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OCR MEI C3 Q3
16 marks Standard +0.3
3 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-3_654_1034_463_531} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
  3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).
OCR MEI C3 Q1
20 marks Standard +0.3
1 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aee8da6a-7d5c-442f-9729-55d81d9a606f-1_427_968_432_584} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the exact coordinates of P .
  2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
  3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
  5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
  6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.
OCR MEI C3 Q2
17 marks Standard +0.3
2 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]{aee8da6a-7d5c-442f-9729-55d81d9a606f-2_418_769_516_673} \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 Q3
19 marks Moderate -0.3
3 The function \(f ( x ) = \ln \left( t + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aee8da6a-7d5c-442f-9729-55d81d9a606f-3_510_895_523_604} \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 \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\).
  4. Sketch the curves \(y = f ( x )\) and \(y = g ( x )\) on the same axes. State the domain of the function \(g ( x )\),
    Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
  5. Differentiate \(\mathrm { g } ( \mathrm { x } )\). 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 Q1
5 marks Standard +0.3
1 Fig. 4 shows a cone with its axis vertical. The angle between the axis and the slant edge is \(45 ^ { \circ }\). Water is poured into the cone at a constant rate of \(5 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the height of the water surface above the vertex O of the cone is \(h \mathrm {~cm}\), and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-1_295_403_542_871} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} Find \(V\) in terms of \(h\). Hence find the rate at which the height of water is increasing when the height is 10 cm .
[0pt] [You are given that the volume \(V\) of a cone of height \(h\) and radius \(r\) is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ].
OCR MEI C3 Q2
5 marks Standard +0.3
2 A spherical balloon of radius \(r \mathrm {~cm}\) has volume \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\). The balloon is inflated at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate of increase of \(r\) when \(r = 8\).
OCR MEI C3 Q3
5 marks Moderate -0.3
3 The driving force \(F\) newtons and velocity \(v \mathrm {~km} \mathrm {~s} ^ { - 1 }\) of a car at time \(t\) seconds are related by the equation \(F = \frac { 25 } { v }\).
  1. Find \(\frac { \mathrm { d } F } { \mathrm {~d} v }\).
  2. Find \(\frac { \mathrm { d } F } { \mathrm {~d} t }\) when \(v = 50\) and \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 1.5\).
OCR MEI C3 Q4
5 marks Standard +0.3
4 Water flows into a bowl at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) (see Fig. 4). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-2_414_379_485_838} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} When the depth of water in the bowl is \(h \mathrm {~cm}\), the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \pi h ^ { 2 }\). Find the rate at which the depth is increasing at the instant in time when the depth is 5 cm .
OCR MEI C3 Q5
18 marks Moderate -0.3
5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \cos ^ { 2 } x } , - \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), together with its asymptotes \(x = \frac { 1 } { 2 } \pi\) and \(x = - \frac { 1 } { 2 } \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-3_921_1398_538_414} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Use the quotient rule to show that the derivative of \(\frac { \sin x } { \cos x }\) is \(\frac { 1 } { \cos ^ { 2 } x }\).
  2. Find the area bounded by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 4 } \pi\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \mathrm { f } \left( x + \frac { 1 } { 4 } \pi \right)\).
  3. Verify that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) cross at \(( 0,1 )\).
  4. State a sequence of two transformations such that the curve \(y = \mathrm { f } ( x )\) is mapped to the curve \(y = \mathrm { g } ( x )\). On the copy of Fig. 9, sketch the curve \(y = \mathrm { g } ( x )\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve.
  5. Use your result from part (ii) to write down the area bounded by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = - \frac { 1 } { 4 } \pi\).
OCR MEI C3 Q6
7 marks Moderate -0.3
6 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 Q7
7 marks Standard +0.3
7 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]{431d496a-a606-4b92-9f5c-e12b074a7ba9-4_363_391_1447_887} \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 Q8
6 marks Moderate -0.8
8 Fig. 4 is a diagram of a garden pond. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-5_295_742_410_693} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} The volume \(V \mathrm {~m} ^ { 3 }\) of water in the pond when the depth is \(h\) metres is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 3 - h ) .$$
  1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} h }\). Water is poured into the pond at the rate of \(0.02 \mathrm {~m} ^ { 3 }\) per minute.
  2. Find the value of \(\frac { \mathrm { d } h } { \mathrm {~d} t }\) when \(h = 0.4\).
OCR MEI C3 Q1
5 marks Standard +0.3
1 Given that \(\mathrm { f } ( x ) = \frac { x + 1 } { x - 1 }\), show that \(\mathrm { ff } ( x ) = x\).
Hence write down the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). What can you deduce about the symmetry of the curve \(y = \mathrm { f } ( x ) ?\)
OCR MEI C3 Q2
5 marks Moderate -0.8
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 Q3
3 marks Moderate -0.8
3 Given that \(\mathrm { f } ( x ) = 1 - x\) and \(\mathrm { g } ( x ) = | x |\), write down the composite function \(\mathrm { gf } ( x )\). On separate diagrams, sketch the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { gf } ( x )\).
OCR MEI C3 Q4
7 marks Moderate -0.3
4 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 \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x \mathrm { r } } { 2 } \right)\) 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]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-2_499_562_779_785} \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 Q5
3 marks Easy -1.2
5 Given that \(\arcsin x = \frac { 1 } { 6 } \pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).
OCR MEI C3 Q6
3 marks Moderate -0.8
6 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 Q7
19 marks Standard +0.3
7 Fig. 9 shows the line \(y = x\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)\). The line and the curve intersect at the origin and at the point \(\mathrm { P } ( a , a )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-3_694_886_430_604} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Show that \(\mathrm { e } ^ { a } = 1 + 2 a\).
  2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac { 1 } { 2 } a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\).
  3. Show that the inverse function of \(\mathrm { f } ( x )\) is \(\mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = \ln ( 1 + 2 x )\). Add a sketch of \(y = \mathrm { g } ( x )\) to the copy of Fig. 9.
  4. Find the derivatives of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }\). Give a geometrical interpretation of this result.
OCR MEI C3 Q8
8 marks Moderate -0.3
8 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-4_736_809_419_653} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Write down the range of the function \(\mathrm { f } ( x )\).
  2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
  3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
OCR MEI C3 Q1
18 marks Challenging +1.2
1 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-1_555_641_573_748} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
  2. Find the gradient of the curve at P . [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
  3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R.
  4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\).
    [0pt] [Hint: consider its reflection in \(y = x\).]
OCR MEI C3 Q2
18 marks Standard +0.3
2 Fig. 8 shows the line \(y = x\) and parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where $$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$ The curves intersect the axes at the points A and B, as shown. The curves and the line \(y = x\) meet at the point C . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-2_811_893_609_655} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the exact coordinates of A and B . Verify that the coordinates of C are \(( 1,1 )\).
  2. Prove algebraically that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).
  3. Evaluate \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\), giving your answer in terms of e .
  4. Use integration by parts to find \(\int \ln x \mathrm {~d} x\). Hence show that \(\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }\).
  5. Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .
OCR MEI C3 Q3
17 marks Moderate -0.3
3 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]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-3_577_815_392_719} \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 Q4
18 marks Standard +0.8
4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-4_820_815_551_715} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the \(y\)-coordinate of P .
  2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
  3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
  4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
OCR MEI C3 Q1
4 marks Standard +0.3
1 Solve the equation \(| 3 - 2 x | = 4 | x |\).