Questions — OCR MEI C2 (454 questions)

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OCR MEI C2 Q2
2 At a place where a river is 7.5 m wide, its depth is measured every 1.5 m across the river. The table shows the results.
Distance across river \(( \mathrm { m } )\)01.534.567.5
Depth of river \(( \mathrm { m } )\)0.62.33.12.81.80.7
Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.
OCR MEI C2 Q3
3 Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-2_579_1385_1035_424} \captionsetup{labelformat=empty} \caption{Fig. 11}
\end{figure}
  1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.
  2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. The curve of the roof may be modelled by \(y = - 0.013 x ^ { 3 } + 0.16 x ^ { 2 } - 0.082 x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.
  3. Use integration to find the area of the cross-section according to this model.
  4. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\).
OCR MEI C2 Q4
4 Fig. 2 shows the coordinates at certain points on a curve. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-3_646_1149_285_530} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Use the trapezium rule with 6 strips to calculate an estimate of the area of the region bounded by this curve and the axes.
OCR MEI C2 Q5
5 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-4_608_908_290_663} \captionsetup{labelformat=empty} \caption{Fig. 10}
\end{figure}
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the equation of the tangent to the curve at the point on the curve where \(x = 2\). Show that this tangent crosses the \(x\)-axis where \(x = \frac { 2 } { 3 }\).
  2. Show that the curve crosses the \(x\)-axis where \(x = 1\) and find the \(x\)-coordinate of the other point of intersection of the curve with the \(x\)-axis.
  3. Find \(\int _ { 1 } ^ { 2 } \left( 7 x - x ^ { 2 } - 6 \right) \mathrm { d } x\). Hence find the area of the region bounded by the curve, the tangent and the \(x\)-axis, shown shaded on Fig. 10.
OCR MEI C2 Q1
1
  1. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-1_650_759_252_762} \captionsetup{labelformat=empty} \caption{Fig. 12}
    \end{figure} Fig. 12 shows part of the curve \(y = x ^ { 4 }\) and the line \(y = 8 x\), which intersect at the origin and the point P .
    (A) Find the coordinates of P , and show that the area of triangle OPQ is 16 square units.
    (B) Find the area of the region bounded by the line and the curve.
  2. You are given that \(\mathrm { f } ( x ) = x ^ { 4 }\).
    (A) Complete this identity for \(\mathrm { f } ( x + h )\). $$\mathrm { f } ( x + h ) = ( x + h ) ^ { 4 } = x ^ { 4 } + 4 x ^ { 3 } h + \ldots$$ (B) Simplify \(\frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
    (C) Find \(\lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
    (D) State what this limit represents.
OCR MEI C2 Q2
5 marks
2 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-2_622_979_232_553} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} Fig. 4 shows a curve which passes through the points shown in the following table.
\(x\)11.522.533.54
\(y\)8.26.45.55.04.74.44.2
Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve, the lines \(x = 1\) and \(x = 4\), and the \(x\)-axis. State, with a reason, whether the trapezium rule gives an overestimate or an underestimate of the area of this region.
[0pt] [5]
OCR MEI C2 Q3
5 marks
3
  1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_512_819_493_700} \captionsetup{labelformat=empty} \caption{Figure 9.1}
    \end{figure} Using this model,
    (A) find the greatest height of the tunnel,
    (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
    [0pt] [5]
  2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_506_942_1703_629} \captionsetup{labelformat=empty} \caption{Not to scale}
    \end{figure} Fig. 9.2 Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
    Hence estimate the volume of earth removed when the tunnel is re-shaped.
OCR MEI C2 Q1
1 Find \(\int 7 x ^ { \frac { 5 } { 2 } } \mathrm {~d} x\).
OCR MEI C2 Q2
2 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 18 } { x ^ { 3 } } + 2\). The curve passes through the point \(( 3,6 )\). Find the
equation of the curve. equation of the curve.
OCR MEI C2 Q3
3 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { \frac { 1 } { 2 } } - 5\). Given also that the curve passes through the point (4, 20), find the equation of the curve.
OCR MEI C2 Q4
4 Find \(\int _ { 2 } ^ { 5 } \left( 2 x ^ { 3 } + 3 \right) \mathrm { d } x\).
OCR MEI C2 Q5
5 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \sqrt { x } - 2\). Given also that the curve passes through the point \(( 9,4 )\), find the equation of the curve.
OCR MEI C2 Q6
6 Find \(\int _ { 2 } ^ { 5 } \left( 1 - \frac { 6 } { x ^ { 3 } } \right) \mathrm { d } x\).
OCR MEI C2 Q7
7 Find \(\int _ { 1 } ^ { 2 } \left( 12 x ^ { 5 } + 5 \right) \mathrm { d } x\).
OCR MEI C2 Q9
9 A curve has gradient given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \sqrt { x }\). Find the equation of the curve, given that it passes through the point \(( 9,105 )\).
OCR MEI C2 Q10
10 Find \(\int _ { 1 } ^ { 2 } \left( \begin{array} { l l } x ^ { 4 } & \frac { 3 } { x ^ { 2 } } + 1 \end{array} \right) \mathrm { d } x\), showing your working.
OCR MEI C2 Q11
11 Find \(\int 30 x ^ { \frac { 3 } { 2 } } \mathrm {~d} x\).
OCR MEI C2 Q12
12 Find \(\int \left( x ^ { 5 } + 10 x ^ { \frac { 3 } { 2 } } \right) \mathrm { d } x\).
OCR MEI C2 Q1
1 Find \(\int \left( 3 x ^ { 5 } + 2 x ^ { - \frac { 1 } { 2 } } \right) \mathrm { d } x\).
OCR MEI C2 Q2
2 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f44e12ce-6725-4922-be03-902a01716980-1_766_1017_517_602} \captionsetup{labelformat=empty} \caption{Fig. 11}
\end{figure} Fig. 11 shows the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\).
  1. Use calculus to find \(\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x\) and state what this represents.
  2. Find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\), giving your answers in surd form. Hence state the set of values of \(x\) for which \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\) is a decreasing function.
OCR MEI C2 Q3
3 Find \(\int \left( x - \frac { 3 } { x ^ { 2 } } \right) \mathrm { d } x\).
OCR MEI C2 Q4
4 Find \(\int \left( 20 x ^ { 4 } + 6 x ^ { - \frac { 3 } { 2 } } \right) \mathrm { d } x\).
OCR MEI C2 Q6
6 Find \(\int \left( x ^ { \frac { 1 } { 2 } } + \frac { 6 } { x ^ { 3 } } \right) \mathrm { d } x\).
OCR MEI C2 Q7
7 Find \(\int \left( x ^ { 3 } + \frac { 1 } { x ^ { 3 } } \right) \mathrm { d } x\).
OCR MEI C2 Q8
8
  1. Differentiate \(12 \sqrt [ 3 ] { x }\).
  2. Integrate \(\frac { 6 } { x ^ { 3 } }\).