Questions — OCR MEI C4 (332 questions)

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OCR MEI C4 Q3
3 Express \(\sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Hence solve the equation \(\sin \theta - 3 \cos \theta = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{54a69773-651f-4e2f-9a3c-06ea7c07098b-4_606_624_236_754} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where \(x\) and \(y\) are in metres.
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }\). Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 3 } \pi\). Hence find the exact coordinates of the highest point A on the path of C .
  2. Express \(x ^ { 2 } + y ^ { 2 }\) in terms of \(\theta\). Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
  3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
  4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures.
OCR MEI C4 Q5
5 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$
OCR MEI C4 Q1
1 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).
  1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
  2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588} \captionsetup{labelformat=empty} \caption{Fig. 5}
    \end{figure}
  3. Find the volume of revolution produced, giving your answer in exact form.
OCR MEI C4 Q2
2 Show that the equation \(\operatorname { cosec } x + 5 \cot x = 3 \sin x\) may be rearranged as $$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$ Hence solve the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\), giving your answers to 1 decimal place.
OCR MEI C4 Q3
3 Using appropriate right-angled triangles, show that \(\tan 45 ^ { \circ } = 1\) and \(\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }\).
Hence show that \(\tan 75 ^ { \circ } = 2 + \sqrt { 3 }\).
OCR MEI C4 Q4
4 Prove that \(\sec ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta \operatorname { cosec } ^ { 2 } \theta\).
OCR MEI C4 Q5
5 Solve the equation \(\operatorname { cosec } ^ { 2 } \theta = 1 + 2 \cot \theta\), for \(- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 Q6
6 Given that \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\), show that \(\cot ^ { 2 } \theta - \cot \theta - 2 = 0\).
Hence solve the equation \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 Q7
7 Given that \(x = 2 \sec \theta\) and \(y = 3 \tan \theta\), show that \(\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1\).
OCR MEI C4 Q8
8 Solve the equation $$\sec ^ { 2 } \theta = 4 , \quad 0 \leqslant \theta \leqslant \pi ,$$ giving your answers in terms of \(\pi\).
OCR MEI C4 Q1
1 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).
OCR MEI C4 Q2
2 Solve, correct to 2 decimal places, the equation \(\cot 2 \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 Q3
3 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]{f5ab2a9d-3366-4b9b-86b0-85d5d923953c-2_695_877_503_242} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f5ab2a9d-3366-4b9b-86b0-85d5d923953c-2_333_799_505_1048} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure} In the following, all lengths are in metres.
  1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
  2. 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 \(\tan \beta = t\).
  3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
  4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
  5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).
OCR MEI C4 Q4
4 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$
OCR MEI C4 Q5
5 Prove that \(\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }\).
OCR MEI C4 Q2
2 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]{c8ea5913-c527-40e7-bfcc-c1c2df544e04-2_452_535_437_781} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find AC and AD in terms of \(\theta\) and \(\phi\).
  2. Hence show that \(\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }\).
OCR MEI C4 Q3
5 marks
3 Verify that the vector \(\mathbf { d } - \mathbf { j } + 4 \mathbf { k }\) 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. [5]
OCR MEI C4 Q4
4 Show that the straight lines with equations \(\mathbf { r } = \begin{array} { r r r } 2 & + \lambda & 0
4 & & 1 \end{array}\) and \(\mathbf { r } = \quad + \mu \quad\) meet.
Find their point of intersection.
OCR MEI C4 Q5
5 The points A , B and C have coordinates \(( 2,0 , - 1 ) , ( 4,3 , - 6 )\) and \(( 9,3 , - 4 )\) respectively.
  1. Show that AB is perpendicular to BC .
  2. Find the area of triangle ABC .
OCR MEI C4 Q6
6
  1. Verify that the lines \(\left. \mathbf { r } = \begin{array} { r } - 5
    3
    4 \end{array} \right) + \lambda \left( \begin{array} { r } 3
    0
    - 1 \end{array} \right)\) and \(\left. \left. \mathbf { r } = \begin{array} { r } - 1
    4
    2 \end{array} \right) + \mu - \begin{array} { r } 2
    - 1
    0 \end{array} \right)\) meet at the point ( \(1,3,2\) ).
  2. Find the acute angle between the lines.
OCR MEI C4 Q1
1 A glass ornament OABCDEFG is a truncated pyramid on a rectangular base (see Fig. 7). All dimensions are in centimetres. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-1_625_1109_416_522} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Write down the vectors \(\overrightarrow { \mathrm { CD } }\) and \(\overrightarrow { \mathrm { CB } }\).
  2. Find the length of the edge CD.
  3. Show that the vector \(4 \mathbf { i } + \mathbf { k }\) is perpendicular to the vectors \(\overrightarrow { \mathrm { CD } }\) and \(\overrightarrow { \mathrm { CB } }\). Hence find the cartesian equation of the plane BCDE.
  4. Write down vector equations for the lines OG and AF . Show that they meet at the point P with coordinates (5, 10, 40). You may assume that the lines CD and BE also meet at the point P .
    The volume of a pyramid is \(\frac { 1 } { 3 } \times\) area of base × height.
  5. Find the volumes of the pyramids POABC and PDEFG . Hence find the volume of the ornament.
OCR MEI C4 Q2
2 In a game of rugby, a kick is to be taken from a point P (see Fig. 7). P is a perpendicular distance \(y\) metres from the line TOA. Other distances and angles are as shown. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-2_511_630_449_750} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Show that \(\theta = \beta - \alpha\), and hence that \(\tan \theta = \frac { 6 y } { 160 + y ^ { 2 } }\). Calculate the angle \(\theta\) when \(y = 6\).
  2. By differentiating implicitly, show that \(\frac { \mathrm { d } \theta } { \mathrm { d } y } = \frac { 6 \left( 160 - y ^ { 2 } \right) } { \left( 160 + y ^ { 2 } \right) ^ { 2 } } \cos ^ { 2 } \theta\).
  3. Use this result to find the value of \(y\) that maximises the angle \(\theta\). Calculate this maximum value of \(\theta\). [You need not verify that this value is indeed a maximum.]
OCR MEI C4 Q3
3 Fig. 6 shows a lean-to greenhouse ABCDHEFG . With respect to coordinate axes Oxyz , the coordinates of the vertices are as shown. All distances are in metres. Ground level is the plane \(z = 0\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-3_796_1296_354_418} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Verify that the equation of the plane through \(\mathrm { A } , \mathrm { B }\) and E is \(x + 6 y + 12 = 0\). Hence, given that F lies in this plane, show that \(a = - 2 \frac { 1 } { 3 }\).
  2. (A) Show that the vector \(\left( \begin{array} { r } 1
    - 6
    0 \end{array} \right)\) is normal to the plane DHC.
    (B) Hence find the cartesian equation of this plane.
    (C) Given that G lies in the plane DHC , find \(b\) and the length FG .
  3. Find the angle EFB . A straight wire joins point H to a point P which is half way between E and \(\mathrm { F } . \mathrm { Q }\) is a point two-thirds of the way along this wire, so that \(\mathrm { HQ } = 2 \mathrm { QP }\).
  4. Find the height of Q above the ground.
OCR MEI C4 Q4
4 A computer-controlled machine can be programmed to make ats by entering the equation of the plane of the cut, and to drill holes by entering the equation of the line of the hole. A \(20 \mathrm {~cm} \times 30 \mathrm {~cm} \times 30 \mathrm {~cm}\) cuboid is to be at and drilled. The cuboid is positioned relative to \(x\)-, \(y ^ { 2 }\) and z-axes as shown in Fig. 8.1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-4_416_702_463_322} \captionsetup{labelformat=empty} \caption{Fig.8.1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-4_420_683_459_1044} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
\end{figure} Fkst, a plane out is made to remove the comer at \(E\). The cut goes through the points \(P . Q\) and \(R\), which are the midpoints of the sides \(\mathrm { ED } , \mathrm { EA }\) and EF respectively.
  1. Write down the coordinates of \(\mathbf { P } , \mathbf { Q }\) and \(\mathbf { R }\). $$\text { Henceshowlhat } \mathbb { N } ^ { 1 } \left\{ \begin{array} { l } 1
    0
    0 \end{array} \right\} \text { and } \left\{ \begin{array} { l } 1
    0 \end{array} \right\}$$ (;) Show that the veeto, \(\binom { \text { I } } { \text { ) } }\) is pc,pcndicula, to the plone through \(\mathrm { P } , \mathrm { Q }\), nd R
    Hence find the Cartesian equation of this plane. A hole is then drilled perpendicular to lriangle \(P Q R\), as shown in Fig. 82. The hole passes through the triangle at the point \(T\) which divides the line \(P S\) in the ratio 2 : \(I\), where \(S\) is the midpoint of \(Q R\).
  2. Write down the coordinates of S , and show that the point T has coordinates \(( - 5,16,25 )\).
  3. Write down a vector equation of the line of the drill hole. Hence determine whether or not this line passes through C .
OCR MEI C4 Q1
1 Fig. 6 shows a lean-to greenhouse ABCDHEFG . With respect to coordinate axes \(\mathrm { O } x y z\), the coordinates of the vertices are as shown. All distances are in metres. Ground level is the plane \(z = 0\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{27c27c79-9aea-45a4-a000-41aac70ff866-1_798_1296_354_418} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Verify that the equation of the plane through \(\mathrm { A } , \mathrm { B }\) and E is \(x + 6 y + 12 = 0\). Hence, given that F lies in this plane, show that \(a = - 2 \frac { 1 } { 3 }\).
  2. (A) Show that the vector \(\left( \begin{array} { r } 1
    - 6
    0 \end{array} \right)\) is normal to the plane DHC.
    (B) Hence find the cartesian equation of this plane.
    (C) Given that G lies in the plane DHC , find \(b\) and the length FG .
  3. Find the angle EFB . A straight wire joins point H to a point P which is half way between E and \(\mathrm { F } . \mathrm { Q }\) is a point two-thirds of the way along this wire, so that \(\mathrm { HQ } = 2 \mathrm { QP }\).
  4. Find the height of Q above the ground.