2019 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 (workings/explanations are after the numerical answers.)

Question 1: $b = -a, c = -7a, d = 15a$
Question 2: $a = 0.682; ~b = 1.891$
Question 3: $f(x) = 2 \{(x - 1)^3 - 5 \}$
Question 4: $x \in [2, 4]$
Question 5: $f^{-1}(x) = \frac{1}{2} \ln (x + 4), D_{f^{-1}(x)} = (-4, \infty );~ x = \ln 3 - 2$
Question 6: $\frac{1}{2} \bigg( 1 - \frac{1}{2n+1} \bigg); ~\frac{1}{42}$
Question 7: $y = \frac{1}{e}, y = 2e x + e;~ 79.6^{\circ}$
Question 8: $k = 13;~ r = -1, f \in \mathbb{R}, f \neq 0, S_n = 0 \text{~or~} f;~ T_{11} = 63$
Question 9: $w + \frac{1}{w} = 2 \cos \theta;~ k = i;~ 1$
Question 10: $y = 0;~ \theta = 0, \pi, 2\pi;~ \theta_1 = 0, \theta_2 = \pi; 6 \pi a^2$
Question 11: $\frac{d \theta}{dt} = - k(\theta - 16), \theta = 16 + 64 \bigg( \frac{1}{4} \bigg)^{t/30};24^{\circ} C~ t = 540 \text{minutes}$
Question 12: $Q \bigg( \frac{8}{11}, \frac{1}{11}, \frac{2}{11} \bigg), R \bigg(- \frac{37}{11}, -\frac{39}{11},- \frac{23}{11} \bigg); ~\cos \theta = \frac{11\sqrt{3}}{21}, \cos \beta = \frac{11\sqrt{510}}{255}; ~\frac{10}{3}\sqrt{3};~ k = 1.86;~ k < 1$

Since the coefficients of the equation are real, $z = 2 + i$ is a root, then $z = 2 -i$ is also a root.
$a[z - (2 + i)][z - (2 - i)](z+3) = az^3 + bz^2 + cz + d$
$a(z^2 - 4z + 5)(z+3) = az^3 + bz^2 + cz + d$
By comparing coefficients,
$b = -a, c = -7a, d = 15a$

(i)
$0 = a^3 + a - 1 \implies a \approx 0.682$

(ii)
$2 \int_{-1}^0 -(x^3 + x -1) dx = \int_a^b x^3 + x - 1 dx$
$3.5 = \bigg[ \frac{1}{4}x^4 + \frac{1}{2}x^2 - x \bigg]_a^b$
Using GC, $b \approx 1.892$

(i)
$f(x) = (2x^3 - 6x^2 + 6x - 12)$
$f(x) = 2(x^3 - 3x^2 + 3x - 6)$
$f(x) = 2(x^3 - 3x^2 + 3x - 1 - 5)$
$f(x) = 2\{ (x-1)^3 - 5\}$
$\therefore p = 2, q = -1, r = - 5$

(ii)
Firstly, translate the graph 1 unit in the positive $x$ direction.
Secondly, translate the graph 5 units in the negative $y$ direction.
Lastly, scale the graph by a factor of 2 units parallel to $y$ axis.

(i)

(ii)
$| 2^x - 10 | \le 6$
$- 6 \le 2^x - 10 \le 6$
$4 \le 2^x \le 16$
$\ln 4 \le x \ln 2 \le \ln 16$
$2 \ln 2 \le x \ln 2 \le 4 \ln 2$
$2 \le x \le 4$
Required interval $= [2, 4]$

(i)
Let $f(x) = y$
$f^{-1}(x) = \frac{1}{2} \ln (x + 4)$
$D_{f^{-1}(x)} = R_f = (-4, \infty)$

(ii)
$fg(x) = 5$
$g(x) = f^{-1}(5)$
$x + 2 = \frac{1}{2} \ln (5 + 4)$
$x = \ln 3 - 2$

(i)
$\frac{1}{4r^2 -1} = \frac{1}{2} \bigg( \frac{1}{2r-1} - \frac{1}{2r+1} \bigg)$
Using method of differences,
$\sum_{r=1}^n \frac{1}{4r^2 -1} = \frac{1}{2} \bigg( 1 - \frac{1}{2n+1} \bigg)$

(ii)
$\sum_{r=11}^{\infty} \frac{1}{4r^2 -1}$
$= \sum_{r=1}^{\infty} \frac{1}{4r^2 -1} - \sum_{r=1}^{10} \frac{1}{4r^2 -1}$
$= \frac{1}{2} - \frac{1}{2} \bigg( 1 - \frac{1}{21} \bigg)$
$= \frac{1}{42}$

(i)
$y = xe^{-x}$
$\frac{dy}{dx} = e^{-x}( 1 - x)$
At $x =1$ , $\frac{dy}{dx} = 0$ , tangent is $y = e^{-1}$ .
At $x = - 1$ , $\frac{dy}{dx} = 2e$ , tangent is $y = 2ex + e$ .

(ii)
Required angle $= \tan ^{-1} (2e) \approx 79.6^{\circ}$

(a)
$a(2)^{k-1} = \frac{64}{2} [ 2a + (64-1)(2a)]$
$2^{k-1} = 32( 128)$
$\implies k = 13$

(b)
$\frac{f(1-r^4)}{1-r} = 0$
$\implies r = - 1, f \in \mathbb{R}, f \neq 0$
$S_n = \frac{f[ 1 - (-1)^n]}{1 - (-1)}$
$S_n = \frac{1}{2} f [ 1 - (-1)^n]$
When $n$ is even, $S_n = 0$ .
When $n$ is odd, $S_n = f$ .

(c)
$\frac{4}{2} (2a + 3d) = 14 \implies 2a + 3d = 7$
$a(a+d)(a+2d)(a+3d) = 0 \implies a = 0, ~ a = -d,~ a = -2d, ~a = -3d$
Since $a$ is negative, then $a= -d \implies d = 7 \implies a = -7$
$T_{11} = -7 + (11-1)(7) = 63$

(i)
Note that $w = \cos \theta + i \sin \theta = e^{i \theta}$
$w + \frac{1}{w} = e^{i \theta} + \frac{1}{e^{i \theta}} = e^{i \theta} + e^{-i \theta} = 2 \cos \theta$
Since $\text{Im}\bigg( w + \frac{1}{w} \bigg) = 0$ , it is a real number.

(ii)
$\frac{w-1}{w+1}$
$= \dfrac{e^{i \theta} - 1}{e^{i \theta} + 1}$
$= \dfrac{e^{i \frac{\theta}{2}}(e^{i \frac{\theta}{2}} - e^{-i \frac{\theta}{2}})}{e^{i \frac{\theta}{2}}(e^{i \frac{\theta}{2}} + e^{-i \frac{\theta}{2}})}$
$= \dfrac{(e^{i \frac{\theta}{2}} - e^{-i \frac{\theta}{2}})}{(e^{i \frac{\theta}{2}} + e^{-i \frac{\theta}{2}})}$
$= \dfrac{ 2 i \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}}$
$= i \tan \frac{\theta}{2}$
$\implies k = i$

(iii)
Note that since $|z| = 1$ , then $z = e^{i \theta}$ and $\frac{1}{z} = \frac{1}{e^{i \theta}} = e^{-i \theta} = z^*$ .
$\bigg| \dfrac{z - 3i}{1 + 3iz} \bigg|$
$\bigg| \dfrac{z(1 - \frac{3i}{z})}{1 + 3iz} \bigg|$
$\bigg| \dfrac{z(1 - 3iz^*)}{1 + 3iz} \bigg|$
Since $(zw)^* = z^* w^*$ ,
$\bigg| \dfrac{z(1 + 3iz)^*}{1 + 3iz} \bigg|$
$=\dfrac{| z| |(1 + 3iz)^*|}{|1 + 3iz|}$
Since $|z^*| = |z|$ ,
$=\dfrac{| z| |1 + 3iz|}{|1 + 3iz|}$
$= 1$

Q10

(i)
Line of symmetry is $y = 0$

(ii)
$y = 0$
$a (2\sin \theta - \sin 2 \theta) = 0$
$2 \sin \theta - 2 \sin \theta \cos \theta = 0$
$\sin \theta = 0 \implies \theta = 0, \pi, 2 \pi$
$1 - \cos \theta = 0 \implies \theta = 0, 2\pi$
$\therefore \theta = 0, \pi, 2 \pi$

(ii)
When $\theta = 0, x = a$
When $\theta = \pi, x = - 3a$
$A = \int_{-3a}^a y dx$
$= \int_{\pi}^0 a (2 \sin \theta - \sin 2 \theta) [a (- 2 \sin \theta + 2 \sin 2 \theta)] d \theta$
$= \int_0^{\pi} a^2 (2 \sin \theta - \sin 2 \theta) (2 \sin \theta - 2 \sin 2 \theta) d \theta$
$= \int_{0}^{\pi} a^2 (4 \sin ^2 \theta - 6 \sin \theta \sin 2 \theta + 2 \sin ^2 2 \theta ) d\theta$
$\therefore \theta_1 = 0, \theta_2 = \pi$

(iii)

$A = \int_{0}^{\pi} a^2 (4 \sin ^2 \theta - 6 \sin \theta \sin 2 \theta + 2 \sin ^2 2 \theta ) d\theta$
$= a^2 \int_{0}^{\pi} 2 - 2\cos 2 \theta - 12 \sin ^2 \theta \cos \theta + 1 - \cos 4 \theta d\theta$
$= a^2 \bigg[ 3 \theta - \sin 2 \theta - 4 \sin ^3 \theta - \frac{1}{4} \sin 4 \theta \bigg]_0^{\pi}$
$= 3 \pi a^2$
Total area enclosed $= 2 \times 3 \pi a^2 = 6\pi a^2$

Q11

(ia)
$\frac{d \theta}{dt} = -k ( \theta - 16)$ where $k$ is a positive constant.
$\int \frac{1}{\theta - 16} d \theta = \int -k dt$
$\ln |\theta - 16 | = - kt + c$
$\theta = 16 + Ae^{-kt}$ where $A = \pm e^c$
When $t = 0, \theta =80 \implies A = 64$
When $t = 30, \theta = 32$ ,
$32 = 16 + 64e^{-30k}$
$\frac{1}{4} = e^{-30k}$
$\implies \big( e^{-k} \big)^{30} = \frac{1}{4}$
$\theta = 16 + 64e^{-kt}$
$\theta = 16 + 64 \big(e^{-k} \big)t$
$\theta = 16 + 64 \big(\frac{1}{4} \big)^{t/30}$

(ib)
When $t = 45, \theta = 24^{\circ}C$

(ii)
$\frac{dT}{dt} = \frac{L}{T}$ , where $L$ is a positive constant.
$\int T dT = \int L dt$
$\frac{1}{2} T^2 = Lt + C$
When $t = 0, T = 0 \implies C = 0$
When $t = 60, T =1, L = \frac{1}{120}$
$\therefore \frac{1}{2}T^2 = \frac{1}{120} t$
When $T = 3, 540$ minutes.

Q12

(i)
$L_{PQ} : \textbf{r} = \begin{pmatrix}2\\2\\4\end{pmatrix} + \lambda \begin{pmatrix} -2\\ -3\\ -6\end{pmatrix}, \lambda \in \mathbb{R}$
$\pi_{top}: \textbf{r} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} = 1$
At $Q$ ,
$\begin{pmatrix}2 - 2 \lambda \\2 - 3 \lambda \\4 - 6 \lambda \end{pmatrix} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} = 1$
$\lambda = \frac{7}{11}$
$\therefore Q \bigg( \frac{8}{11}, \frac{1}{11}, \frac{2}{11} \bigg)$
$L_{RS} : \textbf{r} = \begin{pmatrix}-5\\-6\\-7\end{pmatrix} + \mu \begin{pmatrix} -2\\ -3\\ -6\end{pmatrix}, \mu \in \mathbb{R}$
$\pi_{bottom}: \textbf{r} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} = -9$
At $R$ ,
$\begin{pmatrix}-5 - 2 \mu \\-6 - 3 \mu \\-7 - 6 \mu \end{pmatrix} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} = 1$
$\mu = - \frac{9}{11}$
$R \bigg(- \frac{37}{11}, -\frac{39}{11},- \frac{23}{11} \bigg)$

(ii)
$\cos \theta = \frac{\bigg| \begin{pmatrix} -2\\ -3\\ -6\end{pmatrix} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} \bigg|}{\bigg| \begin{pmatrix} -2\\ -3\\ -6\end{pmatrix} \bigg| \bigg| \begin{pmatrix}1\\1\\1\end{pmatrix} \bigg|}= \frac{11\sqrt{3}}{21}$
$\vec{QR} = \begin{pmatrix} -45/11 \\ -40/11 \\ -25/11\end{pmatrix} = \frac{-5}{11} \begin{pmatrix} 9\\ 8 \\5\end{pmatrix}$
$\cos \beta = \frac{\bigg| \begin{pmatrix} 9\\ 8 \\5\end{pmatrix} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} \bigg|}{\bigg|\begin{pmatrix} 9\\ 8 \\5\end{pmatrix} \bigg| \bigg| \begin{pmatrix}1\\1\\1\end{pmatrix} \bigg|}= \frac{11\sqrt{510}}{255}$

(iii)
Thickness $= \dfrac{1}{\bigg|\begin{pmatrix}1\\1\\1\end{pmatrix} \bigg|} - \dfrac{-9}{\bigg|\begin{pmatrix}1\\1\\1\end{pmatrix} \bigg|} = \frac{10}{\sqrt{3}}$

(iv)
By finding $\theta$ and $\beta$ , we have that $k = 1.86$

(v)
Since $\sin x$ is an increasing function from $0 \le x \le 90^{\circ}$ ,
$0 < \theta < \beta$
$\sin 0< \sin \theta < \sin \beta$
$0< \frac{\sin \theta}{\sin \beta} < 1$
Since $k = \frac{\sin \theta}{ \sin \beta}$ , we have that $0 < k < 1$ .

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KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for almost two decades. 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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2019 A-level H2 Mathematics (9758) Paper 1 Suggested Solutions

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