2021 H2 Math Paper 1 Full Solutions

{f(x)=ax^3+bx^2+cx+d}

{f(1)=5 \textrm{ and } f(-1)=-3:}

\begin{align} a + b + c + d&=5 \\ - a + b - c + d&=-3 \end{align}

{f'(x)=3ax^2 + 2bx + c}

{f'(1)=0:}

\begin{align} 3 a + 2 b + c&=0 \end{align}

{\int f(x) \; \mathrm{d}x = \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 + dx + C'}

\int_0^1 f(x) \; \mathrm{d}x = 6:

\begin{align} \frac{1}{4} a + \frac{1}{3} b + \frac{1}{2} c + d&=6 \end{align}

Solving (1)-(4) simultaneously,

a=4, \; b=-6, \; c=0, \; d=7

\begin{gather*} |x^2-5| = 2x-1 \Rightarrow \\ x^2-5=2x-1 \; \textrm{ or } \; {x^2-5=-(2x-1)} \end{gather*}

\begin{gather*} x^2+5=2x-1 \\ x^2 - 2 x - 4=0 \\ \begin{aligned} x &= \frac{-(-2) \pm \sqrt{(-2)^2 - 4 (1) (-4)}}{2 (1)} \\ &= 1-\sqrt{5} \textrm{ (N.A.) } \; \textrm{ or } \; 1+\sqrt{5} \end{aligned} \end{gather*}

\begin{gather*} x^2+5=-(2x-1) \\ x^2 + 2 x - 6=0 \\ \begin{aligned} x &= \frac{-(2) \pm \sqrt{(2)^2 - 4 (1) (-6)}}{2 (1)} \\ &= -1-\sqrt{7} \textrm{ (N.A.) } \; \textrm{ or } \; -1+\sqrt{7} \end{aligned} \end{gather*}

\textrm{Solutions: } x=1 + \sqrt{5} \; \textrm{ or } \; x=- 1 + \sqrt{7}

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 3

Differentiating implicitly w.r.t.

{x,}

\frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2}y^{-\frac{1}{2}} \frac{\mathrm{d}y}{\mathrm{d}x} = 0

\frac{1}{y^{\frac{1}{2}}} \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x^{\frac{1}{2}}}

\frac{\mathrm{d}y}{\mathrm{d}x} = -\left(\frac{y}{x}\right)^{\frac{1}{2}}

When

{x=1,}

1 + y^{\frac{1}{2}} = 3

{y=4}

\frac{\mathrm{d}y}{\mathrm{d}x} = -2

Gradient of normal

{= \frac{1}{2}}

{y-4 = \frac{1}{2}(x-1)}

Equation of normal:

{y = \frac{1}{2} x + \frac{7}{2}}

\begin{aligned} z &= \frac{\left( \cos \left( \frac{1}{16} \pi \right) + \mathrm{i} \sin \left( \frac{1}{16} \pi \right) \right)^2}{\cos \left( \frac{1}{8} \pi \right) - \mathrm{i} \sin \left( \frac{1}{8} \pi \right)} \\ &= \frac{\left( \mathrm{e}^{\frac{1}{16} \pi \mathrm{i}} \right)^2}{\mathrm{e}^{- \frac{1}{8} \pi \mathrm{i}}} \\ &= \mathrm{e}^{\frac{1}{4} \pi \mathrm{i}} \\ |z| &= 1 \\ \arg(z) &= \frac{1}{4} \pi \\ z^2 &= \mathrm{e}^{\frac{1}{2} \pi \mathrm{i}} \\ &= \mathrm{i} \end{aligned}

\begin{aligned} & (\cos \theta + \mathrm{i}\sin \theta)(1+\cos \theta + \mathrm{i}\sin \theta) \\ &= \mathrm{e}^{\mathrm{i}\theta}(1+\mathrm{e}^{-\mathrm{i}\theta}) \\ &= \mathrm{e}^{\mathrm{i}\theta}+\mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\theta} \\ &= \mathrm{e}^{\mathrm{i}\theta}+1 \\ &= 1 + \cos \theta + \mathrm{i} \sin \theta \end{aligned}

If

{z=\cos \theta + \mathrm{i}\sin \theta,}

then part (i) gives us

{z(1+z^*)=1+z.}

\begin{aligned} & (1+z)^4 + (1+z^*)^4 \\ &= \left( z(1+z^*) \right)^4 + (1+z^*)^4 \\ &= (1+z^*)^4(z^4+1) \\ &= (1+z^*)^4(\mathrm{i}^2+1) \\ &= 0 \end{aligned}

By the cover-up rule,

\begin{aligned} & \frac{x}{(x+2)(x+3)(x+4)} \\ &= -\frac{1}{x+2} + \frac{3}{x+3} - \frac{2}{x+4} \end{aligned}

\begin{aligned} & \sum_{r=1}^n \frac{r}{(r+2)(r+3)(r+4)} \\ & = \sum_{r=1}^n -\frac{1}{r+2} + \frac{3}{r+3} - \frac{2}{r+4} \\ & = \def\arraystretch{1.5} \begin{array}{lclclc} - & \frac{1}{3} &+& \frac{3}{4} &-& \cancel{\frac{2}{5}} \\ - & \frac{1}{4} &+& \cancel{\frac{3}{5}} &-& \cancel{\frac{2}{6}} \\ - & \cancel{\frac{1}{5}} &+& \cancel{\frac{3}{6}} &-& \cancel{\frac{2}{7}} \\ - & && \cdots && \\ - & \cancel{\frac{1}{n}} &+& \cancel{\frac{3}{n+1}} &-& \cancel{\frac{2}{n+2}} \\ - & \cancel{\frac{1}{n+1}} &+& \cancel{\frac{3}{n+2}} &-& {\frac{2}{n+3}} \\ - & \cancel{\frac{1}{n+2}} &+& \frac{3}{n+3} &-& {\frac{2}{n+4}} \end{array} \\ & = \frac{1}{6} + \frac{1}{n+3} - \frac{2}{n+4} \end{aligned}

As

{n\to \infty,}

{\frac{1}{n+3}, \frac{2}{n+4} \to 0}

\sum_{r=1}^n \frac{r}{(r+2)(r+3)(r+4)} = \frac{1}{6}

Smallest value of

{y}

occurs when

{4a^2-x^2}

is largest
This occurs at

{x=2a}

Smallest value of

{y = \frac{1}{2a}}

By completing the square,

y = \frac{1}{4ax-x^2} = \frac{1}{4a^2-(x-2a)^2}

Hence replacing

{x}

in the equation of

{C}

with

{x+2a}

allows us to get

{y=\displaystyle \frac{1}{4a^2-x^2}}

Transformation required: translation of

{2a}

units in the negative

{x}

-direction

\begin{gather*} y = \mathrm{e}^{\sin^{-1}x} \\ \textrm{Differentiating w.r.t. } x \\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1-x^2}} \mathrm{e}^{\sin^{-1}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1-x^2}} y \\ \sqrt{1-x^2} \frac{\mathrm{d}y}{\mathrm{d}x} = y \\ (1-x^2) \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 = y^2 \\ \textrm{Differentiating w.r.t. } x \\ (1-x^2) \left( 2 \frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right) - 2x \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 = 2 y \frac{\mathrm{d}y}{\mathrm{d}x} \\ (1-x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x \frac{\mathrm{d}y}{\mathrm{d}x} + y \end{gather*}

\begin{gather*} \textrm{Differentiating w.r.t. } x \\ (1-x^2) \frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2x \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{\mathrm{d}y}{\mathrm{d}x} \end{gather*}

\begin{gather*} \textrm{When } x=0, \\ y = 1, \frac{\mathrm{d}y}{\mathrm{d}x} = 1 \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 1, \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 2 \end{gather*}

\begin{align*} \mathrm{e}^{\sin^{-1}x} &= 1 + 1x + \frac{1}{2!}x^2 + \frac{2}{3!}x^3 +\ldots \\ &= 1 + x + \frac{1}{2} x^2 + \frac{1}{3} x^3 + \ldots \end{align*}

Let

{A (3,2,0)}

be a point on

{l_1}

and

{B = (1,-3,-1)}

\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 2 \\ - 5 \\ - 1 \end{pmatrix} \end{align*}

\begin{align*} \mathbf{n'} &= \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} - 2 \\ - 5 \\ - 1 \end{pmatrix} \\ &= \begin{pmatrix} 8 \\ 0 \\ - 16 \end{pmatrix} \\ \mathbf{n} &= \begin{pmatrix} 1 \\ 0 \\ - 2 \end{pmatrix} \end{align*}

\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ &= \begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ - 2 \end{pmatrix} \\ &= 3 \end{align*}

Cartesian equation of plane:

{x-2z=3}

\begin{align*} \mathbf{d_1} \cdot \mathbf{d_2} &= \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} \\ &= 2 - 6 + 4 \\ &= 0 \end{align*}

Hence the direction vectors of

{l_1}

and

{l_2}

are perpendicular.
Hence

{l_1}

is perpendicular to

{d_2}

.

\mathbf{r_1}-\mathbf{r_2}= \begin{pmatrix} 7 + 2 \lambda - \mu \\ 1 - 3 \lambda - 2 \mu \\ 3 + \lambda - 4 \mu \end{pmatrix}

\begin{gather*} (\mathbf{r_1} - \mathbf{r_2}) \cdot \mathbf{d_1} = 0 \\ \begin{pmatrix} 7 + 2 \lambda - \mu \\ 1 - 3 \lambda - 2 \mu \\ 3 + \lambda - 4 \mu \end{pmatrix} \cdot \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} = 0 \\ 14 \lambda = -14 \\ \lambda = -1 \end{gather*}

\begin{gather*} (\mathbf{r_1} - \mathbf{r_2}) \cdot \mathbf{d_2} = 0 \\ \begin{pmatrix} 7 + 2 \lambda - \mu \\ 1 - 3 \lambda - 2 \mu \\ 3 + \lambda - 4 \mu \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = 0 \\ - 21 \mu = -21 \\ \mu = 1 \end{gather*}

When

{\lambda = -1, \mathbf{r_1} = \begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}}

When

{\mu = 1, \mathbf{r_2} = \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}}

Hence the common perpendicular meets

{l_1}

and

{l_2}

at points with position vectors

{\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}}

and

{\begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}}

\begin{align*} & \textrm{Length of perpendicular } \\ &= \left|\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix} - \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix} \right| \\ &= \left|\begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix} \right| \\ &= 2 \sqrt{6} \end{align*}

\begin{gather*} f(x) = \mathrm{e}^x \cos x \\ f'(x) = \mathrm{e}^x \cos x - \mathrm{e}^x \sin x \\ \textrm{At stationary point, } f'(x) = 0 \\ \mathrm{e}^x (\cos x - \sin x ) = 0 \\ \cos x = \sin x \\ \tan x = 1 \\ x = \frac{\pi}{4} \\ f(x) = \frac{\sqrt{2}}{2} \mathrm{e}^{\frac{\pi}4} \end{gather*}

\begin{gather*} f'(x) = f(x) - \mathrm{e}^x \sin x \\ f''(x) = f'(x) - \mathrm{e}^x \sin x - \mathrm{e}^x \cos x \\ \textrm{At stationary point, } f'(x) = 0, x = \frac{\pi}{4} \\ f''(x) = -\sqrt{2}\mathrm{e}^{\frac{\pi}{4}} < 0 \\ \textrm{Hence } \left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \mathrm{e}^{\frac{\pi}4} \right) \textrm{ is a maximum point} \end{gather*}

\begin{align*} \int \mathrm{e}^{2x} \cos 2x \; \mathrm{d}x &= \frac{\mathrm{e}^{2x}}{2} \cos 2x - \int \frac{\mathrm{e}^{2x}}{2} (- 2 \sin 2x) \; \mathrm{d}x \\ &= \frac{\mathrm{e}^{2x}}{2} \cos 2x + \int \mathrm{e}^{2x} \sin 2x \; \mathrm{d}x \\ &= \frac{\mathrm{e}^{2x}}{2} \cos 2x + \frac{\mathrm{e}^{2x}}{2} \sin 2x - \int \frac{\mathrm{e}^{2x}}{2} (2 \cos 2x) \; \mathrm{d}x \\ &= \frac{\mathrm{e}^{2x}}{2} \cos 2x + \frac{\mathrm{e}^{2x}}{2} \sin 2x - \int \mathrm{e}^{2x} \cos 2x \; \mathrm{d}x \\ 2 \int \mathrm{e}^{2x} \cos 2x \; \mathrm{d}x &= \frac{\mathrm{e}^{2x}}{2} \cos 2x + \frac{\mathrm{e}^{2x}}{2} \sin 2x + C \\ \int \mathrm{e}^{2x} \cos 2x \; \mathrm{d}x &= \frac{\mathrm{e}^{2x}}{4} \cos 2x + \frac{\mathrm{e}^{2x}}{4} \sin 2x + C \\ &= \frac{1}{4} \mathrm{e}^{2x} (\sin 2x + \cos 2x) + c. \end{align*}

\begin{align*} & \textrm{Volume } \\ &= \pi \int_0^{\frac{\pi}{2}} \left( \mathrm{e}^x \cos x \right)^2 \mathrm{d}x \\ &= \pi \int_0^{\frac{\pi}{2}} \mathrm{e}^{2x} \cos^2 x \mathrm{d}x \\ &= \pi \int_0^{\frac{\pi}{2}} \mathrm{e}^{2x} \left( \frac{\cos 2x + 1}{2} \right) \mathrm{d}x \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \mathrm{e}^{2x} \cos 2x + \mathrm{e}^{2x} \mathrm{d}x \\ &= \frac{\pi}{2} \left[ \frac{1}{4}\mathrm{e}^{2x}(\sin 2x + \cos 2x) + \frac{\mathrm{e}^{2x}}{2} \right]_0^{\frac{\pi}{2}} \\ &= \frac{\pi}{8} \left( \mathrm{e}^{\pi}(\sin \pi + \cos \pi) + 2 \mathrm{e}^{\pi} - (\sin 0 + \cos 0 + 2) \right) \\ &= \frac{\pi}{8} \left( \mathrm{e}^{\pi} - 3 \right) \textrm{ units}^3 \end{align*}

\frac{\mathrm{d}N}{\mathrm{d}t} = kN

\begin{gather*} \int \frac{1}{N} \; \mathrm{d}N = \int k \; \mathrm{d}t \\ \ln |N| = kt + C \\ N = A\mathrm{e}^{kt} \\ \textrm{When } t=0, N=100, \\ A = 100 \\ N = 100 \mathrm{e}^{kt} \end{gather*}

\begin{gather*} \textrm{When } t=2, N=300, \\ 300 = 100 \mathrm{e}^{2k} \\ k = \frac{1}{2} \ln 3 \end{gather*}

\begin{gather*} \textrm{When } N=1000, \\ 1000 = 100 \mathrm{e}^{kt} \\ t = \frac{\ln 10}{k} \\ t = \frac{2 \ln 10}{\ln 3} \textrm{ min} \end{gather*}

\frac{\mathrm{d}N}{\mathrm{d}t} = kN - d

\begin{gather*} \int \frac{1}{kN-d} \; \mathrm{d}N = \int 1 \; \mathrm{d}t \\ \frac{1}{k} \ln |kN-d| = t + C \\ kN - d = A\mathrm{e}^{kt} \\ \textrm{When } t=0, N=100, \\ A = 100k - d \\ (100k-d)\mathrm{e}^{kt} = kN-d \\ \mathrm{e}^{kt} = \frac{kN-d}{100k-d} \\ \mathrm{e}^{kt} = \frac{N\ln 3-2d}{100\ln 3-2d} \\ t = \frac{2}{\ln 3} \left( \frac{N\ln 3-2d}{100\ln 3-2d} \right) \end{gather*}

\begin{gather*} (100k-d)\mathrm{e}^{kt} = kN-d \\ N = \frac{1}{k} \left( (100k-d)\mathrm{e}^{kt} + d \right) \\ N = \frac{2}{\ln 3} \left( (50\ln 3 - d)\mathrm{e}^{\frac{t \ln 3}{2}} + d \right) \end{gather*}

\begin{gather*} \textrm{For bacteria to decrease, } \\ 50 \ln 3 - d < 0 \\ d > 50 \ln 3 \end{gather*}

\begin{gather*} \textrm{When } N=0, \\ t = \frac{2}{\ln 3} \ln \left( \frac{0-2d}{100 \ln 3 - 2d} \right) \\ t = \frac{2}{\ln 3} \ln \left( \frac{-116}{100 \ln 3 - 116} \right) \\ t = 5.35 \textrm{ min} \end{gather*}

\begin{gather*} \textrm{At } P, \\ OP = \alpha (\beta - \sin \beta) \\ \textrm{By symmetry, } \\ SC = OP \\ \textrm{At } C, y=0 \\ \alpha(1- \cos \theta) = 0 \\ \cos \theta = 1 \\ \theta = 0 (\textrm{ at } O) \textrm{ or } \theta = 2 \pi (\textrm{ at } C) \\ OC = \alpha( 2\pi - \sin 2 \pi) \\ OC = 2\alpha \pi \end{gather*}

\begin{align*} PS &= OC - OP - SC \\ &= 2 \alpha \pi - 2\alpha(\beta - \sin \beta) \\ &= 2 \alpha (\pi - \beta - \sin \beta) \end{align*}

\frac{\mathrm{d}x}{\mathrm{d}\theta} = \alpha(1- \cos \theta)

\begin{align*} & \textrm{Area } \\ &= \int_{x_Q}^{x_R} y \; \mathrm{d}x \\ &= \int_\beta^{2\pi - \beta} \alpha(1-\cos \theta) \alpha(1-\cos \theta) \; \mathrm{d}\theta \\ &= \alpha^2 \int_\beta^{2\pi - \beta} 1 - 2 \cos \theta + \cos^2 \theta \; \mathrm{d} \theta \\ &= \alpha^2 \int_\beta^{2\pi - \beta} 1 - 2 \cos \theta + \frac{\cos 2 \theta + 1}{2} \; \mathrm{d} \theta \\ &= \alpha^2 \left[ \theta - 2 \sin \theta + \frac{\sin 2 \theta}{4} + \frac{\theta}{2} \right]_\beta^{2\pi - \beta} \\ &= \alpha^2 \left[ \frac{3\theta}{2} - 2 \sin \theta + \frac{\sin 2 \theta}{4} \right]_\beta^{2\pi - \beta} \\ &= \alpha^2 \left[ \left( \frac{6 \pi - 3\beta}{2} - 2 \sin (2\pi - \beta) + \frac{\sin (4\pi - 2 \beta)}{4} \right) - \left( \frac{3\beta}{2} - 2 \sin \beta + \frac{\sin 2 \beta}{4} \right) \right] \\ &= \alpha^2 \left( 3\pi - 3\beta + 2 \sin (\beta) - \frac{\sin 2 \beta}{4} + 2 \sin \beta - \frac{\sin 2 \beta}{4} \right) \\ &= \alpha^2 \left( 3\pi - 3\beta + 4\sin\beta - \frac{\sin 2 \beta}{2} \right) \\ &= \frac{1}{2} \alpha^2 \left( 6 \pi - 6 \beta + 8 \sin \beta - \sin 2 \beta \right) \end{align*}

\begin{gather*} \frac{1}{2} \alpha^2 \left( 6 \pi - 6 \beta + 8 \sin \beta - \sin 2 \beta \right) = 7.8159a^2 \\ 6 \pi - 6 \beta + 8 \sin \beta - \sin 2 \beta = 15.6318 \\ \textrm{Using GC, } \\ \beta = 1.9000 \\ \beta = 1.90 (3 \textrm{s.f.}) \end{gather*}

\begin{gather*} 2 \alpha (\pi - \beta - \sin \beta) = 50 \\ \alpha = 11.4265 \\ \end{gather*}

Greatest height occurs when

{\theta = \pi:}

\begin{align*} \textrm{Greatest height } &= \alpha (1 - \cos \pi) \\ &= 22.9 \textrm{ m} \end{align*}

Least height occurs when

{\theta = \beta:}

\begin{align*} \textrm{Least height } &= \alpha (1 - \cos \beta) \\ &= 15.1 \textrm{ m} \end{align*}