Questions — OCR MEI C4 (354 questions)

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OCR MEI C4 Q3
5 marks Moderate -0.8
3 A curve is defined parametrically by the equations $$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$ Find the gradient of the curve at the point where \(t = 2\).
OCR MEI C4 Q4
19 marks Standard +0.3
4 Fig. 7a shows the curve with the parametric equations $$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve meets the \(x\)-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-4_509_660_571_714} \captionsetup{labelformat=empty} \caption{Fig. 7a}
\end{figure}
  1. State, with their coordinates, the points on the curve for which \(\theta = - \frac { \pi } { 2 } , \theta = 0\) and \(\theta = \frac { \pi } { 2 }\).
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac { \pi } { 2 }\), and verify that the two tangents to the curve at the origin meet at right angles.
  3. Find the exact coordinates of the turning point Q . When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-4_324_389_1692_857} \captionsetup{labelformat=empty} \caption{Fig. 7b}
    \end{figure}
  4. Express \(\sin ^ { 2 } \theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)\).
  5. Find the volume of the paperweight shape.
  6. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
  7. Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.
OCR MEI C4 Q1
7 marks Moderate -0.3
1 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
OCR MEI C4 Q2
8 marks Standard +0.3
2
  1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
  2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
  3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 Q3
6 marks Standard +0.3
3 Given the equation \(\sin \left( + 45 ^ { \circ } \right) = 2 \cos [\), show that \(\sin + \cos = 22 \cos\). Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$
OCR MEI C4 Q4
7 marks Standard +0.3
4 Solve the equation \(\tan \left( \theta + 45 ^ { \circ } \right) = 1 - 2 \tan \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\).
OCR MEI C4 Q5
7 marks Standard +0.3
5 Given that \(\sin ( \theta + \alpha ) = 2 \sin \theta\), show that \(\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }\). Hence solve the equation \(\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
OCR MEI C4 Q6
7 marks Moderate -0.3
6 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
OCR MEI C4 Q1
16 marks Standard +0.3
1 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-1_427_898_464_683} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
  2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
  3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
  4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).
OCR MEI C4 Q2
7 marks Moderate -0.3
2 Express \(3 \cos \theta + 4 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { - \pi } { 2 }\).
Hence solve the equation \(3 \cos \theta + 4 \sin \theta = 2\) for \(\quad - \pi \leqslant \theta \leqslant \pi\).
OCR MEI C4 Q3
7 marks Standard +0.3
3 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 Q4
18 marks Standard +0.3
4 Part of the track of a roller-coaster is modelled by a curve with the parametric equations $$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$ This is shown in Fig. 8. B is a minimum point, and BC is vertical. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-3_591_1437_433_391} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
  3. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2$$
  4. Express \(\cos \theta - 4 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\). Hence solve the equation \(\cos \theta - 4 \sin \theta = 2\) for \(0 \leqslant \theta \leqslant 2 \pi\).
OCR MEI C4 Q5
7 marks Standard +0.3
5 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).
OCR MEI C4 Q1
8 marks Standard +0.8
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
7 marks Standard +0.3
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
7 marks Moderate -0.8
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 marks Moderate -0.3
4 Prove that \(\sec ^ { 2 } \theta + \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta \operatorname { cosec } ^ { 2 } \theta\).
OCR MEI C4 Q5
6 marks Standard +0.3
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 marks Standard +0.3
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
3 marks Easy -1.2
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
4 marks Moderate -0.8
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
6 marks Standard +0.3
1 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).
OCR MEI C4 Q2
4 marks Moderate -0.3
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
18 marks Challenging +1.2
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
7 marks Standard +0.3
4 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$