2020 H2 Math Paper 1 Full Solutions

{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ - 5 \\ - 2 \end{pmatrix}} \allowbreak {= \begin{pmatrix} - 2 \\ 2 \\ - 6 \end{pmatrix}}\allowbreak {= 2 \begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix}}

Vector normal to

{\pi_1 = \begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix}.}

Let

{\theta}

be the acute angle between

{\pi_1}

and

{\pi_2}

.

\begin{aligned} & \left| \mathbf{n_1} \cdot \mathbf{n_2} \right| = | \mathbf{n_1} | | \mathbf{n_2} | \cos \theta \\ & \left| \begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right| = \left| \begin{pmatrix} - 1 \\ 1 \\ - 3 \end{pmatrix} \right| \left| \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right| \cos \theta \\ & \left| 19 \right| = \sqrt{11} \sqrt{77} \cos \theta \\ & \theta = 49.2^{\circ} \end{aligned}

Differentiating w.r.t.

{x,}

\frac{2 x \cdot x^2 - 2 x (1 + x^2)}{(1 + x^2)^2} + \frac{2 y \frac{\mathrm{d}y}{\mathrm{d}x} \cdot y^2 - 2 y \frac{\mathrm{d}y}{\mathrm{d}x} (1 + y^2)}{(1 + y^2)^2} = 3 x^2y^5 + x^3 \cdot 5 y^4 \frac{\mathrm{d}y}{\mathrm{d}x}

Substituting

{x=1, y=1,}

\begin{aligned} & \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{5}{9} \\ & y - 1 = - \frac{5}{9} ( x - 1 ) \end{aligned}

Equation of tangent:

{5 x + 9 y = 14.}

\begin{aligned} f'(x) &= \frac{3 \cos ( 3 x )}{1 + \sin ( 3 x )} \\ f''(x) &= \frac{ - 9 \sin ( 3 x ) \Big (1+\sin ( 3 x ) \Big) - \Big(3 \cos ( 3 x ) \Big)^2 }{\Big(1+\sin ( 3 x )\Big)^2} \\ &= \frac{ - 9 \sin ( 3 x ) - 9 \sin^2 (3 x) - 9 \cos^2 (3 x) }{\Big(1+\sin ( 3 x )\Big)^2} \\ &= \frac{ - 9 \sin ( 3 x ) - 9 }{\Big(1+\sin ( 3 x )\Big)^2} \\ &= \frac{ - 9 \Big(1+\sin ( 3 x )\Big) }{\Big(1+\sin ( 3 x )\Big)^2} \\ &= \frac{ - 9 }{1+\sin ( 3 x )} \\ \end{aligned}

{k = -9.}

\displaystyle f'''(x) = \frac{- 27 \cos ( 3 x )}{\Big(1+\sin ( 3 x )\Big)^2}

When

{x=0, f(0) = 0, f'(0) = 3,}

{f''(0) = -9, f'''(0) = -27.}

\begin{aligned} f(x) &= 0 + 3 x + \frac{-9}{2!} x^2 + \frac{-27}{3!} x^3 + \ldots \\ &= 3 x - \frac{9}{2} x^2 + \frac{9}{2} x^3 + \ldots \end{aligned}

{z_1 = 2 \mathrm{e}^{ \frac{1}{3} \pi \mathrm{i} }}

,

{z_2 = \sqrt{2} \mathrm{e}^{ - \frac{1}{4} \pi \mathrm{i} }}

,

{z_3 = 2 \mathrm{e}^{ \frac{1}{6} \pi \mathrm{i} }.}

\begin{aligned} \frac{z_1}{z_2 z_3} &= \frac{2 \mathrm{e}^{ \frac{1}{3} \pi \mathrm{i} }}{\sqrt{2} \mathrm{e}^{ - \frac{1}{4} \pi \mathrm{i} } \cdot 2 \mathrm{e}^{ \frac{1}{6} \pi \mathrm{i} }} \\ &= \frac{2 \mathrm{e}^{ \frac{1}{3} \pi \mathrm{i} }}{2 \sqrt{2} \mathrm{e}^{ - \frac{1}{12} \pi \mathrm{i} }} \\ &= {\textstyle \frac{1}{2} \sqrt{2} \mathrm{e}^{ \frac{5}{12} \pi \mathrm{i} }} \\ &= {\textstyle \frac{1}{2} \sqrt{2} \left( \cos \frac{5}{12} \pi + \mathrm{i} \sin \frac{5}{12} \pi \right)}. \end{aligned}

\begin{aligned} & \left| \frac{z_1 z_4}{z_2 z_3} \right| = 1 \\ & \left| \frac{z_1}{z_2 z_3} \right| \left| z_4 \right| = 1 \\ & \frac{1}{2} \sqrt{2} \left| z_4 \right| = 1 \\ & \left| z_4 \right| = \sqrt{2} \end{aligned}

Since

\frac{z_1z_4}{z_2z_3}

is purely imaginary,

\arg \left( \frac{z_1z_4}{z_2z_3} \right) = \frac{\pi}{2} + k\pi

where

{k \in \mathbb{Z}.}

\begin{aligned} & \arg \left( \frac{z_1 z_4}{z_2 z_3} \right) = \frac{\pi}{2} + k\pi \\ & \arg \left( \frac{z_1 }{z_2 z_3} \right) + \arg z_4 = \frac{\pi}{2} + k\pi \\ & \frac{5}{12} \pi + \arg z_4 = \frac{\pi}{2} + k\pi \\ & \arg z_4 = \frac{1}{12} \pi + k\pi \\ \end{aligned}

Since

{-\pi < \theta \leq \pi,}

{k=0}

or

{k=-1.}

z_4 = \sqrt{2} \left( \cos \frac{1}{12} \pi + \mathrm{i} \sin \frac{1}{12} \pi \right) \allowbreak \textrm{ or } {z_4 = \sqrt{2} \left( \cos - \frac{11}{12} \pi + \mathrm{i} \sin - \frac{11}{12} \pi \right).}

\begin{aligned} & \mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{a} \\ & \mathbf{a} \times \mathbf{b} - \mathbf{b} \times \mathbf{a} = \mathbf{0} \\ & \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{b} = \mathbf{0} \\ & 2 \mathbf{a} \times \mathbf{b} = \mathbf{0} \\ & \mathbf{a} \times \mathbf{b} = \mathbf{0} \end{aligned}

Since

{\mathbf{a}}

and

{\mathbf{b}}

are non-zero,

{\mathbf{a}}

is parallel to

{\mathbf{b}.}

{\mathbf{a} = k \mathbf{b}, k \in \mathbb{R}, k \neq 0.}

\begin{aligned} & ( \mathbf{r} - \mathbf{p} ) \times \mathbf{q} = \mathbf{0} \\ & ( \mathbf{r} - \mathbf{p} ) = k \mathbf{q} \\ & \mathbf{r} = \mathbf{p} + k \mathbf{q} \end{aligned}

The set of all possible positions of the point

{R}

forms a line passing through the point

{P}

and parallel to the direction vector

{\mathbf{q}.}

\begin{aligned} & ( \mathbf{r} - \mathbf{p} ) \cdot \mathbf{q} = \mathbf{0} \\ & \mathbf{r} \cdot \mathbf{q} - \mathbf{p} \cdot \mathbf{q} = 0 \\ & \mathbf{r} \cdot \mathbf{q} = \mathbf{p} \cdot \mathbf{q} \\ & \mathbf{r} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} = \begin{pmatrix} - 1 \\ 2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} \\ & \mathbf{r} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} = - 5 \\ & \begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 3 \\ - 5 \\ 2 \end{pmatrix} = - 5 \end{aligned}

{3 x - 5 y + 2 z = - 5.}

The set of all possible positions of the point

{R}

forms a plane passing through the point

{P}

and perpendicular to the normal vector

{\mathbf{q}.}

Substituting

{z=k+k\mathrm{i}}

into the equation,

{(k+k\mathrm{i})^2 (2 + \mathrm{i}) - 8 \mathrm{i} (k+k\mathrm{i}) + t = 0}

{k^2(1+\mathrm{i})^2 (2 + \mathrm{i}) - 8 \mathrm{i} k (1+\mathrm{i}) + t = 0}

{k^2(- 2 + 4 \mathrm{i}) + k (8 - 8 \mathrm{i}) + t = 0}

{(- 2 k^2 + 8 k+t) + (4 k^2 - 8 k)\mathrm{i} = 0}

Comparing imaginary parts,

{4 k^2 - 8 k=0}

Since

{k}

is non-zero,

{k = 2.}

Comparing real parts,

{- 2 k^2 + 8 k+t=0}

{t = - 8.}

Let

{\alpha}

be the other root of the equation.

{z^2 (2 + \mathrm{i}) - 8 \mathrm{i} z - 8}

{= (2 + \mathrm{i}) (z-2-2\mathrm{i})(z-\alpha)}

{= (2 + \mathrm{i}) \big(z^2 + (2+2\mathrm{i})z - \alpha z + \alpha (2+2\mathrm{i}) \big)}

Comparing constants,

{\alpha (2 + \mathrm{i})(2+2\mathrm{i}) = -8}

Other root

\alpha = \frac{-8}{(2 + \mathrm{i})(2+2\mathrm{i})} = - \frac{2}{5} + \frac{6}{5} \mathrm{i}.

\int 2 - \sin ( 4 x ) \, \mathrm{d}x = 2 x + \frac{1}{4} \cos ( 4 x ) + C.

\begin{aligned} \int x \sin 4x \, \mathrm{d}x &= x \frac{-\cos 4x}{4} + \int \frac{\cos 4x}{4} \, \mathrm{d} x \\ &= - \frac{x \cos 4x}{4} + \frac{\sin 4x}{16} + C. \end{aligned}

\begin{aligned} \int_0^{ \frac{1}{2} \pi } x f(x) \, \mathrm{d}x &= \int_0^{ \frac{1}{2} \pi } 2 x - x \sin 4x \, \mathrm{d}x \\ &= \left[ x^2 + \frac{x \cos 4x}{4} - \frac{\sin 4x}{16} \right]_0^{\frac{1}{2} \pi} \\ &= \frac{1}{4} \pi^2 + \frac{\pi \cos 2 \pi}{8} - \frac{\sin 2 \pi}{16} - \left( 0 + 0 - \sin 0 \right) \\ &= \frac{1}{4} \pi^2 + \frac{1}{8} \pi. \end{aligned}

\begin{aligned} \int \sin^2 4x \, \mathrm{d}x &= \int \frac{1 - \cos 8x }{2} \, \mathrm{d}x \\ &= \frac{x}{2} - \frac{\sin 8x}{16} + C. \end{aligned}

\begin{aligned} \int_0^{ \frac{1}{2} \pi } \Big( f(x) \Big)^2 \, \mathrm{d}x &= \int_0^{ \frac{1}{2} \pi } 4 - 4 \sin 4x + \sin^2 4x \, \mathrm{d}x \\ &= \left[ 4x + \cos 4x + \frac{x}{2} - \frac{\sin 8x}{16} \right]_0^{\frac{1}{2}\pi} \\ &= \frac{9}{4} \pi + \cos 2 \pi - \frac{\sin 4 \pi}{16} - \left( 0 + \cos 0 - \frac{\sin 0}{16} \right) \\ &= \frac{9}{4} \pi. \end{aligned}

{a = 4}

{a + 4d = 10}

{d = \frac{3}{2}}

{u_{30} = a + 29d = \frac{95}{2}.}

\begin{aligned} S_{50} - S_{20} &= \frac{50}{2} \left( 4 + 49 \cdot \frac{3}{2} \right) - \frac{20}{2} \left( 4 + 19 \cdot \frac{3}{2} \right) \\ &= \frac{3345}{2}. \end{aligned}

{a = 4}

{a r^4 = 1.6384}

{r = \frac{4}{5}}

S_\infty = \frac{4}{1-\frac{4}{5}} = 20.

\begin{aligned} S_n &> 19.6 \\ \frac{4\left( 1- \frac{4}{5}^n \right)}{1 - \frac{4}{5}} &> 19.6 \\ 1 - \frac{4}{5}^n &> 0.98 \\ 0.8^n &< 0.02 \end{aligned}

n \ln 0.8 < \ln 0.02

Since

{\ln 0.8 < 0,}

\begin{aligned} n &> \frac{\ln 0.02}{\ln 0.8} \\ n &> 17.531 \end{aligned}

{\textrm{Smallest } n = 18.}

{y=2x}

and

{y=-\frac{1}{2}x}

are perpendicular since their gradients are

{2}

and

{-\frac{1}{2}}

respectively and

{m_1m_2 = 2(-\frac{1}{2}) = -1.}

{y=2x}

makes an angle

{\tan^{-1}(2)}

clockwise to the positive

{x\textrm{-axis}}

and

{y=-\frac{1}{2}x}

makes an angle

{-\tan^{-1}(-\frac{1}{2})}

anticlockwise to the positive

{x\textrm{-axis}.}

Hence

\tan^{-1}(2)-\tan^{-1}(-\frac{1}{2}) = \frac{1}{2} \pi.

{\frac{1}{x^2+1} = \frac{k}{3x+4}}

{kx^2 - 3x + k - 4 = 0}

For

{C_1}

and

{C_2}

to intersect,
Discriminant

{=b^2-4ac \geq 0}

{9 - 4(k)(k-4) \geq 0}

{4k^2 - 16k - 9 \leq 0}

{-\frac{1}{2} \leq k \leq 4.5}

Since

{k>0,}

{0 < k \leq 4.5.}

Area required

{= \displaystyle \int_{\textstyle -\frac{1}{2}}^{2} \frac{1}{x^2+1} - \frac{2}{3x+4} \, \mathrm{d}x}

{= \Big[ \tan^{-1} x - \frac{2}{3} \ln | 3x + 4 | \Big]_{-\frac{1}{2}}^2}

= \tan^{-1} (2) - \tan^{-1} (-{\textstyle \frac{1}{2}}) \allowbreak {- \frac{2}{3} \ln 10 + \frac{2}{3} \ln \frac{5}{2}}

{= \frac{\pi}{2} - \frac{2}{3} \ln 4}

{= \left( \frac{\pi}{2} - \frac{4}{3} \ln 2 \right) \textrm{ units}^2.}

\frac{\mathrm{d}P}{\mathrm{d}t} = - \frac{3}{100} P.

\int \frac{1}{P} \, \mathrm{d} P = \int - \frac{3}{100} \, \mathrm{d}t

{\ln | P | = - \frac{3}{100} t + C}

{P = A \mathrm{e}^{- \frac{3}{100} t}.}

{\textrm{As } t \to \infty, P \to 0.}

Hence over many years the number of sheep will approach 0 and there will be no more sheep.

\frac{\mathrm{d}P}{\mathrm{d}t} = - \frac{3}{100} P + n

\frac{\mathrm{d}P}{\mathrm{d}t} = - \frac{3}{100} \left( P - - \frac{100}{3} n \right)

\int \frac{1}{P-- \frac{100}{3} n} \, \mathrm{d} P = \int - \frac{3}{100} \, \mathrm{d}t

{\ln | P - - \frac{100}{3} n | = - \frac{3}{100} t + C}

{P = A \mathrm{e}^{- \frac{3}{100} t} + \frac{100}{3} n.}

As

{n \to \infty, P \to \frac{100}{3} n.}

{\frac{100}{3} n = 500}

{n = 15.}

\begin{aligned} \tan \theta &= \tan( \angle AKD - \angle AKC ) \\ &= \frac{\tan \angle AKD - \tan \angle AKC}{1 + \tan \angle AKD \cdot \tan AKC} \\ &= \frac{\frac{a+4}{x} - \frac{a}{x} }{ 1 + \frac{a+4}{x} \cdot \frac{a}{x} } \\ &= \frac{ \frac{4}{x} }{ 1 + \frac{a^2 + 4a }{x^2} } \\ &= \frac{x}{x^2 + 4a + a^2} \end{aligned}

Differentiating w.r.t.

{x,}

{\frac{\mathrm{d}}{\mathrm{d}x} \left( \tan \theta \right) = \frac{4(x^2+4a+a^2) - (4x)(2x)}{(x^2+4a+a^2)}}

At maximum

{\tan \theta, \frac{\mathrm{d}}{\mathrm{d}x} \left( \tan \theta \right) = 0}

{4x^2 + 16 a + 4a^2 - 8x^2 = 0}

{x^2 = 4 a + a^2}

{x = \sqrt{4 a + a^2}}

since

{x > 0}

\tan \theta = \frac{4 \sqrt{4 a + a^2} }{4a + a^2 + 4a + a^2}

= \frac{2}{\sqrt{4a + a^2}}.

The optimal point and optimal angle may not correspond to the highest chance of scoring.

\tan KDA = \frac{x}{a+4}

= \frac{\sqrt{4a+a^2}}{a+4}

= \sqrt{\frac{a}{a+4}}.

As

{a \gg 4, \frac{a}{4+a} \approx 1}

{\tan \angle KDA \approx 1}

{\angle KDA \approx \frac{\pi}{2} \textrm{ rad}.}

Since

{XY = 50, CD = 4, XC = DY,}

{0 < a \leq 23}

{0 < 4a \leq 92}

and

{0 < a^2 \leq 529}

{0 < 4a + a^2 \leq 621}

{0 < \sqrt{4a + a^2} \leq \sqrt{621}}

{\frac{1}{\sqrt{4a + a^2}} \geq \frac{1}{\sqrt{621}} }

{\frac{2}{\sqrt{4a + a^2}} \geq \frac{2}{\sqrt{621}} }

{\tan \theta \geq \frac{2}{\sqrt{621}} }

{\tan^{-1} \frac{2}{\sqrt{621}} \leq \theta < \frac{\pi}{2}}

{0.0801 \leq \theta < \frac{\pi}{2}}