Questions — OCR MEI C3 (366 questions)

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OCR MEI C3 2008 January Q1
1 Differentiate \(\sqrt [ 3 ] { 1 + 6 x ^ { 2 } }\).
OCR MEI C3 2008 January Q2
2 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for all real numbers \(x\) by $$\mathrm { f } ( x ) = x ^ { 2 } , \quad \mathrm {~g} ( x ) = x - 2$$
  1. Find the composite functions \(\mathrm { fg } ( x )\) and \(\mathrm { gf } ( x )\).
  2. Sketch the curves \(y = \mathrm { f } ( x ) , y = \mathrm { fg } ( x )\) and \(y = \mathrm { gf } ( x )\), indicating clearly which is which.
OCR MEI C3 2008 January Q3
3 The profit \(\pounds P\) made by a company in its \(n\)th year is modelled by the exponential function $$P = A \mathrm { e } ^ { b n }$$ In the first year (when \(n = 1\) ), the profit was \(\pounds 10000\). In the second year, the profit was \(\pounds 16000\).
  1. Show that \(\mathrm { e } ^ { b } = 1.6\), and find \(b\) and \(A\).
  2. What does this model predict the profit to be in the 20th year?
OCR MEI C3 2008 January Q4
4 When the gas in a balloon is kept at a constant temperature, the pressure \(P\) in atmospheres and the volume \(V \mathrm {~m} ^ { 3 }\) are related by the equation $$P = \frac { k } { V }$$ where \(k\) is a constant. [This is known as Boyle's Law.]
When the volume is \(100 \mathrm {~m} ^ { 3 }\), the pressure is 5 atmospheres, and the volume is increasing at a rate of \(10 \mathrm {~m} ^ { 3 }\) per second.
  1. Show that \(k = 500\).
  2. Find \(\frac { \mathrm { d } P } { \mathrm {~d} V }\) in terms of \(V\).
  3. Find the rate at which the pressure is decreasing when \(V = 100\).
OCR MEI C3 2008 January Q5
5
  1. Verify the following statement: $$\text { ' } 2 ^ { p } - 1 \text { is a prime number for all prime numbers } p \text { less than } 11 \text { '. }$$
  2. Calculate \(23 \times 89\), and hence disprove this statement: $$\text { ' } 2 ^ { p } - 1 \text { is a prime number for all prime numbers } p ^ { \prime } \text {. }$$
OCR MEI C3 2008 January Q6
6 Fig. 6 shows the curve \(\mathrm { e } ^ { 2 y } = x ^ { 2 } + y\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a66c2da5-46b4-4467-933e-179be04b03b1-3_739_1339_349_404} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { 2 \mathrm { e } ^ { 2 y } - 1 }\).
  2. Hence find to 3 significant figures the coordinates of the point P , shown in Fig. 6, where the curve has infinite gradient. Section B (36 marks)
OCR MEI C3 2008 January Q7
7 A curve is defined by the equation \(y = 2 x \ln ( 1 + x )\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
  2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
  3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
  4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).
OCR MEI C3 2008 January Q8
8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a66c2da5-46b4-4467-933e-179be04b03b1-4_581_816_354_662} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
  2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
  3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point \(( 0,1 )\). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
  4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
  5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
OCR MEI C3 2005 June Q1
1 Solve the equation \(| 3 x + 2 | = 1\).
OCR MEI C3 2005 June Q2
2 Given that \(\arcsin x = \frac { 1 } { 6 } \pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).
OCR MEI C3 2005 June Q3
3 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for the domain \(x > 0\) as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function \(\mathrm { fg } ( x )\) in terms of \(\ln x\).
State the transformation which maps the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { fg } ( x )\).
OCR MEI C3 2005 June Q4
4 The temperature \(T ^ { \circ } \mathrm { C }\) of a liquid at time \(t\) minutes is given by the equation $$T = 30 + 20 \mathrm { e } ^ { - 0.05 t } , \quad \text { for } t \geqslant 0 .$$ Write down the initial temperature of the liquid, and find the initial rate of change of temperature.
Find the time at which the temperature is \(40 ^ { \circ } \mathrm { C }\).
OCR MEI C3 2005 June Q5
5 Using the substitution \(u = 2 x + 1\), show that \(\int _ { 0 } ^ { 1 } \frac { x } { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } ( 2 - \ln 3 )\).
OCR MEI C3 2005 June Q6
6 A curve has equation \(y = \frac { x } { 2 + 3 \ln x }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence find the exact coordinates of the stationary point of the curve.
OCR MEI C3 2005 June Q7
7 Fig. 7 shows the curve defined implicitly by the equation $$y ^ { 2 } + y = x ^ { 3 } + 2 x ,$$ together with the line \(x = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-3_465_378_534_808} \captionsetup{labelformat=empty} \caption{Not to scale}
\end{figure} Fig. 7 Find the coordinates of the points of intersection of the line and the curve.
Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.
OCR MEI C3 2005 June Q8
8 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-3_421_789_1748_610} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the \(x\)-coordinate of P .
  2. Show that Q lies on the line \(y = x\).
  3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
  4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).
OCR MEI C3 2005 June Q9
9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-4_540_943_477_550} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
  2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
  3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
  4. Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(\mathrm { g } ( x )\). Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
  5. Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).
OCR MEI C3 2006 January Q1
1 Given that \(y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }\).
OCR MEI C3 2006 January Q2
2 A population is \(P\) million at time \(t\) years. \(P\) is modelled by the equation $$P = 5 + a \mathrm { e } ^ { - b t }$$ where \(a\) and \(b\) are constants.
The population is initially 8 million, and declines to 6 million after 1 year.
  1. Use this information to calculate the values of \(a\) and \(b\), giving \(b\) correct to 3 significant figures.
  2. What is the long-term population predicted by the model?
OCR MEI C3 2006 January Q3
3
  1. Express \(2 \ln x + \ln 3\) as a single logarithm.
  2. Hence, given that \(x\) satisfies the equation $$2 \ln x + \ln 3 = \ln ( 5 x + 2 )$$ show that \(x\) is a root of the quadratic equation \(3 x ^ { 2 } - 5 x - 2 = 0\).
  3. Solve this quadratic equation, explaining why only one root is a valid solution of $$2 \ln x + \ln 3 = \ln ( 5 x + 2 ) .$$
OCR MEI C3 2006 January Q4
4 Fig. 4 shows a cone. The angle between the axis and the slant edge is \(30 ^ { \circ }\). Water is poured into the cone at a constant rate of \(2 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the radius of the water surface is \(r \mathrm {~cm}\) and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-3_369_401_431_831} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. Write down the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
  2. Show that \(V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }\), and find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
    [0pt] [You may assume that the volume of a cone of height \(h\) and radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
  3. Use the results of parts (i) and (ii) to find the value of \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when \(r = 2\).
OCR MEI C3 2006 January Q5
5 A curve is defined implicitly by the equation $$y ^ { 3 } = 2 x y + x ^ { 2 }$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( x + y ) } { 3 y ^ { 2 } - 2 x }\).
  2. Hence write down \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(x\) and \(y\).
OCR MEI C3 2006 January Q6
6 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 + 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
  1. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x - 1 } { 2 } \right)\) and state the domain of this function. Fig. 6 shows a sketch of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-4_506_561_705_751} \captionsetup{labelformat=empty} \caption{Fig. 6}
    \end{figure}
  2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .
OCR MEI C3 2006 January Q7
7 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-5_531_1262_671_536} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
  2. Find the exact coordinates of B .
  3. Show that the tangents at A and C are perpendicular to each other.
  4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE . \section*{[Question 8 is printed overleaf.]}
OCR MEI C3 2007 January Q1
1 Fig. 1 shows the graphs of \(y = | x |\) and \(y = | x - 2 | + 1\). The point P is the minimum point of \(y = | x - 2 | + 1\), and Q is the point of intersection of the two graphs. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{666dc19e-f293-4738-8530-fce90df23d17-2_490_844_493_607} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure}
  1. Write down the coordinates of P .
  2. Verify that the \(y\)-coordinate of Q is \(1 \frac { 1 } { 2 }\).