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HL Paper 2

A wind turbine is designed so that the rotation of the blades generates electricity. The turbine is built on horizontal ground and is made up of a vertical tower and three blades.

The point $A$ is on the base of the tower directly below point $B$ at the top of the tower. The height of the tower, $AB$ , is $90 m$ . The blades of the turbine are centred at $B$ and are each of length $40 m$ . This is shown in the following diagram.

The end of one of the blades of the turbine is represented by point $C$ on the diagram. Let $h$ be the height of $C$ above the ground, measured in metres, where $h$ varies as the blade rotates.

Find the

The blades of the turbine complete $12$ rotations per minute under normal conditions, moving at a constant rate.

The height, $h$ , of point $C$ can be modelled by the following function. Time, $t$ , is measured from the instant when the blade $[BC]$ first passes $[AB]$ and is measured in seconds.

$h (t) = 90 - 40 \cos (72 t °), t \geq 0$

maximum value of $h$ .

[1]

a.i.

minimum value of $h$ .

[1]

a.ii.

Find the time, in seconds, it takes for the blade $[BC]$ to make one complete rotation under these conditions.

[1]

b.i.

Calculate the angle, in degrees, that the blade $[BC]$ turns through in one second.

[2]

b.ii.

Write down the amplitude of the function.

[1]

c.i.

Find the period of the function.

[1]

c.ii.

Sketch the function $h (t)$ for $0 \leq t \leq 5$ , clearly labelling the coordinates of the maximum and minimum points.

[3]

Find the height of $C$ above the ground when $t = 2$ .

[2]

e.i.

Find the time, in seconds, that point $C$ is above a height of $100 m$ , during each complete rotation.

[3]

e.ii.

The wind speed increases and the blades rotate faster, but still at a constant rate.

Given that point $C$ is now higher than $110 m$ for $1$ second during each complete rotation, find the time for one complete rotation.

[5]

The cross-sectional view of a tunnel is shown on the axes below. The line $[AB]$ represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by $y = - 0.1 x^{3} + 0.8 x^{2}, 2 \leq x \leq 8$ , relative to an origin $O$ .

Point $A$ has coordinates $(2, 0)$ , point $B$ has coordinates $(2, 2.4)$ , and point $C$ has coordinates $(8, 0)$ .

Find the height of the tunnel when

Find $\frac{d y}{d x}$ .

[2]

a.i.

Hence find the maximum height of the tunnel.

[4]

a.ii.

$x = 4$ .

[2]

b.i.

$x = 6$ .

[1]

b.ii.

Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.

[3]

Write down the integral which can be used to find the cross-sectional area of the tunnel.

[2]

d.i.

Hence find the cross-sectional area of the tunnel.

[2]

d.ii.

A student investigating the relationship between chemical reactions and temperature finds the Arrhenius equation on the internet.

$k = A e^{- \frac{c}{T}}$

This equation links a variable $k$ with the temperature $T$ , where $A$ and $c$ are positive constants and $T > 0$ .

The Arrhenius equation predicts that the graph of $\ln k$ against $\frac{1}{T}$ is a straight line.

Write down

The following data are found for a particular reaction, where $T$ is measured in Kelvin and $k$ is measured in ${cm}^{3} {mol}^{- 1} s^{- 1}$ :

Find an estimate of

Show that $\frac{d k}{d T}$ is always positive.

[3]

Given that $\lim_{T \to \infty} k = A$ and $\lim_{T \to 0} k = 0$ , sketch the graph of $k$ against $T$ .

[3]

(i) the gradient of this line in terms of $c$ ;

(ii) the $y$ -intercept of this line in terms of $A$ .

[4]

Find the equation of the regression line for $\ln k$ on $\frac{1}{T}$ .

[2]

$c$ .

It is not required to state units for this value.

[1]

e.i.

$A$ .

It is not required to state units for this value.

[2]

e.ii.

Consider the expression $f\left( x \right) = {\text{tan}}\left( {x + \frac{\pi }{4}} \right){\text{cot}}\left( {\frac{\pi }{4} - x} \right)$ .

The expression $f\left( x \right)$ can be written as $g\left( t \right)$ where $t = {\text{tan}}\,x$ .

Let $\alpha$ , β be the roots of $g\left( t \right) = k$ , where 0 < $k$ < 1.

Sketch the graph of $y = f\left( x \right)$ for $- \frac{{5\pi }}{8} \leqslant x \leqslant \frac{\pi }{8}$ .

[2]

a.i.

With reference to your graph, explain why $f$ is a function on the given domain.

[1]

a.ii.

Explain why $f$ has no inverse on the given domain.

[1]

a.iii.

Explain why $f$ is not a function for $- \frac{{3\pi }}{4} \leqslant x \leqslant \frac{\pi }{4}$ .

[1]

a.iv.

Show that $g\left( t \right) = {\left( {\frac{{1 + t}}{{1 - t}}} \right)^2}$ .

[3]

Sketch the graph of $y = g\left( t \right)$ for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.

[3]

Find $\alpha$ and β in terms of $k$ .

[5]

d.i.

Show that $\alpha$ + β < −2.

[2]

d.ii.

Charlotte decides to model the shape of a cupcake to calculate its volume.

From rotating a photograph of her cupcake she estimates that its cross-section passes through the points $(0, 3.5), (4, 6), (6.5, 4), (7, 3)$ and $(7.5, 0)$ , where all units are in centimetres. The cross-section is symmetrical in the $x$ -axis, as shown below:

She models the section from $(0, 3.5)$ to $(4, 6)$ as a straight line.

Charlotte models the section of the cupcake that passes through the points $(4, 6), (6.5, 4), (7, 3)$ and $(7.5, 0)$ with a quadratic curve.

Charlotte thinks that a quadratic with a maximum point at $(4, 6)$ and that passes through the point $(7.5, 0)$ would be a better fit.

Believing this to be a better model for her cupcake, Charlotte finds the volume of revolution about the $x$ -axis to estimate the volume of the cupcake.

Find the equation of the line passing through these two points.

[2]

Find the equation of the least squares regression quadratic curve for these four points.

[2]

b.i.

By considering the gradient of this curve when $x = 4$ , explain why it may not be a good model.

[1]

b.ii.

Find the equation of the new model.

[4]

Write down an expression for her estimate of the volume as a sum of two integrals.

[4]

d.i.

Find the value of Charlotte’s estimate.

[1]

d.ii.

An environmental scientist is asked by a river authority to model the effect of a leak from a power plant on the mercury levels in a local river. The variable $x$ measures the concentration of mercury in micrograms per litre.

The situation is modelled using the second order differential equation

$\frac{d^{2} x}{d t^{2}} + 3 \frac{d x}{d t} + 2 x = 0$

where $t \geq 0$ is the time measured in days since the leak started. It is known that when $t = 0, x = 0$ and $\frac{d x}{d t} = 1$ .

If the mercury levels are greater than $0.1$ micrograms per litre, fishing in the river is considered unsafe and is stopped.

The river authority decides to stop people from fishing in the river for $10 %$ longer than the time found from the model.

Show that the system of coupled first order equations:

$\frac{d x}{d t} = y$

$\frac{d y}{d t} = - 2 x - 3 y$

can be written as the given second order differential equation.

[2]

Find the eigenvalues of the system of coupled first order equations given in part (a).

[3]

Hence find the exact solution of the second order differential equation.

[5]

Sketch the graph of $x$ against $t$ , labelling the maximum point of the graph with its coordinates.

[2]

Use the model to calculate the total amount of time when fishing should be stopped.

[3]

Write down one reason, with reference to the context, to support this decision.

[1]

The voltage $v$ in a circuit is given by the equation

$v\left( t \right) = 3\,{\text{sin}}\left( {100\pi t} \right)$ , $t \geqslant 0$ where $t$ is measured in seconds.

The current $i$ in this circuit is given by the equation

$i\left( t \right) = 2\,{\text{sin}}\left( {100\pi \left( {t + 0.003} \right)} \right)$ .

The power $p$ in this circuit is given by $p\left( t \right) = v\left( t \right) \times i\left( t \right)$ .

The average power ${p_{av}}$ in this circuit from $t = 0$ to $t = T$ is given by the equation

${p_{av}}\left( T \right) = \frac{1}{T}\int_0^T {p\left( t \right){\text{d}}t}$ , where $T > 0$ .

Write down the maximum and minimum value of $v$ .

[2]

Write down two transformations that will transform the graph of $y = v\left( t \right)$ onto the graph of $y = i\left( t \right)$ .

[2]

Sketch the graph of $y = p\left( t \right)$ for 0 ≤ $t$ ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.

[3]

Find the total time in the interval 0 ≤ $t$ ≤ 0.02 for which $p\left( t \right)$ ≥ 3.

[3]

Find ${p_{av}}$ (0.007).

[2]

With reference to your graph of $y = p\left( t \right)$ explain why ${p_{av}}\left( T \right)$ > 0 for all $T$ > 0.

[2]

Given that $p\left( t \right)$ can be written as $p\left( t \right) = a\,{\text{sin}}\left( {b\left( {t - c} \right)} \right) + d$ where $a$ , $b$ , $c$ , $d$ > 0, use your graph to find the values of $a$ , $b$ , $c$ and $d$ .

[6]

Consider $f(x) = - 1 + \ln \left( {\sqrt {{x^2} - 1} } \right)$

The function $f$ is defined by $f(x) = - 1 + \ln \left( {\sqrt {{x^2} - 1} } \right),{\text{ }}x \in D$

The function $g$ is defined by $g(x) = - 1 + \ln \left( {\sqrt {{x^2} - 1} } \right),{\text{ }}x \in \left] {1,{\text{ }}\infty } \right[$ .

Find the largest possible domain $D$ for $f$ to be a function.

[2]

Sketch the graph of $y = f(x)$ showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.

[3]

Explain why $f$ is an even function.

[1]

Explain why the inverse function ${f^{ - 1}}$ does not exist.

[1]

Find the inverse function ${g^{ - 1}}$ and state its domain.

[4]

Find $g'(x)$ .

[3]

Hence, show that there are no solutions to $g'(x) = 0$ ;

[2]

g.i.

Hence, show that there are no solutions to $({g^{ - 1}})'(x) = 0$ .

[2]

g.ii.

At an archery tournament, a particular competition sees a ball launched into the air while an archer attempts to hit it with an arrow.

The path of the ball is modelled by the equation

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 5 \\ 0 \end{matrix}) + t (\begin{matrix} u_{x} \\ u_{y} - 5 t \end{matrix})$

where $x$ is the horizontal displacement from the archer and $y$ is the vertical displacement from the ground, both measured in metres, and $t$ is the time, in seconds, since the ball was launched.

$u_{x}$ is the horizontal component of the initial velocity
$u_{y}$ is the vertical component of the initial velocity.

In this question both the ball and the arrow are modelled as single points. The ball is launched with an initial velocity such that $u_{x} = 8$ and $u_{y} = 10$ .

An archer releases an arrow from the point $(0, 2)$ . The arrow is modelled as travelling in a straight line, in the same plane as the ball, with speed $60 {m s}^{- 1}$ and an angle of elevation of $10 °$ .

Find the initial speed of the ball.

[2]

a.i.

Find the angle of elevation of the ball as it is launched.

[2]

a.ii.

Find the maximum height reached by the ball.

[3]

Assuming that the ground is horizontal and the ball is not hit by the arrow, find the $x$ coordinate of the point where the ball lands.

[3]

For the path of the ball, find an expression for $y$ in terms of $x$ .

[3]

Determine the two positions where the path of the arrow intersects the path of the ball.

[4]

Determine the time when the arrow should be released to hit the ball before the ball reaches its maximum height.

[4]

Sketch the graphs $y = {\text{si}}{{\text{n}}^3}\,x + {\text{ln}}\,x$ and $y = 1 + {\text{cos}}\,x$ on the following axes for 0 < $x$ ≤ 9.

[2]

Hence solve ${\text{si}}{{\text{n}}^3}\,x + {\text{ln}}\,x - {\text{cos}}\,x - 1 < 0$ in the range 0 < $x$ ≤ 9.

[4]

Consider the function $f(x) = \frac{{\sqrt x }}{{\sin x}},{\text{ }}0 < x < \pi$ .

Consider the region bounded by the curve $y = f(x)$ , the $x$ -axis and the lines $x = \frac{\pi }{6},{\text{ }}x = \frac{\pi }{3}$ .

Show that the $x$ -coordinate of the minimum point on the curve $y = f(x)$ satisfies the equation $\tan x = 2x$ .

[5]

a.i.

Determine the values of $x$ for which $f(x)$ is a decreasing function.

[2]

a.ii.

Sketch the graph of $y = f(x)$ showing clearly the minimum point and any asymptotic behaviour.

[3]

Find the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y = - x$ .

[4]

This region is now rotated through $2\pi$ radians about the $x$ -axis. Find the volume of revolution.

[3]

Beth goes for a run. She uses a fitness app to record her distance, $s$ km, and time, $t$ minutes. A graph of her distance against time is shown.

Beth runs at a constant speed of 2.3 ms^–1 for the first 8 minutes.

Between 8 and 20 minutes, her distance can be modeled by a cubic function, $s = a{t^3} + b{t^2} + ct + d$ . She reads the following data from her app.

Hence find

Calculate her distance after 8 minutes. Give your answer in km, correct to 3 decimal places.

[2]

Find the value of $a$ , $b$ , $c$ and $d$ .

[5]

the distance she runs in 20 minutes.

[2]

c.i.

her maximum speed, in ms^–1.

[4]

c.ii.

Consider the function $f$ defined by $f(x) = 3x\arccos (x)$ where $- 1 \leqslant x \leqslant 1$ .

Sketch the graph of $f$ indicating clearly any intercepts with the axes and the coordinates of any local maximum or minimum points.

[3]

State the range of $f$ .

[2]

Solve the inequality $\left| {3x\arccos (x)} \right| > 1$ .

[4]

Jorge is carefully observing the rise in sales of a new app he has created.

The number of sales in the first four months is shown in the table below.

Jorge believes that the increase is exponential and proposes to model the number of sales $N$ in month $t$ with the equation

$N = A e^{r t}, A, r \in ℝ$

Jorge plans to adapt Euler’s method to find an approximate value for $r$ .

With a step length of one month the solution to the differential equation can be approximated using Euler’s method where

$N (n + 1) \approx N (n) + 1 \times N' (n), n \in ℕ$

Jorge decides to take the mean of these values as the approximation of $r$ for his model. He also decides the graph of the model should pass through the point $(2, 52)$ .

The sum of the square residuals for these points for the least squares regression model is approximately $6.555$ .

Show that Jorge’s model satisfies the differential equation

$\frac{d N}{d t} = r N$

[2]

Show that $r \approx \frac{N (n + 1) - N (n)}{N (n)}$

[3]

Hence find three approximations for the value of $r$ .

[3]

Find the equation for Jorge’s model.

[3]

Find the sum of the square residuals for Jorge’s model using the values $t = 1, 2, 3, 4$ .

[2]

Comment how well Jorge’s model fits the data.

[1]

f.i.

Give two possible sources of error in the construction of his model.

[2]

f.ii.

Consider the curve $y = \sqrt{x}$ .

The shape of a piece of metal can be modelled by the region bounded by the functions $f$ , $g$ , the $x$ -axis and the line segment $[AB]$ , as shown in the following diagram. The units on the $x$ and $y$ axes are measured in metres.

The piecewise function $f$ is defined by

$f (x) = {\begin{matrix} \sqrt{x} & 0 \leq x \leq 0.16 \\ 1.25 x + 0.2 & 0.16 < x \leq 0.5 \end{matrix}$

The graph of $g$ is obtained from the graph of $f$ by:

a stretch scale factor of $\frac{1}{2}$ in the $x$ direction,
followed by a stretch scale factor $\frac{1}{2}$ in the $y$ direction,
followed by a translation of $0.2$ units to the right.

Point $A$ lies on the graph of $f$ and has coordinates $(0.5, 0.825)$ . Point $B$ is the image of $A$ under the given transformations and has coordinates $(p, q)$ .

The piecewise function $g$ is given by

$g (x) = {\begin{matrix} h (x) & 0.2 \leq x \leq a \\ 1.25 x + b & a < x \leq p \end{matrix}$

The area enclosed by $y = g (x)$ , the $x$ -axis and the line $x = p$ is $0.0627292 m^{2}$ correct to six significant figures.

Find $\frac{d y}{d x}$ .

[2]

a.i.

Hence show that the equation of the tangent to the curve at the point $(0.16, 0.4)$ is $y = 1.25 x + 0.2$ .

[2]

a.ii.

Find the value of $p$ and the value of $q$ .

[2]

Find an expression for $h (x)$ .

[2]

c.i.

Find the value of $a$ .

[1]

c.ii.

Find the value of $b$ .

[2]

c.iii.

Find the area enclosed by $y = f (x)$ , the $x$ -axis and the line $x = 0.5$ .

[3]

d.i.

Find the area of the shaded region on the diagram.

[4]

d.ii.