2018 A-level H2 Mathematics (9758) Paper 1 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

Numerical Answers (click the questions for workings/explanation)

Question 1: $\frac{1}{x^2} (1 - \text{ln}x)$ ; $1 - \frac{2}{e}$
Question 2: $x= \frac{1}{2}$ or $x=3$ ; $\frac{125}{6}\pi \text{units}^3$
Question 3: $y = 3 -x^2$
Question 4: $x = - 1 \pm \sqrt{3}, 0, -1$ ; $- 1 - \sqrt{3} \textless x \textless -1$ or $0 \textless x \textless -1 + \sqrt{3}$
*Question 5: $b = -1$ , $f^{-1}(x) = 1 + \frac{a+1}{x-1}, x \in \mathbb{R}, x \neq 1, a \neq -1$
Question 6: $\lambda = \pm 2 \sqrt{31}$
Question 7: $N(-\frac{1}{17}, 0)$
Question 8: $A = 5, u_3 = 40$ ; $a =7.5, b =-5, c = -5$ ; $15(2^n - 1) - 7.5n - 2.5n^2$
Question 9: $k= 2$ ; 8 units
Question 10: $I = \frac{3A}{2e}$
Question 11: $102.43;$ 1215.71; first day of June 2018; $b = \frac{4}{3}; b = 1.23$

(i)
$y = \frac{1}{x} \text{ln}x$

$\frac{dy}{dx} = -\frac{1}{x^2} \text{ln}x + \frac{1}{x} \frac{1}{x}$

$\frac{dy}{dx} = \frac{1}{x^2} (1-\text{ln}x)$

(ii)
$\frac{1}{x}\text{ln}x = \int \frac{1}{x^2} -\frac{1}{x^2}\text{ln}x ~dx$

$\int \frac{1}{x^2}\text{ln}x ~dx = \int \frac{1}{x^2} ~dx- \frac{1}{x}\text{ln}x$

$\int \frac{1}{x^2}\text{ln}x ~dx = -\frac{1}{x} - \frac{1}{x}\text{ln}x + C$

$\int_1^e \frac{1}{x^2}\ln x ~dx$

$= \bigg[ - \frac{1}{x} -\frac{1}{x}\text{ln}x \bigg]_1^e$

$= 1 - \frac{2}{e}$

(i)
$\frac{3}{x} + 2x = 7$

$2x^2 -7x + 3 = 0$

$(2x-1)(x-3) = 0$

$x = \frac{1}{2}$ or $x = 3$

(ii)
$\pi \int_{0.5}^3 (-2x+7)^2 - \bigg( (\frac{3}{x} \bigg)^2 ~dx$

$= \pi \int_{0.5}^3 (-2x+7)^2 - (\frac{9}{x^2} ~dx$

$= \pi \bigg[ \frac{(-2x+7)^3}{(-2)(3)} + (\frac{9}{x}) \bigg]_{0.5}^3$

$= \frac{125}{6} \pi \text{units}^3$

(i)
$x \frac{dy}{dx} = 2y - 6$

$y = ux^2$

$\frac{dy}{dx} = \frac{du}{dx}x^2 + 2ux$

$\Rightarrow x (\frac{du}{dx}x^2 + 2ux) = 2(ux^2) - 6$

$\frac{du}{dx} = - \frac{6}{x^3}$

$\therefore f(x) = - \frac{6}{x^3}$

(ii)
$u = \frac{3}{x^2} + C$

$\frac{y}{x^2} = \frac{3}{x^2} + C$

$y = 3 + Cx^2$

When $x=1, y =2, C=-1$

$\therefore y = 3 - x^2$

(i)
$\vert 2x^2 + 3x -2 \vert = 2 -x$

$2x^2 + 3x -2 = 2 -x$

$2x^2 + 4x - 4 = 0$

$x = \frac{-4 \pm \sqrt{4^2 -4(2)(-4)}}{2(2)} = -1 \pm \sqrt{3}$

$2x^2 + 3x -2 = x - 2$

$2x^2 + 2x = 0$

$x = 0$ or $x = -1$

$\therefore, x = -1 \pm \sqrt{3}, -1, 0$

(ii)

Graph of 4ii

Note: The graphs should cut the same y intercept.

$- 1 - \sqrt{3} \textless x \textless -1$ or $0 \textless x \textless -1 + \sqrt{3}$

$f(x) = \frac{x+a}{x+b} = 1 + \frac{a-b}{x+b}$

Since $f(x)$ has asymptote at $y=1$ and for $ff = g$ , $f$ must be self inverse, we have that $x =1$ must be an asymptote for $f$ .

Thus, $b = -1$ .

Let $f(x) = y$

$f(x) = 1 + \frac{a+1}{x-1}$

$\frac{a+1}{x-1} = y-1$

$x - 1 = \frac{a+1}{y-1}$

$x = 1 + \frac{a+1}{y-1}$

$f^{-1}(x) = 1 + \frac{a+1}{x-1}, x \in \mathbb{R}, x \neq 1, a \neq -1$

(i)
$\textbf{a} \times 3\textbf{b} = 2 \textbf{a} \times \textbf{c}$

$\textbf{a} \times 3\textbf{b} - 2 \textbf{a} \times \textbf{c} = \textbf{0}$

$\textbf{a} \times ( 3\textbf{b} - 2 \textbf{c} )= \textbf{0}$

$\Rightarrow 3\textbf{b} - 2 \textbf{c}$ is parallel to $\textbf{a}$

$\therefore 3\textbf{b} - 2 \textbf{c} = \lambda \textbf{a}$ for $\lambda$ is a scalar.

Note to students: Must show zero vector, i.e., $\textbf{0}$

(ii)
$\textbf{b} \cdot \textbf{c} = \vert \textbf{b} \vert \vert \textbf{c} \vert \text{cos} 60^{\circ}$

$\textbf{b} \cdot \textbf{c} = (4)(1) \frac{1}{2} = 2$

From $3\textbf{b} - 2 \textbf{c} = \lambda \textbf{a}$

$\vert 3\textbf{b} - 2 \textbf{c} \vert ^2= \lambda ^2 \vert \textbf{a} \vert ^2$

$(3\textbf{b} - 2 \textbf{c}) \cdot (3\textbf{b} - 2 \textbf{c}) = \lambda ^2 (1)^2$

$9 \textbf{b} \cdot \textbf{b} - 12 \textbf{b} \cdot \textbf{c} + 4 \textbf{c} \cdot \textbf{c} = \lambda ^2$

$9 \vert \textbf{b} \vert ^2 - 12 (2) + 4 \vert \textbf{c} \vert ^2 = \lambda ^2$

$9 (4)^2 - 12 (2) + 4 (1) ^2 = \lambda ^2$

$\lambda = \pm \sqrt{124} = \pm 2 \sqrt{31}$

(i)
$2x^2 - 8y^2 = x^2 + xy^2$

Differentiating wrt $x$ ,

$4x - 16 y \frac{dy}{dx} = 2x + y^2 + x(2y \frac{dy}{dx})$

$\frac{dy}{dx}( 2xy + 16y) = 4x - 2x - y^2$

$\frac{dy}{dx} = \frac{2x-y^2}{2xy+16y}$

(ii)
When $x =1$ , $2-8y^2 = 1 + y^2 \Rightarrow y = \pm \frac{1}{3}$

$\frac{dy}{dx} = \frac{2 - \frac{1}{9}}{ 2(\pm \frac{1}{3}) + 16 (\pm \frac{1}{3})}$

$\frac{dy}{dx} = \pm \frac{17}{54}$

$\text{Tangent}_1: y - \frac{1}{3} = \frac{17}{54} (x - 1)$ —(1)

$\text{Tangent}_2: y + \frac{1}{3} = -\frac{17}{54} (x - 1)$ —(2)

(1) – (2),

$- \frac{2}{3} = \frac{17}{54} (x - 1) + \frac{17}{54} (x - 1)$

$- \frac{2}{3} = \frac{17}{27}(x-1)$

$x = - \frac{1}{17}$

$y = \frac{17}{54} (- \frac{1}{17} - 1) + \frac{1}{3} = 0$

$\therefore N(-\frac{1}{17}, 0)$ .

(i)
$u_2 = 2u_1 + A$

$15 = 2(5) + A \Rightarrow A = 5$

$u_3 = 2u_2 + 2A = 40$

(ii)
$u_n = a(2^n) + bn + c$

$u_1 = 2a + b + c = 5$ —(1)

$u_2 = 4a + 2b + c = 15$ —(2)

$u_3 = 8a + 3b + c = 40$ —(3)

Using GC, $a = 7.5, b = -5, c = -5$ .

(iii)
$\sum_{r=1}^n u_r$

$= \sum_{r=1}^n \bigg[ 7.5 (2^r) - 5r - 5 \bigg]$

$= 7.5 \frac{2(2^n - 1)}{2-1} - 5 \frac{n}{2}(1+n) - 5n$

$= 15 (2^n - 1) - 7.5n - 2.5n^2$

(i)
$x = 2 \theta - \text{sin} \theta \Rightarrow \frac{dx}{d\theta} = 2 - 2\text{cos}2\theta = 4 \text{sin}^2 \theta$

$y = 2 \text{sin}^2 \theta \Rightarrow \frac{dy}{d\theta} = 4 \text{sin}\theta \text{cos} \theta$

$\frac{dy}{dx} = \frac{4 \text{sin}\theta \text{cos} \theta }{4 \text{sin}^2 \theta}$

$\frac{dy}{dx} = \frac{\text{cos}\theta}{\text{sin}\theta} = \text{cot}\theta$

(ii)
Normal: $y - 2 \text{sin}^2 \alpha = - \frac{1}{\text{cot} \alpha} [x - (2 \alpha - \text{sin}2\alpha)]$

When $y=0$ ,

$- 2 \text{sin}^2 \alpha = - \text{tan}\alpha [x - 2 \alpha + \text{sin}2\alpha]$

$2 \text{sin}\alpha \text{cos}\alpha = x - 2 \alpha + 2 \text{sin}\alpha \text{cos}\alpha$

$x = 2\alpha$

$\therefore k = 2$

(iii)
$L = \int_0^{\pi} \sqrt{(4\text{sin}^2 \theta)^2 + (4 \text{sin} \theta \text{cos} \theta)^2} ~d \theta$

$L = \int_0^{\pi} \sqrt{16 \text{sin}^2 \theta (\text{sin}^2 \theta + \text{cos}^2 \theta)} ~d\theta$

$L = \int_0^{\pi} 4 \text{sin}\theta ~d \theta$

$L = \bigg[ - 4 \text{cos}\theta \bigg]_0^{\pi}$

$L = -4(-1) - [-4(1)]$

$L = 8$ units.

Note that students need to show the workings, and do it H2 Math way, not H2 F Math.

Q10

(i)
If $V$ is a constant,

$L \frac{dI}{dt} + RI + \frac{q}{C} = V$

$L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{1}{C} \frac{dq}{dt} = 0$

Since $I = \frac{dq}{dt}$ ,

$L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{I}{C} = 0$

(ii)

Since $I = Ate^{-\frac{Rt}{2L}}$ is a given solution, it will satisfy the given differential equation. Differentiating wrt $x$ ,

$\frac{dI}{dt} = Ae^{-\frac{Rt}{2L}} + At(-\frac{R}{2L})e^{-\frac{Rt}{2L}}$

$\frac{dI}{dt} = Ae^{-\frac{Rt}{2L}} - \frac{R}{2L} I$

Differentiate wrt $x$ ,

$\frac{d^2I}{dt^2} = A(-\frac{R}{2L})e^{-\frac{Rt}{2L}} - \frac{R}{2L} \frac{dI}{dt}$

$\frac{d^2I}{dt^2} = A(-\frac{R}{2L})e^{-\frac{Rt}{2L}} - \frac{R}{2L} [Ae^{-\frac{Rt}{2L}} - \frac{R}{2L} I]$

$\frac{d^2I}{dt^2} = -\frac{AR}{L}e^{-\frac{Rt}{2L}} + \frac{R^2}{4L^2} I$

We are given that

$L \frac{d^2I}{dt²} + R \frac{dI}{dt} + \frac{I}{C} = 0$

$L [-\frac{AR}{L}e^{-\frac{Rt}{2L}} + \frac{R^2}{4L^2} I ] + R [Ae^{-\frac{Rt}{2L}} - \frac{R}{2L} I] + \frac{I}{C} = 0$

$-AR e^{-\frac{Rt}{2L}} +\frac{R^2}{4L}I + ARe^{-\frac{Rt}{2L}} - \frac{R^2}{2L} I + \frac{I}{C} = 0$

$\frac{R^2}{4L}I - \frac{R^2}{2L} I + \frac{I}{C} = 0$

$\frac{R^2}{4L} - \frac{R^2}{2L} + \frac{1}{C} = 0$

$\frac{R^2}{4L} = \frac{1}{C}$

$C = \frac{4L}{R^2}$

An alternative Method

$I = Ate^{- \frac{Rt}{2L}} = Ate^{Bt}$ , where $B = -\frac{R}{2l}$

$\frac{dq}{dt} = Ate^{Bt}$

$q = A \int te^{Bt} ~dt$

$q = A ( \frac{1}{B}e^{Bt}t - \int \frac{1}{B} e^{Bt} ~dt)$

$q = \frac{A}{B} (te^{Bt} - \frac{1}{B}e^{Bt})$

$L \frac{dI}{dt} + RI + \frac{q}{C} = V$

Integrating wrt $t$ , note that since $I = \frac{dq}{dt} \Rightarrow \int I ~dt = q$ ,

$L I + Rq + \frac{1}{C} \int q ~dt = Vt$

$L I + Rq + \frac{1}{C} \int \frac{A}{B} (te^{Bt} - \frac{1}{B}e^{Bt}) ~dt = Vt$

$L I + Rq + \frac{A}{BC} \int (te^{Bt} - \frac{1}{B}e^{Bt}) ~dt = Vt$

$L I + Rq + \frac{A}{BC} (\frac{1}{B}e^{Bt}t - 2 \int \frac{1}{B}e^{Bt}~dt) = Vt$

$L I + R \frac{A}{B} (te^{Bt} - \frac{1}{B}e^{Bt}) + \frac{A}{BC} ( \frac{1}{B}e^{Bt}t - \frac{2}{B^2}e^{Bt}) = Vt$

Sub $t=0, I = 0$

$0 + R\frac{A}{B} (0 - \frac{1}{B}) + \frac{A}{BC} (0 - \frac{2}{B^2}) = 0$

$- \frac{R}{B^2} - \frac{2}{B^3 C} = 0$

$C = - \frac{2}{RB}$

$C = - \frac{2}{R( - \frac{R}{2L})}$

$C = \frac{4L}{R^2}$

(iii)
$I = At e^{-\frac{2}{3}t}$

$\frac{dI}{dt} = Ae^{-\frac{2}{3}t} - \frac{2}{3} Ate^{-\frac{2}{3}t}$

$\frac{dI}{dt^2} = Ae^{-\frac{2}{3}t} (1 - \frac{2}{3}t)$

Let $\frac{dI}{dt} = 0 \Rightarrow t = 1.5$

$\frac{d^2I}{dt^2} = - \frac{2}{3}Ae^{-\frac{2}{3}t} (1- \frac{2}{3}t) + Ae^{-\frac{2}{3}t} (-\frac{2}{3})$

When $t = 1.5$

$\frac{d^2I}{dt^2} = Ae^{-1}(- \frac{2}{3}) \textless 0$

Thus, I is maximum at $\frac{3A}{2e}$ when $t = 1.5$

(iv)

10iv Graph of I against t

Q11

(ia)
$100(1.002)^{12} = 102.43$

It is worth 102.43 dollars.

(ib)
$\frac{100(1.002)(1.002^{12} - 1)}{1.002 - 1} = 1215.71$

Total amount is 1215.71 dollars.

(ic)
Total amount at end of $n$ months $= \frac{100(1.002)(1.002^{n} - 1)}{1.002 - 1} = 50100 (1.002^n - 1)$

Using GC,
when $n = 29$ , total amount $= 2988.65$

when $n = 30$ , total amount $= 3094.825$

Since he deposit 100 dollars at the start of every month, on the first day of June 2018, his account will exceed 3000 dollars.

(iia)
It is worth (100 + 12b) dollars

(iib)
$24 \times 100 + \frac{24}{2}(b + 24b) = 2800 \Rightarrow b = \frac{4}{3}$

(iii)
For plan P, total amount at the end of 60 months $= \frac{100(1.01)(1.01^{60} - 1)}{1.01 - 1} = 8248.636655$

$8248.636655 = 60 \times 100 + \frac{60}{2}(b + 60b)$

$b = 1.23$

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KS Comments

Fairly interesting paper. It was tedious at some point. It is good to note that some H2 Further Mathematics content or techniques were brought down and tested in this paper. Students should be comfortable with unseen questions and being in unfamiliar territories while applying known or learnt concepts well.

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KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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