Questions C4 - A-Level Maths

Show that AB is perpendicular to BC .
Find the area of triangle ABC .

OCR MEI C4 2011 January Q5

5 Show that $\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta$.

OCR MEI C4 2011 January Q6

Find the point of intersection of the line $\mathbf { r } = \left( \begin{array} { r } - 8
- 2
6 \end{array} \right) + \lambda \left( \begin{array} { r } - 3
0
1 \end{array} \right)$ and the plane $2 x - 3 y + z = 11$.
Find the acute angle between the line and the normal to the plane. Section B (36 marks)

OCR MEI C4 2011 January Q7

7 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after $t$ seconds it is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Its terminal (long-term) velocity is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A model of the particle's motion is proposed. In this model, $v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)$.

Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
Verify that $v$ satisfies the differential equation $\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v$. In a second model, $v$ satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when $t = 0 , v = 0$.
Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give $v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }$. [You are not required to show this result.]
Verify that this model also gives a terminal velocity of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Which of the two models fits the data better?

OCR MEI C4 2011 January Q8

8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at $\alpha$ to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all $\beta$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{f657e167-e6f8-4df2-901b-067c32835877-04_684_872_461_278} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure} In the following, all lengths are in metres.

Find AC in terms of $\alpha$, and hence show that $\mathrm { GF } = 10 \sec \alpha \tan \beta$.
Show that $\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )$. $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that $\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }$. [You are not required to derive this result.]
You are now given that $\alpha = 45 ^ { \circ }$ and that $\tan \beta = t$.
Find CE and CD in terms of $t$. Hence show that $\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }$.
Show that $\mathrm { GF } = 10 \sqrt { 2 } t$. For a certain value of $\beta , \mathrm { DE } = 2 \mathrm { GF }$.
Show that $t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }$. Hence find this value of $\beta$.

OCR MEI C4 2012 January Q1

1 Express $\frac { x + 1 } { x ^ { 2 } ( 2 x - 1 ) }$ in partial fractions.

OCR MEI C4 2012 January Q3

3 Express $3 \sin x + 2 \cos x$ in the form $R \sin ( x + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.
Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve $y = \mathrm { f } ( x )$, where $$f ( x ) = 3 \sin x + 2 \cos x , 0 \leqslant x \leqslant \pi .$$

OCR MEI C4 2012 January Q4

Complete the table of values for the curve $y = \sqrt { \cos x }$.
$x$ 0 $\frac { \pi } { 8 }$ $\frac { \pi } { 4 }$ $\frac { 3 \pi } { 8 }$ $\frac { \pi } { 2 }$
$y$ 0.9612 0.8409
Hence use the trapezium rule with strip width $h = \frac { \pi } { 8 }$ to estimate the value of the integral $\int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \cos x } \mathrm {~d} x$, giving your answer to 3 decimal places. Fig. 4 shows the curve $y = \sqrt { \cos x }$ for $0 \leqslant x \leqslant \frac { \pi } { 2 }$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-02_457_750_1446_653} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
State, with a reason, whether the trapezium rule with a strip width of $\frac { \pi } { 16 }$ would give a larger or smaller estimate of the integral.

OCR MEI C4 2012 January Q5

5 Verify that the vector $2 \mathbf { i } - \mathbf { j } + 4 \mathbf { k }$ is perpendicular to the plane through the points $\mathrm { A } ( 2,0,1 ) , \mathrm { B } ( 1,2,2 )$ and $\mathrm { C } ( 0 , - 4,1 )$. Hence find the cartesian equation of the plane.

OCR MEI C4 2012 January Q6

6 Given the binomial expansion $( 1 + q x ) ^ { p } = 1 - x + 2 x ^ { 2 } + \ldots$, find the values of $p$ and $q$. Hence state the set of values of $x$ for which the expansion is valid.

OCR MEI C4 2012 January Q7

7 Show that the straight lines with equations $\mathbf { r } = \left( \begin{array} { l } 4
2
4 \end{array} \right) + \lambda \left( \begin{array} { l } 3
0
1 \end{array} \right)$ and $\mathbf { r } = \left( \begin{array} { r } - 1
4
9 \end{array} \right) + \mu \left( \begin{array} { r } - 1
1
3 \end{array} \right)$ meet.
Find their point of intersection.

OCR MEI C4 2012 January Q8

8 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$ $\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)$ is a point on the curve with parameter $t$. TS is the tangent to the curve at P , and PR is the line through P parallel to the $x$-axis. Q is the point $( 2,0 )$. The angles that PS and QP make with the positive $x$-direction are $\theta$ and $\phi$ respectively. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-03_969_1262_733_388} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

By considering the gradient of the tangent TS, show that $\tan \theta = \frac { 1 } { t }$.
Find the gradient of the line QP in terms of $t$. Hence show that $\phi = 2 \theta$, and that angle TPQ is equal to $\theta$.
[0pt] [The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the $x$-axis.
Show that the curve has cartesian equation $y ^ { 2 } = 8 x$. Hence find the volume of revolution of the curve, giving your answer as a multiple of $\pi$.

OCR MEI C4 2012 January Q9

1 marks

9 \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{81433914-a56f-4765-af34-990a0127f98b-04_269_453_255_806} \captionsetup{labelformat=empty} \caption{Fig. 9}

\end{figure} Fig. 9 shows a hemispherical bowl, of radius 10 cm , filled with water to a depth of $x \mathrm {~cm}$. It can be shown that the volume of water, $V \mathrm {~cm} ^ { 3 }$, is given by $$V = \pi \left( 10 x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 } \right) .$$ Water is poured into a leaking hemispherical bowl of radius 10 cm . Initially, the bowl is empty. After $t$ seconds, the volume of water is changing at a rate, in $\mathrm { cm } ^ { 3 } \mathrm {~s} ^ { - 1 }$, given by the equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = k ( 20 - x ) ,$$ where $k$ is a constant.

Find $\frac { \mathrm { d } V } { \mathrm {~d} x }$, and hence show that $\pi x \frac { \mathrm {~d} x } { \mathrm {~d} t } = k$.
Solve this differential equation, and hence show that the bowl fills completely after $T$ seconds, where $$T = \frac { 50 \pi } { k } .$$ Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of $k x \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
Show that, $t$ seconds later, $\pi ( 20 - x ) \frac { \mathrm { d } x } { \mathrm {~d} t } = - k$.
Solve this differential equation. Hence show that the bowl empties in $3 T$ seconds.

OCR MEI C4 2013 January Q2

2 Find the first four terms of the binomial expansion of $\sqrt [ 3 ] { 1 - 2 x }$. State the set of values of $x$ for which the expansion is valid.

OCR MEI C4 2013 January Q3

3 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$

Find the exact value of the gradient of the curve at the point where $\theta = \frac { 1 } { 6 } \pi$.
Show that the cartesian equation of the curve is $y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }$.

OCR MEI C4 2013 January Q4

4 Fig. 4 shows the curve $y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }$, and the region between the curve, the $x$-axis, the $y$-axis and the line $x = 2$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-02_650_727_1176_653} \captionsetup{labelformat=empty} \caption{Fig. 4}

\end{figure}

Find the exact volume of revolution when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.
1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.
  $x$ 0 0.5 1 1.5 2
  $y$ 1.9283 2.8964 4.5919
2. The trapezium rule for $\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x$ with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

OCR MEI C4 2013 January Q5

5 Solve the equation $2 \sec ^ { 2 } \theta = 5 \tan \theta$, for $0 \leqslant \theta \leqslant \pi$.

OCR MEI C4 2013 January Q6

6 In Fig. 6, $\mathrm { ABC } , \mathrm { ACD }$ and AED are right-angled triangles and $\mathrm { BC } = 1$ unit. Angles CAB and CAD are $\theta$ and $\phi$ respectively. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-03_440_524_504_753} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure}

Find AC and AD in terms of $\theta$ and $\phi$.
Hence show that $\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }$. Section B (36 marks)

OCR MEI C4 2013 January Q7

7 A tent has vertices ABCDEF with coordinates as shown in Fig. 7. Lengths are in metres. The $\mathrm { O } x y$ plane is horizontal. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9bceee25-35bd-448b-a4a2-1a5667be5f11-03_547_987_1580_539} \captionsetup{labelformat=empty} \caption{Fig. 7}

\end{figure}

Find the length of the ridge of the tent DE , and the angle this makes with the horizontal.
Show that the vector $\mathbf { i } - 4 \mathbf { j } + 5 \mathbf { k }$ is normal to the plane through $\mathrm { A } , \mathrm { D }$ and E . Hence find the equation of this plane. Given that B lies in this plane, find $a$.
Verify that the equation of the plane BCD is $x + z = 8$. Hence find the acute angle between the planes ABDE and BCD .

OCR MEI C4 2013 January Q8

1 marks

8 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h ,$$ where $h$ is its height in metres and the time $t$ is in years. It is assumed that the tree is grown from seed, so that $h = 0$ when $t = 0$.

Write down the value of $h$ for which $\frac { \mathrm { d } h } { \mathrm {~d} t } = 0$, and interpret this in terms of the growth of the tree.
Verify that $h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)$ satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, $h = 0$ when $t = 0$.
Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } } .$$
What does this solution indicate about the long-term height of the tree?
After a year, the tree has grown to a height of 2 m . Which model fits this information better?