Questions — OCR MEI C3 (366 questions)

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OCR MEI C3 Q1
1 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2437cecc-f084-4e49-ab36-1c132ba13267-1_480_1058_364_578} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure} Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } + 2 }\).
  1. Show algebraically that \(\mathrm { f } ( x )\) is an even function, and state how this property relates to the curve \(y = \mathrm { f } ( x )\).
  2. Find \(\mathrm { f } ^ { \prime } ( x )\).
  3. Show that \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { \left( \mathrm { e } ^ { x } + 1 \right) ^ { 2 } }\).
  4. Hence, using the substitution \(u = \mathrm { e } ^ { x } + 1\), or otherwise, find the exact area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, and the lines \(x = 0\) and \(x = 1\).
  5. Show that there is only one point of intersection of the curves \(y = \mathrm { f } ( x )\) and \(y = \frac { 1 } { 4 } \mathrm { e } ^ { x }\), and find its coordinates.
OCR MEI C3 Q2
2 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).
OCR MEI C3 Q3
3
  1. Differentiate \(x \cos 2 x\) with respect to \(x\).
  2. Integrate \(x \cos 2 x\) with respect to \(x\).
OCR MEI C3 Q4
4 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }\).
The curve has asymptotes \(x = 0\) and \(x = a\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2437cecc-f084-4e49-ab36-1c132ba13267-2_652_795_876_717} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Find \(a\). Hence write down the domain of the function.
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the coordinates of the turning point of the curve, and write down the range of the function. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
  3. (A) Show algebraically that \(\mathrm { g } ( x )\) is an even function.
    (B) Show that \(\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )\).
    (C) Hence prove that the curve \(y = \mathrm { f } ( x )\) is symmetrical, and state its line of symmetry.
OCR MEI C3 Q1
1 Fig. 9 shows the curve \(y = \frac { x ^ { 2 } } { 3 x - 1 }\).
P is a turning point, and the curve has a vertical asymptote \(x = a\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bce065bf-a56c-4686-8fa7-cb18cb95012e-1_835_1474_591_372} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Write down the value of \(a\).
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ( 3 x - 2 ) } { ( 3 x - 1 ) ^ { 2 } }\).
  3. Find the exact coordinates of the turning point P . Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point.
  4. Using the substitution \(u = 3 x - 1\), show that \(\int \frac { x ^ { 2 } } { 3 x - 1 } \mathrm {~d} x = \frac { 1 } { 27 } \int \left( u + 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 1\).
OCR MEI C3 Q2
2 Differentiate \(\sqrt [ 3 ] { 1 + 6 x ^ { 2 } }\).
OCR MEI C3 Q3
3 Show that the curve \(y = x ^ { 2 } \ln x\) has a stationary point when \(x = \frac { 1 } { \sqrt { \mathrm { e } } }\).
OCR MEI C3 Q4
4 The equation of a curve is \(y = \frac { x ^ { 2 } } { 2 x + 1 }\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( x + 1 ) } { ( 2 x + 1 ) ^ { 2 } }\).
  2. Find the coordinates of the stationary points of the curve. You need not determine their nature.
  3. Differentiate \(\sqrt { 1 + 2 x }\).
  4. Show that the derivative of \(\ln \left( 1 - \mathrm { e } ^ { - x } \right)\) is \(\frac { 1 } { \mathrm { e } ^ { x } - 1 }\).
OCR MEI C3 Q6
6 The function \(\mathrm { f } ( x ) = \frac { \sin x } { 2 - \cos x }\) has domain \(- \pi \leqslant x \leqslant \pi\).
Fig. 8 shows the graph of \(y = \mathrm { f } ( x )\) for \(0 \leqslant x \leqslant \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bce065bf-a56c-4686-8fa7-cb18cb95012e-3_557_844_602_646} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find \(\mathrm { f } ( - x )\) in terms of \(\mathrm { f } ( x )\). Hence sketch the graph of \(y = \mathrm { f } ( x )\) for the complete domain \(- \pi \leqslant x \leqslant \pi\).
  2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 \cos x - 1 } { ( 2 - \cos x ) ^ { 2 } }\). Hence find the exact coordinates of the turning point P . State the range of the function \(\mathrm { f } ( x )\), giving your answer exactly.
  3. Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int _ { 0 } ^ { \pi } \frac { \sin x } { 2 \cos x } \mathrm {~d} x\).
  4. Sketch the graph of \(y = \mathrm { f } ( 2 x )\).
  5. Using your answers to parts (iii) and (iv), write down the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 2 \cos 2 x } \mathrm {~d} x\).
OCR MEI C3 Q7
7 Fig. 3 shows the curve defined by the equation \(y = \arcsin ( x - 1 )\), for \(0 \leqslant x \leqslant 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bce065bf-a56c-4686-8fa7-cb18cb95012e-4_681_542_498_794} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Find \(x\) in terms of \(y\), and show that \(\frac { \mathrm { d } x } { \mathrm {~d} y } = \cos y\).
  2. Hence find the exact gradient of the curve at the point where \(x = 1.5\).
OCR MEI C3 Q8
8 A curve has equation \(y = \frac { x } { 2 + 3 \ln x }\). Find \(\frac { d y } { d x }\). Hence find the exact coordinates of the stationary point of the curve.
OCR MEI C3 Q1
1 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-1_961_1473_445_320} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
  2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
  3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
  4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.
OCR MEI C3 Q2
2 Fig. 9 shows the curve \(y = x \mathrm { e } ^ { - 2 x }\) together with the straight line \(y = m x\), where \(m\) is a constant, with \(0 < m < 1\). The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-2_433_979_472_591} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Show that the \(x\)-coordinate of P is \(- \frac { 1 } { 2 } \ln m\).
  2. Find, in terms of \(m\), the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the \(x\)-axis.
  3. Show that \(m = \mathrm { e } ^ { - 2 }\), and find the exact coordinates of P .
  4. Find the exact area of the shaded region between the line OP and the curve. END OF QUESTION PAPER
OCR MEI C3 Q3
3
  1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
    Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
    \end{figure}
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
  3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).
OCR MEI C3 Q1
1 Fig. 8 shows the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{65ac8807-cd93-450f-adb5-dc6864f8470c-1_720_834_578_681} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Verify that the coordinates of P are \(( 1,0 )\).
  2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
  3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.
OCR MEI C3 Q2
2 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{65ac8807-cd93-450f-adb5-dc6864f8470c-2_595_1230_445_496} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Find the coordinates of P .
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
  3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
  4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
  5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
    (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
    (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?
OCR MEI C3 Q3
3 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 Q1
1 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).
  1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-1_628_935_728_657} \captionsetup{labelformat=empty} \caption{Fig. 9}
    \end{figure}
  2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
    (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
    (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
  3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
  4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).
OCR MEI C3 Q2
2 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-2_487_875_487_624} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
  3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
  4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.
OCR MEI C3 Q3
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
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
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
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
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
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 \(r = 8\).