Q1 [AJC MYE 18 Q10 (modified)]
By using the substitution x = 2 \sin \theta, find \displaystyle \int \frac{2x+x^2}{\sqrt{4-x^2}} {\; \mathrm{d}x}.
Q2 [AJC Prelims 18 Q4]
Find \displaystyle\int \frac{x}{\sqrt{3+2x-x^2}} {\; \mathrm{d}x}.
Find \displaystyle\int \frac{x^2}{3+2x-x^2} {\; \mathrm{d}x} by partial fractions.
Q3 [CJC Prelims 18 Q4]
Find \displaystyle\int \frac{2-x}{4+x^2} {\; \mathrm{d}x}.
Use the substitution x = \tan \theta to find \displaystyle\int \frac{1}{\sqrt{1+x^2}} {\; \mathrm{d}x}.
Write down \displaystyle\int x^2 {\mathrm{e}}^{x^3} {\; \mathrm{d}x}. Hence find \displaystyle\int x^5 {\mathrm{e}}^{x^3} {\; \mathrm{d}x}.
Q4 [IJC Prelims 18 Q2]
Find \displaystyle\int x^2 \sin (nx) {\; \mathrm{d}x}, where n is a constant.
Q5 [JJC Prelims 17 Q2]
Find \displaystyle\int \cos (3 \theta) \sin (2 \theta) {\; \mathrm{d}x}.
Q6 [IJC Prelims 17 Q2]
Find \displaystyle\int \cos^{-1} (nx) {\; \mathrm{d}x}, where n is a constant.
Q7 [AJC MYE 18 Q9]
The plane \Pi contains the origin O and is parallel to the vectors -{\mathbf{i}}+ {\mathbf{k}} and {\mathbf{i}}+2{\mathbf{j}}.
Find an equation of the plane \Pi in scalar product form.
By finding the foot of perpendicular, N, of point P to the plane \Pi, show that the position vector of the mirror image of P in \Pi is -{\mathbf{i}}+3{\mathbf{j}}-2{\mathbf{k}}.
Find the exact length of projection of PQ on to the plane \Pi.
Hence, or otherwise, find the exact area of the triangle PNQ.
Q8 [DHS Prelims 18 Q9 (modified)]
The line l_1 passes through the point A with the position vector \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} and is parallel to \begin{pmatrix} t \\ t^2+1 \\ 3 \end{pmatrix}, while the cartesian equation of the plane p is given by tx-2y+z=-3, where t is a real constant. It is known that l_1 and p have no point in common.
Show that t=-1.
Find the distance between l_1 and p.
Find the vector equation of the line of reflection of l_1 in p.
The line l_2 has cartesian equation \displaystyle x+2=\frac{y-1}{-2}=\frac{z}{2}. Find the angle between l_2 and p.
Q9 [JJC Prelims 18 Q10]
Building contractors are constructing a rock climbing wall at the corner wall of a gymnasium. Points (x, y, z) are defined relative to a ground anchor point at (0,0,0), where units are metres.
Support beams are laid in straight lines and the thickness of the support beams and rock climbing wall can be neglected.
The three support beams of the rock climbing wall, S_1, S_2 and S_3 start at the ground anchor point and go in the direction \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} respectively. The support beams S_1 and S_2 are on the ground level.
The vertices A , B and C of the rock climbing wall lie on the support beams S_1, S_2 and S_3 respectively. The rock climbing wall lies on the plane \pi with vector equation {\mathbf{r}} = { \begin{pmatrix} 1 \\ 5 \\ -1 \end{pmatrix} } + \lambda { \begin{pmatrix} 2 \\ 3 \\ -12 \end{pmatrix} } + \mu { \begin{pmatrix} 1 \\ 2 \\ -7 \end{pmatrix} }, where \lambda, \mu \in \mathbb{R}.
Find the cartesian equation of the plane \pi and hence show that the coordinates of A are (4,0,0).
Determine if this building safety standard is met.
Find the coordinates of N and the exact length of this support beam.
\displaystyle -4\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2} - \frac{x\sqrt{4-x^2}}{2}.
\displaystyle\sin^{-1} \frac{x-1}{2} - \sqrt{3+2x-x^2} + c.
-x-\frac{9}{4}\ln(3-x)+\frac{1}{4}\ln(1+x)+c.
\displaystyle\tan^{-1}\frac{x}{2} - \frac{1}{2}\ln (4+x^2) + c.
\displaystyle\ln \left| \sqrt{1+x^2} + x \right|.
\displaystyle\frac{x^2 \cos (nx)}{n} + \frac{2x \sin (nx) - 2 \sin (nx)}{n^2} + c.
\frac{1}{2} \cos \theta - \frac{1}{10} \cos(5\theta) + c.
\displaystyle\frac{1}{n} \left( \cos^{-1}(nx) - \sqrt{1-n^2x^2} \right) + c.
{\mathbf{r}} \cdot { \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} } = 0.
{\overrightarrow{ON}} = { \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} }, {\overrightarrow{OP'}} = { \begin{pmatrix} -1 \\ 3 \\ -2 \end{pmatrix} }.
\displaystyle\frac{\sqrt{26}}{3}.
\displaystyle\frac{\sqrt{26}}{2}.
\displaystyle\frac{2}{\sqrt{6}}.
\displaystyle{\mathbf{r}} = \frac{1}{3} { \begin{pmatrix} 1 \\ 5 \\ 8 \end{pmatrix} } + \lambda { \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} }, \lambda \in \mathbb{R}.
5.1^\circ.
Yes (angle of 74.5^\circ).
\displaystyle N \left( \frac{18}{7},
\frac{12}{7}, \frac{6}{7} \right).
\displaystyle\frac{6\sqrt{14}}{7}.