Questions C4 (1162 questions)

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OCR MEI C4 Q1
1 Using partial fractions, find \(\int \frac { x } { ( x + 1 ) ( 2 x + 1 ) } \mathrm { d } x\).
  1. Express \(\cos \theta + \sqrt { 3 } \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(\alpha\) is acute, expressing \(\alpha\) in terms of \(\pi\).
  2. Write down the derivative of \(\tan \theta\). Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { ( \cos \theta + \sqrt { 3 } \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { \sqrt { 3 } } { 4 }\).
OCR MEI C4 Q3
3 In a chemical process, the mass \(\boldsymbol { M }\) grams of a chemical at time \(\boldsymbol { t }\) minutes is modelled by the differential equation $$\underset { d t } { d \underline { \underline { M } } } - \underset { \mathrm { z } \left( \left. \right| ^ { \prime } + \mathrm { z } ^ { 2 } \right) ^ { \prime } } { M _ { - } }$$
  1. Find \({ } _ { \mathrm { f } } \overline { 1 } ; \overline { 12 } d t\)
    (li) Find constants \(\boldsymbol { A } , \boldsymbol { B }\) and \(\boldsymbol { C }\) such lhat $$\begin{array} { c c } 1 & B t + C
    \hdashline t \left( I + t ^ { 2 } \right) & I \end{array} . + \begin{array} { c c } I & I + 1 ^ { 2 } . \end{array}$$
  2. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { J \sqrt { + , 2 } } ,$$ where \(\boldsymbol { K }\) is a constant.
  3. When \(\boldsymbol { t } = \mathrm { I } , \boldsymbol { M } = 25\). Calculate \(\boldsymbol { K }\) What is the mass of the chemical in the long term?
OCR MEI C4 Q4
4 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 \(t = 0\).
  1. 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.
  2. 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 \(t = 0\).
  3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } }$$
  4. What does this solution indicate about the long-term height of the tree?
  5. After a year, the tree has grown to a height of 2 m . Which model fits this information better?
OCR MEI C4 Q5
4 marks
5
  1. Find the first three non-zero terms of the binomial expansion of \(\frac { 1 } { \sqrt { 4 } x ^ { 2 } }\) for \(| x | < 2\). [4]
  2. Use this result to find an approximation for \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to
    4 significant figures.
  3. Given that \(\int \frac { 1 } { \sqrt { 4 x ^ { 2 } } } \mathrm {~d} x = \arcsin \left( \frac { 1 } { 2 } x \right) + c\), evaluate \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to 4 significant figures.
OCR MEI C4 Q1
1 Fig. 3 shows the curve \(y = x ^ { 3 } + \sqrt { ( \sin x ) }\) for \(0 \leqslant x \leqslant \frac { \pi } { 4 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-1_587_540_393_768} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), giving your answer to 3 decimal places.
  2. Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say.
OCR MEI C4 Q2
2 Fig. 2 shows the curve \(y = \overline { 1 + x ^ { 2 } }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-2_577_941_549_636} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. The following table gives some values of \(x\) and \(y\).
    \(x\)00.250.50.751
    \(y\)11.03081.251.4142
    Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
  2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
  3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.
OCR MEI C4 Q3
3 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]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-3_656_736_482_665} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. 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.
      \(x\)00.511.52
      \(y\)1.92832.89644.5919
    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 Q4
4
  1. Complete the table of values for the curve \(y = \sqrt { \cos x }\).
    \(X\)0\(\frac { \pi } { 8 }\)\(\frac { \pi } { 4 }\)\(\frac { 3 \pi } { 8 }\)\(\frac { \pi } { 2 }\)
    \(y\)0.96120.8409
    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]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-4_459_751_799_638} \captionsetup{labelformat=empty} \caption{Fig. 4}
    \end{figure}
  2. 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 Q5
2 marks
5
  1. Use the trapezium rule with four strips to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\), showing your working. Fig. 1 shows a sketch of \(y = \sqrt { 1 + \mathrm { e } ^ { x } }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-5_533_1074_441_565} \captionsetup{labelformat=empty} \caption{Fig. 1}
    \end{figure}
  2. Suppose that the trapezium rule is used with more strips than in part (i) to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\). State, with a reason but no further calculation, whether this would give a larger or smaller estimate.
    [0pt] [2]
OCR MEI C4 Q6
6 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.
\(x\)11.52
\(\sqrt { 1 + \mathrm { e } ^ { - x } }\)1.16961.10601.0655
  1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
  2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
  3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.
OCR MEI C4 Q1
1 Fig. 6 shows the region enclosed by the curve \(y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }\) and the line \(y = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-1_426_664_277_714} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\).
OCR MEI C4 Q2
2 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]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_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]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_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.
OCR MEI C4 Q3
3 Fig. 6 shows the region enclosed by part of the curve \(y = 2 x ^ { 2 }\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at \(\mathrm { P } ( 1,2 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-3_640_923_399_613} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed.
[0pt] [You may use the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.]
OCR MEI C4 Q4
4 The part of the curve \(y = 4 - x ^ { 2 }\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-3_534_595_1831_785} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
OCR MEI C4 Q5
5 Fig. 2 shows the curve \(y = \sqrt { } 1 + \mathrm { e } ^ { 2 x }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-4_434_873_306_665} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} The region bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Show that the volume of the solid of revolution produced is \(\frac { 1 } { 2 } \pi \left( 1 + \mathrm { e } ^ { 2 } \right)\).
OCR MEI C4 Q6
6 Fig. 3 shows the curve \(y = \ln x\) and part of the line \(y = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-4_244_548_1346_768} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis.
  1. Show that the volume of the solid of revolution formed is given by \(\int _ { 0 } ^ { 2 } \pi \mathrm { e } ^ { 2 y } \mathrm {~d} y\).
  2. Evaluate this, leaving your answer in an exact form.
OCR MEI C4 Q7
7
  1. Show that \(\int x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = - \frac { 1 } { 4 } \mathrm { e } ^ { - 2 x } ( 1 + 2 x ) + c\). A vase is made in the shape of the volume of revolution of the curve \(y = x ^ { 1 / 2 } \mathrm { e } ^ { - x }\) about the \(x\)-axis between \(x = 0\) and \(x = 2\) (see Fig. 5). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-5_718_751_638_654} \captionsetup{labelformat=empty} \caption{Fig. 5}
    \end{figure}
  2. Show that this volume of revolution is \(\frac { 1 } { 4 } \pi \left( 1 \frac { 5 } { \mathrm { e } ^ { 4 } } \right)\).
OCR MEI C4 Q8
8 Fig. 4 shows a sketch of the region enclosed by the curve . \(\mathrm { J } 1 + \mathrm { e } - 2 \mathrm { x }\), the x -axis, the y -axis and the line \(\boldsymbol { x } = 1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-6_442_628_314_745} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} Find the volume of the solid generated when this region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { x }\)-axis. Give your answer in an exact form.
OCR MEI C4 Q1
1 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]{252453c9-9afa-435c-b64b-5ea37ec69eed-1_962_1248_673_420} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac { 1 } { t }\).
  2. 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.
  3. 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 Q2
2 Fig. 3 shows part of the curve \(y = 1 + x ^ { 2 }\), together with the line \(y = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-2_558_716_302_687} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} The region enclosed by the curve, the \(y\)-axis and the line \(y = 2\) is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find the volume of the solid generated, giving your answer in terms of \(\pi\).
OCR MEI C4 Q3
3 Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u , y = u + \frac { 1 } { u } , \quad 1 \leqslant u \leqslant 10$$ The point A lies on the \(x\)-axis and AC is parallel to the \(y\)-axis. The tangent to the curve at C makes an angle \(\theta\) with AC, as shown. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-3_505_585_598_766} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Find the lengths \(\mathrm { OA } , \mathrm { OB }\) and AC .
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\). Hence find the angle \(\theta\).
  3. Show that the cartesian equation of the curve is \(y = \mathrm { e } ^ { \frac { 1 } { 5 } x } + \mathrm { e } ^ { - \frac { 1 } { 5 } x }\). An object is formed by rotating the region OACB through \(360 ^ { \circ }\) about \(\mathrm { O } x\).
  4. Find the volume of the object.
OCR MEI C4 Q4
4 Fig. 2 shows the curve \(y = \sqrt { 1 + x ^ { 2 } }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-4_572_939_551_638} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. The following table gives some values of \(x\) and \(y\).
    \(x\)00.250.50.751
    \(y\)11.03081.251.4142
    Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
  2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
  3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.
OCR MEI C4 Q5
5 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]{252453c9-9afa-435c-b64b-5ea37ec69eed-5_657_732_375_667} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. 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.
      \(x\)00.511.52
      \(y\)1.92832.89644.5919
    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 Q6
6 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-6_812_801_517_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
  2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
  3. (A) Show that \(y = x \cos \theta\).
    (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
    (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
  4. Find the volume of the balloon.
OCR MEI C4 Q1
1 You are given that \(\mathrm { f } ( x ) = \cos x + \lambda \sin x\) where \(\lambda\) is a positive constant.
  1. Express \(\mathrm { f } ( x )\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving \(R\) and \(\alpha\) in terms of \(\lambda\).
  2. Given that the maximum value (as \(x\) varies) of \(\mathrm { f } ( x )\) is 2 , find \(R , \lambda\) and \(\alpha\), giving your answers in exact form.