Questions — OCR MEI C3 (366 questions)

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AQA AS Paper 1 AS Paper 2 C1 C2 C3 C4 D1 D2 FP1 FP2 FP3 Further AS Paper 1 Further AS Paper 2 Discrete Further AS Paper 2 Mechanics Further AS Paper 2 Statistics Further Paper 1 Further Paper 2 Further Paper 3 Discrete Further Paper 3 Mechanics Further Paper 3 Statistics M1 M2 M3 Paper 1 Paper 2 Paper 3 S1 S2 S3 CAIE FP1 FP2 Further Paper 1 Further Paper 2 Further Paper 3 Further Paper 4 M1 M2 P1 P2 P3 S1 S2 Edexcel AEA AS Paper 1 AS Paper 2 C1 C12 C2 C3 C34 C4 CP AS CP1 CP2 D1 D2 F1 F2 F3 FD1 FD1 AS FD2 FD2 AS FM1 FM1 AS FM2 FM2 AS FP1 FP1 AS FP2 FP2 AS FP3 FS1 FS1 AS FS2 FS2 AS M1 M2 M3 M4 M5 P1 P2 P3 P4 PMT Mocks Paper 1 Paper 2 Paper 3 S1 S2 S3 S4 OCR AS Pure C1 C2 C3 C4 D1 D2 FD1 AS FM1 AS FP1 FP1 AS FP2 FP3 FS1 AS Further Additional Pure Further Additional Pure AS Further Discrete Further Discrete AS Further Mechanics Further Mechanics AS Further Pure Core 1 Further Pure Core 2 Further Pure Core AS Further Statistics Further Statistics AS H240/01 H240/02 H240/03 M1 M2 M3 M4 Mechanics 1 PURE Pure 1 S1 S2 S3 S4 Stats 1 OCR MEI AS Paper 1 AS Paper 2 C1 C2 C3 C4 D1 D2 FP1 FP2 FP3 Further Extra Pure Further Mechanics A AS Further Mechanics B AS Further Mechanics Major Further Mechanics Minor Further Numerical Methods Further Pure Core Further Pure Core AS Further Pure with Technology Further Statistics A AS Further Statistics B AS Further Statistics Major Further Statistics Minor M1 M2 M3 M4 Paper 1 Paper 2 Paper 3 S1 S2 S3 S4 SPS SPS ASFM SPS ASFM Mechanics SPS ASFM Pure SPS ASFM Statistics SPS FM SPS FM Mechanics SPS FM Pure SPS FM Statistics SPS SM SPS SM Mechanics SPS SM Pure SPS SM Statistics WJEC Further Unit 1 Further Unit 2 Further Unit 3 Further Unit 4 Further Unit 5 Further Unit 6 Unit 1 Unit 2 Unit 3 Unit 4
OCR MEI C3 Q8
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
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
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
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
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
1 Solve the equation \(| 3 - 2 x | = 4 | x |\).
OCR MEI C3 Q2
2 Express \(1 < x < 3\) im th \(\quad | x - a | < b\), where \(a\) and \(b\) are to be determined.
OCR MEI C3 Q3
3 Fig. 1 shows the graphs of \(y = | x |\) and \(y = a | x + b |\), where \(a\) and \(b\) are constants. The intercepts of \(y = a | x + b |\) with the \(x\)-and \(y\)-axes are \(( - 1,0 )\) and \(\left( 0 , \frac { 1 } { 2 } \right)\) respectively. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{125b76c1-5ab3-4645-a3c4-cf167a04f453-1_617_950_909_582} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure}
  1. Find \(a\) and \(b\).
  2. Find the coordinates of the two points of intersection of the graphs.
OCR MEI C3 Q4
4 Solve the inequality \(| 2 x + 1 | \geqslant 4\).
OCR MEI C3 Q5
4 marks
5 Solve the equation \(| 2 x - 1 | = | x |\).
[0pt] [4]
OCR MEI C3 Q6
6 Given that \(\mathrm { f } ( x ) = | x |\) and \(\mathrm { g } ( x ) = x + 1\), sketch the graphs of the composite functions \(y = \mathrm { fg } ( x )\) and \(y = \operatorname { gf } ( x )\), indicating clearly which is which.
OCR MEI C3 Q7
7 Solve the inequality \(| x - 1 | < 3\).
OCR MEI C3 Q8
8 Fig. 4 shows a sketch of the graph of \(y = 2 | x - 1 |\). It meets the \(x\) - and \(y\)-axes at ( \(a , 0\) ) and ( \(0 , b\) ) respectively. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{125b76c1-5ab3-4645-a3c4-cf167a04f453-2_478_546_1299_834} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} Find the values of \(a\) and \(b\).
OCR MEI C3 Q9
9 Solve the inequality \(| 2 x - 1 | \leqslant 3\).
OCR MEI C3 Q10
10 Fig. 1 shows the graphs of \(y = | x |\) and \(y = | x - 2 | + 1\). The point P is the minimum point of \(y = | x - 2 | + 1\), and Q is the point of intersection of the two graphs. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{125b76c1-5ab3-4645-a3c4-cf167a04f453-3_491_833_503_657} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure}
  1. Write down the coordinates of P .
  2. Verify that the \(y\)-coordinate of Q is \(1 \frac { 1 } { 2 }\).
OCR MEI C3 Q11
11 Solve the equation \(| 3 x - 2 | = x\).
OCR MEI C3 Q12
12 Solve the equation \(| 3 x + 2 | = 1\).
OCR MEI C3 Q1
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
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 Given that \(\arcsin x = \arccos y\), prove that \(x ^ { 2 } + y ^ { 2 } = 1\). [Hint: let \(\arcsin x = \theta\).]
OCR MEI C3 Q4
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
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
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
7 Sketch the curve \(y = 2 \arccos x\) for \(- 1 \leqslant x \leqslant 1\).
OCR MEI C3 Q8
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.