Questions — OCR MEI (4456 questions)

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OCR MEI C3 Q1
5 marks Standard +0.3
1 Find the exact value of \(\int ^ { 2 } x ^ { 3 } \ln x \mathrm {~d} x\).
OCR MEI C3 Q2
18 marks Standard +0.3
2 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\) \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{93ee09be-f014-4dd7-a8da-8646837b17a5-1_471_674_761_719} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
  2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
  3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
  4. \(( A )\) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
    (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.
OCR MEI C3 Q3
5 marks Standard +0.3
3 Evaluate \(\int _ { 0 } ^ { 3 } x ( x + 1 ) ^ { - \frac { 1 } { 2 } } \mathrm {~d} x\), giving your answer as an exact fraction.
OCR MEI C3 Q4
5 marks Standard +0.3
4 Show that \(\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos \frac { 1 } { 2 } x \mathrm {~d} x = \frac { \sqrt { 2 } } { 2 } \pi + 2 \sqrt { 2 } - 4\).
[0pt] [5]
OCR MEI C3 Q5
18 marks Standard +0.3
5 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]{93ee09be-f014-4dd7-a8da-8646837b17a5-2_606_732_867_710} \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 Q1
18 marks Standard +0.3
1 Fig. 8 shows a sketch of part of the curve \(y = x \sin 2 x\), where \(x\) is in radians.
The curve crosses the \(x\)-axis at the point P . The tangent to the curve at P crosses the \(y\)-axis at Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-1_706_920_489_606} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that the \(x\)-coordinates of the turning points of the curve satisfy the equation \(\tan 2 x + 2 x = 0\).
  2. Find, in terms of \(\pi\), the \(x\)-coordinate of the point P . Show that the tangent PQ has equation \(2 \pi x + 2 y = \pi ^ { 2 }\).
    Find the exact coordinates of Q.
  3. Show that the exact value of the area shaded in Fig. 8 is \(\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)\).
OCR MEI C3 Q2
18 marks Standard +0.8
2
  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]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_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 Q3
8 marks Standard +0.3
3
  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 Q4
8 marks Moderate -0.3
4 Evaluate the following integrals, giving your answers in exact form. \begin{displayquote}
  1. \(\int _ { 0 } ^ { 1 } \frac { 2 x } { x ^ { 2 } + 1 } \mathrm {~d} x\)
  2. \(\int _ { 0 } ^ { 1 } \frac { 2 x } { x + 1 } \mathrm {~d} x\) \end{displayquote}
OCR MEI C3 Q2
17 marks Standard +0.3
2 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]{72893fd5-bc8e-433b-8358-f7979b2da636-2_717_830_606_693} \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 Q3
19 marks Standard +0.3
3 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]{72893fd5-bc8e-433b-8358-f7979b2da636-3_594_1230_514_494} \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 Q1
5 marks Standard +0.3
1 Evaluate \(\int _ { 1 } ^ { 2 } x ^ { 2 } \ln x \mathrm {~d} x\), giving your answer in an exact form.
OCR MEI C3 Q2
16 marks Standard +0.3
2 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]{00c12cc4-f7ee-4219-8d34-a1854284f65d-1_647_1027_832_534} \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 Q3
20 marks Standard +0.3
3 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]{00c12cc4-f7ee-4219-8d34-a1854284f65d-2_425_974_478_591} \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 Q1
18 marks Standard +0.3
1 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]{74cc215f-bd55-489d-aa4b-0f67c2c8de52-1_529_1259_657_602} \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 .
OCR MEI C3 Q2
6 marks Standard +0.3
2 Show that \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } x \sin 2 x \mathrm {~d} x = \frac { 3 \sqrt { 3 } \pi } { 24 }\).
OCR MEI C3 Q3
17 marks Standard +0.3
3 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]{74cc215f-bd55-489d-aa4b-0f67c2c8de52-2_420_780_549_655} \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 \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).
  1. Differentiate \(x \cos 2 x\) with respect to \(x\).
  2. Integrate \(x \cos 2 x\) with respect to \(x\).
OCR MEI C4 Q4
5 marks Standard +0.3
4 Fig. 4 shows a sketch of the region enclosed by the curve \(\sqrt { 1 + \mathrm { e } ^ { - 2 x } }\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{88eadcf5-4335-4016-baf5-d3f74513bbb8-02_517_755_1576_649} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} Find the volume of the solid generated when this region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Give your answer in an exact form.
OCR MEI C4 2007 January Q1
4 marks Moderate -0.8
1 Solve the equation \(\frac { 1 } { x } + \frac { x } { x + 2 } = 1\).
OCR MEI C4 2007 January Q2
5 marks Moderate -0.8
2 Fig. 2 shows part of the curve \(y = \sqrt { 1 + x ^ { 3 } }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5dcd4f44-4c61-4384-be1b-a8d63cb6b5aa-2_540_648_662_712} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Use the trapezium rule with 4 strips to estimate \(\int _ { 0 } ^ { 2 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), giving your answer correct to 3 significant figures.
  2. Chris and Dave each estimate the value of this integral using the trapezium rule with 8 strips. Chris gets a result of 3.25, and Dave gets 3.30. One of these results is correct. Without performing the calculation, state with a reason which is correct.
    [0pt] [2]
OCR MEI C4 2007 January Q3
7 marks Moderate -0.8
3
  1. Use the formula for \(\sin ( \theta + \phi )\), with \(\theta = 45 ^ { \circ }\) and \(\phi = 60 ^ { \circ }\), to show that \(\sin 105 ^ { \circ } = \frac { \sqrt { 3 } + 1 } { 2 \sqrt { 2 } }\).
  2. In triangle ABC , angle \(\mathrm { BAC } = 45 ^ { \circ }\), angle \(\mathrm { ACB } = 30 ^ { \circ }\) and \(\mathrm { AB } = 1\) unit (see Fig. 3). Fig. 3 Using the sine rule, together with the result in part (i), show that \(\mathrm { AC } = \frac { \sqrt { 3 } + 1 } { \sqrt { 2 } }\).
OCR MEI C4 2007 January Q4
7 marks Standard +0.3
4 Show that \(\frac { 1 + \tan ^ { 2 } \theta } { 1 - \tan ^ { 2 } \theta } = \sec 2 \theta\).
Hence, or otherwise, solve the equation \(\frac { 1 + \tan ^ { 2 } \theta } { 1 - \tan ^ { 2 } \theta } = 2\), for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
OCR MEI C4 2007 January Q5
5 marks Moderate -0.5
5 Find the first four terms in the binomial expansion of \(( 1 + 3 x ) ^ { \frac { 1 } { 3 } }\).
State the range of values of \(x\) for which the expansion is valid.
OCR MEI C4 2007 January Q6
8 marks Standard +0.3
6
  1. Express \(\frac { 1 } { ( 2 x + 1 ) ( x + 1 ) }\) in partial fractions.
  2. A curve passes through the point \(( 0,2 )\) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { ( 2 x + 1 ) ( x + 1 ) }$$ Show by integration that \(y = \frac { 4 x + 2 } { x + 1 }\). Section B (36 marks)
OCR MEI C4 2007 January Q7
20 marks Standard +0.3
7 Fig. 7 shows the curve with parametric equations $$x = \cos \theta , y = \sin \theta - \frac { 1 } { 8 } \sin 2 \theta , 0 \leqslant \theta < 2 \pi$$ The curve crosses the \(x\)-axis at points \(\mathrm { A } ( 1,0 )\) and \(\mathrm { B } ( - 1,0 )\), and the positive \(y\)-axis at C . D is the maximum point of the curve, and E is the minimum point. The solid of revolution formed when this curve is rotated through \(360 ^ { \circ }\) about the \(x\)-axis is used to model the shape of an egg. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5dcd4f44-4c61-4384-be1b-a8d63cb6b5aa-4_744_1207_776_431} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Show that, at the point \(\mathrm { A } , \theta = 0\). Write down the value of \(\theta\) at the point B , and find the coordinates of C .
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence show that, at the point D, $$2 \cos ^ { 2 } \theta - 4 \cos \theta - 1 = 0 .$$
  3. Solve this equation, and hence find the \(y\)-coordinate of D , giving your answer correct to 2 decimal places. The cartesian equation of the curve (for \(0 \leqslant \theta \leqslant \pi\) ) is $$y = \frac { 1 } { 4 } ( 4 - x ) \sqrt { 1 - x ^ { 2 } } .$$
  4. Show that the volume of the solid of revolution of this curve about the \(x\)-axis is given by $$\frac { 1 } { 16 } \pi \int _ { - 1 } ^ { 1 } \left( 16 - 8 x - 15 x ^ { 2 } + 8 x ^ { 3 } - x ^ { 4 } \right) \mathrm { d } x .$$ Evaluate this integral.