Mathematics Extension 1

Particle $A$ is projected from the origin with initial speed $v\ \mathrm{m\,s^{-1}}$ at an angle $\theta$ with the horizontal plane. At the same time, particle $B$ is projected horizontally with initial speed $u\ \mathrm{m\,s^{-1}}$ from a point that is $H$ metres above the origin, as shown in the diagram.

The position vector of particle $A$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The position vector of particle $B$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The angle $\theta$ is chosen so that $\tan\theta=2$.

The two particles collide.

(i) By first showing that $\cos\theta=\frac1{\sqrt5}$, verify that .

Particle $A$ is projected from the origin with initial speed $v\ \mathrm{m\,s^{-1}}$ at an angle $\theta$ with the horizontal plane. At the same time, particle $B$ is projected horizontally with initial speed $u\ \mathrm{m\,s^{-1}}$ from a point that is $H$ metres above the origin, as shown in the diagram.

The position vector of particle $A$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The position vector of particle $B$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The angle $\theta$ is chosen so that $\tan\theta=2$.

The two particles collide.

(ii) Show that the particles collide at time

When the particles collide, their velocity vectors are perpendicular.

Particle $A$ is projected from the origin with initial speed $v\ \mathrm{m\,s^{-1}}$ at an angle $\theta$ with the horizontal plane. At the same time, particle $B$ is projected horizontally with initial speed $u\ \mathrm{m\,s^{-1}}$ from a point that is $H$ metres above the origin, as shown in the diagram.

The position vector of particle $A$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The position vector of particle $B$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The angle $\theta$ is chosen so that $\tan\theta=2$.

The two particles collide.

(iii) Show that

Particle $A$ is projected from the origin with initial speed $v\ \mathrm{m\,s^{-1}}$ at an angle $\theta$ with the horizontal plane. At the same time, particle $B$ is projected horizontally with initial speed $u\ \mathrm{m\,s^{-1}}$ from a point that is $H$ metres above the origin, as shown in the diagram.

The position vector of particle $A$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The position vector of particle $B$, $t$ seconds after it is projected, is given by

(Do NOT prove this.)

The angle $\theta$ is chosen so that $\tan\theta=2$.

The two particles collide.

Prior to the collision, the trajectory of particle $A$ was a parabola. (Do NOT prove this.)

(iv) Find the height of the vertex of that parabola above the horizontal plane. Give your answer in terms of $H$.

Let $f(x)=2x+\ln x$, for $x>0$.

(i) Explain why the inverse of $f(x)$ is a function.

Let $f(x)=2x+\ln x$, for $x>0$.

(ii) Let $g(x)=f^{-1}(x)$. By considering the value of $f(1)$, or otherwise, evaluate $g'(2)$.

Consider the hyperbola $y=\dfrac{1}{x}$ and the circle $(x-c)^2+y^2=c^2$, where $c$ is a constant.

(i) Show that the $x$-coordinates of any points of intersection of the hyperbola and circle are zeros of the polynomial $P(x)=x^4-2cx^3+1$.

Given a non-zero vector $\begin{pmatrix}p\\q\end{pmatrix}$, it is known that the vector $\begin{pmatrix}q\\-p\end{pmatrix}$ is perpendicular to $\begin{pmatrix}p\\q\end{pmatrix}$ and has the same magnitude. (Do NOT prove this.)

Points $A$ and $B$ have position vectors $\overrightarrow{OA}=\begin{pmatrix}a_1\\a_2\end{pmatrix}$ and , respectively.

Using the given information, or otherwise, show that the area of triangle $OAB$ is $\frac{1}{2}|a_1b_2-a_2b_1|$.

Consider the hyperbola $y=\dfrac{1}{x}$ and the circle $(x-c)^2+y^2=c^2$, where $c$ is a constant.

(ii) The graphs of $y=x^4-2cx^3+1$ for $c=0.8$ and are shown.

By considering the given graphs, or otherwise, find the exact value of
$c>0$ such that the hyperbola $y=\frac{1}{x}$ and the circle
$(x-c{)}^{}$ intersect at only one point.

The point $P$ lies on the circle centred at $I(r,0)$ with radius $r>0$, such that $\overrightarrow{IP}$ makes an angle of $t$ to the horizontal.

The point $Q$ lies on the circle centred at $J(-R,0)$ with radius $R>0$, such that $\overrightarrow{JQ}$ makes an angle of to the horizontal.

Note that $\overrightarrow{OP}=\overrightarrow{OI}+\overrightarrow{IP}$ and .

Using part (i), or otherwise, find the values of $t$, where $-\pi\leq t\leq\pi$, that maximise the area of triangle $OPQ$.