Questions — OCR MEI C3 (386 questions)

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OCR MEI C3 Q1
18 marks Standard +0.8
1 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0b4c4935-998c-404f-8fed-9b39b849168e-1_752_864_443_617} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Write down the value of \(a\).
  2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
  3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).
OCR MEI C3 Q2
18 marks Standard +0.3
2 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0b4c4935-998c-404f-8fed-9b39b849168e-2_606_729_485_699} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Verify that the \(x\)-coordinate of P is 3 .
  2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
  3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).
OCR MEI C3 Q3
18 marks Standard +0.3
3 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), which has a \(y\)-intercept at \(\mathrm { P } ( 0,3 )\), a minimum point at \(\mathrm { Q } ( 1,2 )\), and an asymptote \(x = - 1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0b4c4935-998c-404f-8fed-9b39b849168e-3_904_937_425_589} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Find the coordinates of the images of the points P and Q when the curve \(y = \mathrm { f } ( x )\) is transformed to
    (A) \(y = 2 \mathrm { f } ( x )\),
    (B) \(y = \mathrm { f } ( x + 1 ) + 2\). You are now given that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1\).
  2. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence find the coordinates of the other turning point on the curve \(y = \mathrm { f } ( x )\).
  3. Show that \(\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }\).
  4. Find \(\int _ { a } ^ { b } \left( x - 2 + \frac { 4 } { x } \right) \mathrm { d } x\) in terms of \(a\) and \(b\). Hence, by choosing suitable values for \(a\) and \(b\), find the exact area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\).
OCR MEI C3 Q4
8 marks Standard +0.3
4
  1. Differentiate \(\frac { \ln x } { x ^ { 2 } }\), simplifying your answer.
  2. Using integration by parts, show that \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c\).
OCR MEI C3 Q1
7 marks Standard +0.3
1 Fig. 4 shows the curve \(y = \mathrm { f } ( x )\), where $$f ( x ) = a + \cos b x , 0 \leqslant x \leqslant 2 \pi ,$$ and \(a\) and \(b\) are positive constants. The curve has stationary points at \(( 0,3 )\) and \(( 2 \pi , 1 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7825ba53-67eb-4050-a671-85e37a30150a-1_420_616_538_759} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. Find \(a\) and \(b\).
  2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range.
OCR MEI C3 Q2
19 marks Standard +0.3
2 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]{7825ba53-67eb-4050-a671-85e37a30150a-2_681_880_461_606} \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 Q3
8 marks Moderate -0.3
3 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]{7825ba53-67eb-4050-a671-85e37a30150a-3_743_818_414_644} \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 Q4
8 marks Standard +0.3
4 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \nless \leqslant 1\).
Fig. 6 also shows the curve \(y = \mathrm { g } ( x )\), where \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
P is the point on the curve \(y = \mathrm { f } ( x )\) with \(x\)-coordinate \(\frac { 1 } { 2 }\).
[diagram]
  1. Find the \(y\)-coordinate of P , giving your answer in terms of \(\pi\). The point Q is the reflection of P in \(y = x\).
  2. Find \(\mathrm { g } ( x )\) and its derivative \(\mathrm { g } ^ { \prime } ( x )\). Hence determine the exact gradient of the curve \(y = \mathrm { g } ( x )\) at the point Q . Write down the exact gradient of \(y = \mathrm { f } ( x )\) at the point P .
OCR MEI C3 Q5
18 marks Standard +0.3
5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\). The endpoints of the curve are \(\mathrm { P } ( - \pi , 1 )\) and \(\mathrm { Q } ( \pi , 3 )\), and \(\mathrm { f } ( x ) = a + \sin b x\), where \(a\) and \(b\) are constants. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7825ba53-67eb-4050-a671-85e37a30150a-5_661_1259_461_478} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Using Fig. 9, show that \(a = 2\) and \(b = \frac { 1 } { 2 }\).
  2. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,2 ). Show that there is no point on the curve at which the gradient is greater than this.
  3. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range. Write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 2,0 )\).
  4. Find the area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\).
OCR MEI C3 Q1
19 marks Standard +0.2
1 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 1 } { x ^ { 2 } + 1 }\) for the domain \(0 \leqslant x \leqslant 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b7588524-8a5e-42af-8b52-29cdddc09eeb-1_976_1208_450_514} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\), and hence that \(\mathrm { f } ( x )\) is an increasing function for \(x > 0\).
  2. Find the range of \(\mathrm { f } ( x )\).
  3. Given that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 6 - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 3 } }\), find the maximum value of \(\mathrm { f } ^ { \prime } ( x )\). The function \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
  4. Write down the domain and range of \(\mathrm { g } ( x )\). Add a sketch of the curve \(y = \mathrm { g } ( x )\) to a copy of Fig. 9 .
  5. Show that \(\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }\).
OCR MEI C3 Q2
7 marks Standard +0.3
2 The variables \(x\) and \(y\) satisfy the equation \(x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }\). Both \(x\) and \(y\) are functions of \(t\).
  2. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(x = 1 , y = 8\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\).
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]{b7588524-8a5e-42af-8b52-29cdddc09eeb-2_577_820_1114_675} \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 Q1
6 marks Standard +0.8
1 Fig. 1 shows part of the curve \(y = \mathrm { e } ^ { 2 x } \cos x\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{01bdea17-c698-44ae-a45a-7da4de631de4-1_669_1032_459_538} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} Find the coordinates of the turning point P .
OCR MEI C3 Q2
5 marks Standard +0.3
2 Find the exact gradient of the curve \(y = \ln ( 1 - \cos 2 x )\) at the point with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\).
OCR MEI C3 Q3
6 marks Standard +0.3
3
  1. Given that \(y = \mathrm { e } ^ { - x } \sin 2 x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  2. Hence show that the curve \(y = \mathrm { e } ^ { - x } \sin 2 x\) has a stationary point when \(x = \frac { 1 } { 2 } \arctan 2\).
OCR MEI C3 Q4
17 marks Standard +0.3
4 Fig. 8 shows parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where \(\mathrm { f } ( x ) = \tan x\) and \(\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{01bdea17-c698-44ae-a45a-7da4de631de4-2_687_888_419_609} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Describe a sequence of two transformations which maps the curve \(y = \mathrm { f } ( x )\) to the curve \(y = \mathrm { g } ( x )\). [4] It can be shown that \(\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }\).
  2. Show that \(\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }\). Hence verify that the gradient of \(y = \mathrm { g } ( x )\) at the point \(\left( \frac { 1 } { 4 } \pi , 1 \right)\) is the same as that of \(y = \mathrm { f } ( x )\) at the origin.
  3. By writing \(\tan x = \frac { \sin x } { \cos x }\) and using the substitution \(u = \cos x\), show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u\). Evaluate this integral exactly.
  4. Hence find the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the lines \(x = \frac { 1 } { 4 } \pi\) and \(x = \frac { 1 } { 2 } \pi\).
OCR MEI C3 Q5
3 marks Standard +0.3
5 Differentiate \(x ^ { 2 } \tan 2 x\).
OCR MEI C3 Q6
4 marks Moderate -0.8
6 Given that \(y = \sqrt [ 3 ] { 1 + x ^ { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
OCR MEI C3 Q7
5 marks Moderate -0.3
7 Given that \(y = x ^ { 2 } \sqrt { 1 + 4 x }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( 5 x + 1 ) } { \sqrt { 1 + 4 x } }\).
OCR MEI C3 Q1
18 marks Standard +0.8
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
3 marks Easy -1.2
2 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \cos 3 x \mathrm {~d} x\).
OCR MEI C3 Q3
7 marks Standard +0.3
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
18 marks Standard +0.3
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
18 marks Standard +0.3
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
18 marks Challenging +1.2
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