terms of x
$\displaystyle y = \tan(\frac{1}{2}e^{2x} + \frac{\pi}{4} - \frac{1}{2})$

•¹ substitute formulae
$\displaystyle \sum_{r=1}^n (r^2 + \frac{1}{3}r) = \frac{n(n+1)(2n+1)}{6} + \frac{1}{3}(\frac{n(n+1)}{2})$

•² factorise fully
$\displaystyle =\frac{n(n+1)((2n+1)+1)}{6} = \frac{n(n+1)^2}{3}$

•³ substitute 2p and 9
$\displaystyle \frac{2p(2p+1)^2}{3}$ and $\displaystyle \frac{9(9+1)^2}{3}$

•⁴ obtain expression
$\displaystyle \frac{2p(2p+1)^2}{3} - \frac{9(9+1)^2}{3}$ leading to $\displaystyle \frac{2p(2p+1)^2}{3} - 300$

•¹ take logarithms of both sides and apply rule
$\displaystyle \ln y = (2x^3+1)\ln x$

•² differentiate LHS
$\displaystyle \frac{1}{y}\frac{dy}{dx}$

•³ evidence use of product rule and one term correct
$\displaystyle 6x^2\ln x$ or $\displaystyle \frac{2x^3+1}{x}$

•⁴ complete differentiation
$\displaystyle \dots 6x^2\ln x + \frac{2x^3+1}{x}$

•⁵ write $\displaystyle \frac{dy}{dx}$ in terms of x
$\displaystyle \frac{dy}{dx} = x^{2x^3+1}(6x^2\ln x + \frac{2x^3+1}{x})$

•¹ show half-turn symmetry and indicate (1,2)
Sketch showing odd function symmetry passing through $\displaystyle (1,2)$ and $\displaystyle (-1,-2).$

•² demonstrate graph approaching parallel asymptote through (0,3)
Sketch showing asymptotic behaviour towards line passing through $\displaystyle (0,3)$ and $\displaystyle (0,-3).$

•³ apply modulus function to graph obtained in (a)
Sketch showing reflection of negative y-values into positive y-values.

•⁴ illustrate asymptotes meeting on the y-axis
Sketch showing "V" shaped asymptotes meeting at $\displaystyle (0,3).$

•⁵ state range
$\displaystyle \frac{1}{2} < f'(x) \le 2$

•¹ write down contrapositive statement
If n is odd then $\displaystyle n^2$ is odd.

•² write down appropriate form for n
$\displaystyle n = 2k+1$, $\displaystyle k \in \mathbb{Z}$

•³ show $\displaystyle n^2$ is odd
$\displaystyle n^2 = 2(2k^2+2k)+1$ which is odd.

•⁴ communicate
Contrapositive statement is true therefore original statement is true.

•¹ construct auxiliary equation
$\displaystyle m^2 - 6m + 9 = 0$

•² solve auxiliary equation and state CF
$\displaystyle y = Ae^{3x} + Bxe^{3x}$

•³ state PI
$\displaystyle y = C\sin x + D\cos x$

•⁴ obtain first and second derivatives of PI
$\displaystyle \frac{dy}{dx} = C\cos x - D\sin x$ and $\displaystyle \frac{d^2y}{dx^2} = -C\sin x - D\cos x$

•⁵ substitute
$\displaystyle -C\sin x - D\cos x - 6(C\cos x - D\sin x) + 9(C\sin x + D\cos x) = 8\sin x + 19\cos x$

•⁶ derive equations
$\displaystyle 8C + 6D = 8$ and $\displaystyle -6C + 8D = 19$

•⁷ obtain both constants of PI
$\displaystyle C = -\frac{1}{2}$, $\displaystyle D = 2$

•⁸ differentiate general solution
$\displaystyle \frac{dy}{dx} = 3Ae^{3x} + Be^{3x} + 3Bxe^{3x} - \frac{1}{2}\cos x - 2\sin x$

•⁹ determine first constant of general solution
$\displaystyle A = 5$ or $\displaystyle B = -14$

•¹⁰ determine second constant and state particular solution
$\displaystyle y = 5e^{3x} - 14xe^{3x} - \frac{1}{2}\sin x + 2\cos x$

•¹ obtain direction vector
$\displaystyle d = \begin{pmatrix} 2 \\ 6 \\ -1 \end{pmatrix}$ or multiple thereof.

•² state parametric equations
$\displaystyle x = 2\lambda + 7$, $\displaystyle y = 6\lambda + 8$, $\displaystyle z = -\lambda + 1$

•³ identify vectors
Any two from $\displaystyle \overrightarrow{PQ} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}$, $\displaystyle \overrightarrow{PR} = \begin{pmatrix} -5 \\ 6 \\ -8 \end{pmatrix}$, $\displaystyle \overrightarrow{QR} = \begin{pmatrix} -4 \\ 5 \\ -6 \end{pmatrix}$

•⁴ evidence of strategy for finding normal
$\displaystyle \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & -2 \\ -5 & 6 & -8 \end{vmatrix}$

•⁵ calculate normal
$\displaystyle n = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}$

•⁶ obtain equation
$\displaystyle 4x + 2y - z = 1$

•⁷ substitute into equation of plane
$\displaystyle 4(2\lambda + 7) + 2(6\lambda + 8) - (-\lambda + 1) = 1$

•⁸ find $\displaystyle \lambda$
$\displaystyle \lambda = -2$

•⁹ determine coordinates of H
$\displaystyle H(3, -4, 3)$

•¹ state form of integral
$\displaystyle V = \pi\int x^2 dy$ or $\displaystyle V = \pi\int (f(y))^2 dy$

•² rearrange and substitute for $\displaystyle x^2$
$\displaystyle V = \pi\int (9 - \frac{9}{4}y^2) dy$

•³ calculate limits to match variable
$\displaystyle y = 0$, $\displaystyle y = 2$

•⁴ integrate
$\displaystyle V = \pi[9y - \frac{3y^3}{4}]_0^2$

•⁵ evaluate
$\displaystyle V = 12\pi$

•¹ state second root
$\displaystyle 2 - i$

•² obtain two linear factors
$\displaystyle z - (2+i)$, $\displaystyle z - (2-i)$

•³ obtain quadratic factor
$\displaystyle z^2 - 4z + 5$

•⁴ set up algebraic division or equivalent
$\displaystyle z^4 - 6z^3 + 16z^2 - 22z + q \div (z^2 - 4z + 5)$ leading to quotient $\displaystyle z^2 - 2z + 3$

•⁵ complete algebraic division/state value of q
$\displaystyle q = 15$

•⁶ obtain remaining two roots
$\displaystyle 1 \pm \sqrt{2}i$

•⁷ show all four solutions on an Argand diagram
Plot $\displaystyle (2,1)$, $\displaystyle (2,-1)$, $\displaystyle (1, \sqrt{2})$, and $\displaystyle (1, -\sqrt{2})$ in correct relative positions.

•¹ evidence of use of product rule to find either $\displaystyle \frac{dx}{dt}$ or $\displaystyle \frac{dy}{dt}$ with one term correct
eg $\displaystyle \frac{dx}{dt} = \cos t + \dots$

•² obtain $\displaystyle \frac{dx}{dt}$ or $\displaystyle \frac{dy}{dt}$
$\displaystyle \frac{dx}{dt} = \cos t - t\sin t$

•³ obtain remaining derivative
$\displaystyle \frac{dy}{dt} = \sin t + t\cos t$

•⁴ state formula for instantaneous speed
$\displaystyle \text{speed} = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$ stated or implied.

•⁵ obtain expression
$\displaystyle \sqrt{(\cos t - t\sin t)^2 + (\sin t + t\cos t)^2} = \sqrt{1+t^2}$

•⁶ evidence of valid strategy to find value of t and obtain at least one non-zero solution
$\displaystyle 0 = t\sin t$ and eg $\displaystyle t = \pi$

•⁷ choose correct value for t and calculate speed
$\displaystyle t = 3\pi$, $\displaystyle \text{speed} = \sqrt{1+9\pi^2}$