Questions - OCR MEI - A-Level Maths

Sketch the curve.
Use calculus to find the equation of the tangent to the curve at A .
Show that the equation of the normal to the curve at A is $x + 6 y = 53$. Find also, using an algebraic method, the $x$-coordinate of the point at which this normal crosses the curve again. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{65dd0efe-5c99-4814-b741-16e368c3469e-3_641_791_240_714} \captionsetup{labelformat=empty} \caption{Fig. 10}
\end{figure} A is the point with coordinates $( 1,4 )$ on the curve $y = 4 x ^ { 2 }$. B is the point with coordinates $( 0,1 )$, as shown in Fig. 10.
The line through A and B intersects the curve again at the point C . Show that the coordinates of C are $\left( - \frac { 1 } { 4 } , \frac { 1 } { 4 } \right)$.
Use calculus to find the equation of the tangent to the curve at A and verify that the equation of the tangent at C is $y = - 2 x - \frac { 1 } { 4 }$.
The two tangents intersect at the point D . Find the $y$-coordinate of D .

OCR MEI C2 Q5

5 marks Moderate -0.8

5 Find the equation of the tangent to the curve $y = 6 \sqrt { x }$ at the point where $x = 16$.

OCR MEI C2 Q1

12 marks Standard +0.3

1 Fig. 12 is a sketch of the curve $y = 2 x ^ { 2 } - 11 x + 12$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{44b860fb-040f-4d3f-94d8-42eae41c0e2d-1_468_940_285_830} \captionsetup{labelformat=empty} \caption{Fig. 12}

\end{figure}

Show that the curve intersects the $x$-axis at $( 4,0 )$ and find the coordinates of the other point of intersection of the curve and the $x$-axis.
Find the equation of the normal to the curve at the point $( 4,0 )$. Show also that the area of the triangle bounded by this normal and the axes is 1.6 units ${ } ^ { 2 }$.
Find the area of the region bounded by the curve and the $x$-axis.

OCR MEI C2 Q2

11 marks Moderate -0.8

2 A curve has equation $y = x ^ { 3 } - 6 x ^ { 2 } + 12$.

Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
Find, in the form $y = m x + c$, the equation of the normal to the curve at the point $( 2 , - 4 )$.

OCR MEI C2 Q3

13 marks Standard +0.3

3 A cubic curve has equation $y = x ^ { 3 } - 3 x ^ { 2 } + 1$.

Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.
Show that the tangent to the curve at the point where $x = - 1$ has gradient 9 . Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P . Show that the area of the triangle bounded by the normal at P and the $x$ - and $y$-axes is 8 square units.

OCR MEI C2 Q4

13 marks Moderate -0.3

4 Fig. 10 shows a sketch of the graph of $y = 7 x - x ^ { 2 } - 6$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{12e190fc-437f-499d-9c27-da49a7546755-2_604_912_1100_638} \captionsetup{labelformat=empty} \caption{Fig. 10}

\end{figure}

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 $x = \frac { 2 } { 3 }$.
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.
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. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{12e190fc-437f-499d-9c27-da49a7546755-3_643_1034_267_549} \captionsetup{labelformat=empty} \caption{Fig. 11}
\end{figure} The equation of the curve shown in Fig. 11 is $y = x ^ { 3 } - 6 x + 2$.
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
Find, in exact form, the range of values of $x$ for which $x ^ { 3 } - 6 x + 2$ is a decreasing function.
Find the equation of the tangent to the curve at the point $( - 1,7 )$. Find also the coordinates of the point where this tangent crosses the curve again.

OCR MEI C2 Q1

10 marks Moderate -0.8

1 Oskar is designing a building. Fig. 12 shows his design for the end wall and the curve of the roof. The units for $x$ and $y$ are metres. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-1_735_1246_335_441} \captionsetup{labelformat=empty} \caption{Fig. 12}

\end{figure}

Use the trapezium rule with 5 strips to estimate the area of the end wall of the building.
Oskar now uses the equation $y = - 0.001 x ^ { 3 } - 0.025 x ^ { 2 } + 0.6 x + 9$, for $0 \leqslant x \leqslant 15$, to model the curve of the roof.
(A) Calculate the difference between the height of the roof when $x = 12$ given by this model and the data shown in Fig. 12.
(B) Use integration to find the area of the end wall given by this model.

OCR MEI C2 Q2

4 marks Easy -1.2

2 Fig. 7 shows a curve and the coordinates of some points on it. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-2_639_1037_294_517} \captionsetup{labelformat=empty} \caption{Fig. 7}

\end{figure} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive $x$ - and $y$-axes.

OCR MEI C2 Q3

12 marks Moderate -0.3

3 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
\end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
\end{figure} With $x$ - and $y$-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by $y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x$.
(A) The actual ditch is 0.6 m deep when $x = 0.2$. Calculate the difference between the depth given by the model and the true depth for this value of $x$.
(B) Find $\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x$. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

OCR MEI C2 Q4

13 marks Moderate -0.3

\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_765_757_203_764} \captionsetup{labelformat=empty} \caption{Fig. 11.1}
\end{figure} A boat travels from P to Q and then to R . As shown in Fig. 11.1, Q is 10.6 km from P on a bearing of $045 ^ { \circ }$. R is 9.2 km from P on a bearing of $113 ^ { \circ }$, so that angle QPR is $68 ^ { \circ }$. Calculate the distance and bearing of R from Q .
Fig. 11.2 shows the cross-section, EBC, of the rudder of a boat. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_527_1474_1452_404} \captionsetup{labelformat=empty} \caption{Fig. 11.2}
\end{figure} BC is an arc of a circle with centre A and radius 80 cm . Angle $\mathrm { CAB } = \frac { 2 \pi } { 3 }$ radians.
EC is an arc of a circle with centre D and radius $r \mathrm {~cm}$. Angle CDE is a right angle.
1. Calculate the area of sector ABC .
2. Show that $r = 40 \sqrt { 3 }$ and calculate the area of triangle CDA.
3. Hence calculate the area of cross-section of the rudder. \begin{figure}[h]
  \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-5_695_1012_271_600} \captionsetup{labelformat=empty} \caption{Fig. 12}
  \end{figure} A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water.
4. Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough. Hence estimate the volume of water in the trough.
5. A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation $y = 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15$, for $0 \leqslant x \leqslant 0.5$. Calculate $\int _ { 0 } ^ { 0.5 } \left( 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15 \right) \mathrm { d } x$ and state what this represents.
  Hence find the volume of water in the trough as given by this model.

OCR MEI C2 Q1

12 marks Moderate -0.3

1 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10 -second intervals. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{f56da008-e7f5-45b9-8db8-e2ba09ab0161-1_732_753_302_700} \captionsetup{labelformat=empty} \caption{Fig. 10}

\end{figure} The measured speeds are shown below.

Time $( t$ seconds $)$	0	10	20	30	40	50	60
Speed $\left( v \mathrm {~m} \mathrm {~s} ^ { 1 } \right)$	28	19	14	11	12	16	22

Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line $t = 60$ and the axes. [This area represents the distance travelled by the car.]
Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area. The speed of the car may be modelled by $v = 28 - t + 0.015 t ^ { 2 }$.
Show that the difference between the value given by the model when $t = 10$ and the measured value is less than $3 \%$ of the measured value.
According to this model, the distance travelled by the car is $$\int _ { 0 } ^ { 60 } \left( 28 \quad t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

OCR MEI C2 Q2

3 marks Easy -1.2

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 $( \mathrm { m } )$	0	1.5	3	4.5	6	7.5
Depth of river $( \mathrm { m } )$	0.6	2.3	3.1	2.8	1.8	0.7

Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.

OCR MEI C2 Q3

12 marks Moderate -0.8

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}

Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.
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.
Use integration to find the area of the cross-section according to this model.
Comment on the accuracy of this model for the height of the hall when $x = 7.5$.

OCR MEI C2 Q4

4 marks Moderate -0.8

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

13 marks Standard +0.3

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}

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 $x = \frac { 2 } { 3 }$.
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.
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

12 marks Moderate -0.8

\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.
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 Moderate -0.8

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.

$x$	1	1.5	2	2.5	3	3.5	4
$y$	8.2	6.4	5.5	5.0	4.7	4.4	4.2

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

12 marks Moderate -0.3

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]
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.