Questions — OCR MEI (4301 questions)

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OCR MEI C3 Q12
3 marks Easy -1.2
12 Solve the equation \(| 3 x + 2 | = 1\).
OCR MEI C3 Q1
4 marks Moderate -0.8
1 Solve each of the following equations, giving your answers in exact form.
  1. \(6 \arcsin x - \pi = 0\).
  2. \(\arcsin x = \arccos x\).
OCR MEI C3 Q2
4 marks Moderate -0.8
2 The curves in parts (i) and (ii) have equations of the form \(y = a + b \sin c x\), where \(a , b\) and \(c\) are constants. For each curve, find the values of \(a , b\) and \(c\).

  1. \includegraphics[max width=\textwidth, alt={}, center]{11877196-83d9-4283-9eef-e617bea50c63-1_449_681_834_408}

  2. \includegraphics[max width=\textwidth, alt={}, center]{11877196-83d9-4283-9eef-e617bea50c63-1_376_681_1344_408}
OCR MEI C3 Q3
3 marks Standard +0.3
3 Given that \(\arcsin x = \arccos y\), prove that \(x ^ { 2 } + y ^ { 2 } = 1\). [Hint: let \(\arcsin x = \theta\).]
OCR MEI C3 Q4
8 marks Moderate -0.8
4
  1. State the period of the function \(\mathrm { f } ( x ) = 1 + \cos 2 x\), where \(x\) is in degrees.
  2. State a sequence of two geometrical transformations which maps the curve \(y = \cos x\) onto the curve \(y = \mathrm { f } ( x )\).
  3. Sketch the graph of \(y = \mathrm { f } ( x )\) for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\).
OCR MEI C3 Q5
8 marks Moderate -0.3
5 Fig. 7 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + 2 \arctan x , x \in \mathbb { R }\). The scales on the \(x\) - and \(y\)-axes are the same. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{11877196-83d9-4283-9eef-e617bea50c63-2_855_838_1028_688} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Find the range of f , giving your answer in terms of \(\pi\).
  2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and add a sketch of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) to the copy of Fig. 7.
OCR MEI C3 Q6
17 marks Standard +0.8
6 Fig. 8 shows part of the curve \(y = x \cos 3 x\). The curve crosses the \(x\)-axis at \(\mathrm { O } , \mathrm { P }\) and Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{11877196-83d9-4283-9eef-e617bea50c63-3_553_1178_622_529} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the exact coordinates of P and Q .
  2. Find the exact gradient of the curve at the point P . Show also that the turning points of the curve occur when \(x \tan 3 x = \frac { 1 } { 3 }\).
  3. Find the area of the region enclosed by the curve and the \(x\)-axis between O and P , giving your answer in exact form.
OCR MEI C3 Q7
3 marks Moderate -0.8
7 Sketch the curve \(y = 2 \arccos x\) for \(- 1 \leqslant x \leqslant 1\).
OCR MEI C3 Q8
8 marks Standard +0.3
8 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{11877196-83d9-4283-9eef-e617bea50c63-4_379_722_467_715} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find the range of the function \(\mathrm { f } ( x )\), giving your answer in terms of \(\pi\).
  2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the origin.
  3. Hence write down the gradient of \(y = \frac { 1 } { 2 } \arctan x\) at the origin.
OCR MEI C3 Q9
3 marks Easy -1.2
9 Given that \(\arcsin x = \frac { 1 } { 6 } \pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).
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 Moderate -0.3
1 Find the exact value of \(\int _ { 0 } ^ { 2 } \sqrt { 1 + 4 x } \mathrm {~d} x\), showing your working.
OCR MEI C3 Q2
23 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]{a55b82e6-3fcb-4283-bd36-06a17a9a7536-1_804_888_1061_662} \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 Q4
4 marks Moderate -0.5
4 Find \(\int x \mathrm { e } ^ { 3 x } \mathrm {~d} x\).
OCR MEI C3 Q5
4 marks Standard +0.3
5 Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 2 } \mathrm {~d} x = \frac { 1 } { 2 } \ln 6\).