Questions — OCR MEI C3 (386 questions)

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OCR MEI C3 Q3
6 marks Moderate -0.8
3 Differentiate the following functions.
  1. \(\quad y = \left( x ^ { 2 } + 3 \right) ^ { 5 }\)
  2. \(y = \frac { \sin 2 x } { x }\)
OCR MEI C3 Q4
5 marks Moderate -0.8
4 A curve has equation \(y ^ { 2 } = 5 x - 4\).
Find the gradient of the curve at the points where \(x = 8\).
OCR MEI C3 Q5
4 marks Moderate -0.8
5 Given that \(x\) and \(t\) are related by the formula \(x = x _ { 0 } \mathrm { e } ^ { - 3 t }\), show that \(t = \ln \left( \frac { a } { x } \right) ^ { b }\) where \(a\) and \(b\) are to be determined.
OCR MEI C3 Q6
8 marks Moderate -0.3
6
  1. Find \(\int ( 2 x - 3 ) ^ { 7 } \mathrm {~d} x\).
  2. Use the substitution \(u = x ^ { 2 } + 1\), or otherwise, to find \(\int _ { 1 } ^ { 2 } x \left( x ^ { 2 } + 1 \right) ^ { 3 } \mathrm {~d} x\).
OCR MEI C3 Q7
5 marks Easy -1.3
7 The functions \(f , g\) and \(h\) are defined as follows. $$\mathrm { f } ( x ) = 2 x \quad \mathrm {~g} ( x ) = x ^ { 2 } \quad \mathrm {~h} ( x ) = x + 2$$ Find each of the following as functions of \(x\).
  1. \(\mathrm { f } ^ { 2 } ( x )\),
  2. \(\operatorname { fgh } ( x )\),
  3. \(\mathrm { h } ^ { - 1 } ( x )\).
OCR MEI C3 Q8
18 marks Standard +0.3
8 A curve has equation \(y = ( x + 2 ) \mathrm { e } ^ { - x }\).
  1. Find the coordinates of the points where the curve cuts the axes.
  2. Find the coordinates of the stationary point, S , on the curve.
  3. By evaluating \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at S , determine whether the stationary point is a maximum or a minimum.
  4. Sketch the curve in the domain \(- 3 < x < 3\).
  5. Find where the normal to the curve at the point \(( 0,2 )\) cuts the curve again.
  6. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).
OCR MEI C3 Q1
2 marks Moderate -0.5
1 John asserts that the expression \(n ^ { 2 } + n + 11\) is prime for all positive integer values of \(n\). Show that John is wrong in his assertion.
OCR MEI C3 Q2
4 marks Easy -1.2
2
  1. Show that \(\mathrm { f } ( x ) = \left| x ^ { 3 } \right|\) is an even function.
  2. It is suggested that the function \(\mathrm { g } ( x ) = ( x - 1 ) ^ { 3 }\) is odd. Prove that this is false.
OCR MEI C3 Q4
5 marks Moderate -0.3
4 The volume of a sphere, \(V \mathrm {~cm} ^ { 3 }\) is given by the formula \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) where \(r \mathrm {~cm}\) is the radius.
The radius of a sphere increases at a constant rate of 2 cm per second.
Find the rate of increase of \(V\) when \(r = 10 \mathrm {~cm}\).
OCR MEI C3 Q5
6 marks Moderate -0.8
5 The equation of a circle is \(x ^ { 2 } + y ^ { 2 } = 25\).
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { y }\).
  2. Hence find the equation of the normal to the circle at the point ( 3,4 ).
OCR MEI C3 Q6
8 marks Moderate -0.3
6
  1. Find \(\int x \cos 2 x d x\).
  2. Using the substitution \(u = x ^ { 2 } + 1\), or otherwise, find the exact value of \(\int _ { 2 } ^ { 3 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x\).
OCR MEI C3 Q7
7 marks Standard +0.8
7 Fig. 7 shows the graphs of the curves \(y = \mathrm { e } ^ { - x }\) and \(y = \mathrm { e } ^ { - x } \sin x\) for \(0 \leq x \leq \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-3_407_793_1085_740} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} The maximum point on \(y = \mathrm { e } ^ { - x } \sin x\) is at A , and the curves touch at B . \(\mathrm { A } ^ { \prime }\) and \(\mathrm { B } ^ { \prime }\) are the points on the \(x\)-axis such that \(\mathrm { A } ^ { \prime } \mathrm { A }\) and \(\mathrm { B } ^ { \prime } \mathrm { B }\) are parallel to the \(y\)-axis.
Show that \(\mathrm { OA } ^ { \prime } = \mathrm { A } ^ { \prime } \mathrm { B } ^ { \prime }\).
OCR MEI C3 Q8
18 marks Standard +0.3
8 Fig. 8 shows part of the graph of the function \(y = 5 x ( 2 x - 1 ) ^ { 3 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-4_508_803_450_703} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of S , the turning point of the curve.
  2. Find the area of the shaded region enclosed between the curve and the \(x\)-axis.
  3. Given that \(\mathrm { f } ( x ) = 5 x ( 2 x - 1 ) ^ { 3 }\), show that \(\mathrm { f } ( x + 0.5 ) = 40 x ^ { 3 } ( x + 0.5 )\).
  4. Find \(\int _ { - \frac { 1 } { 2 } } ^ { 0 } 40 x ^ { 3 } ( x + 0.5 ) \mathrm { d } x\).
  5. Explain, with the aid of a sketch, the connection between your answer to parts (ii) and (iv).
OCR MEI C3 Q1
4 marks Moderate -0.5
1 Find \(\int \sqrt [ 3 ] { 2 x - 1 } \mathrm {~d} x\).
OCR MEI C3 Q2
18 marks Standard +0.3
2 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]{6ea594c5-52ba-4467-a098-cb66004b5a38-1_959_1469_748_317} \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 Q3
3 marks Moderate -0.8
3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } ( 1 - \sin 3 x ) \mathrm { d } x\), giving your answer in exact form.
OCR MEI C3 Q4
18 marks Challenging +1.2
4 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]{6ea594c5-52ba-4467-a098-cb66004b5a38-2_431_977_728_602} \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.
OCR MEI C3 Q5
5 marks Standard +0.3
5 Using a suitable substitution or otherwise, show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { \sin 2 x } { 3 + \cos 2 x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).
OCR MEI C3 Q1
18 marks Standard +0.3
1 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }\), with its turning point P . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-1_722_817_450_642} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Write down the coordinates of the intercepts of \(y = \mathrm { f } ( x )\) with the \(x\) - and \(y\)-axes.
  2. Find the exact coordinates of the turning point P .
  3. Show that the exact area of the region enclosed by the curve and the \(x\) - and \(y\)-axes is \(\frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 } - 3 \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
  4. Express \(\mathrm { g } ( x )\) in terms of \(x\). Sketch the curve \(y = \mathrm { g } ( x )\) on the copy of Fig. 8, indicating the coordinates of its intercepts with the \(x\) - and \(y\)-axes and of its turning point.
  5. Write down the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\) and the \(x\) - and \(y\)-axes.
OCR MEI C3 Q2
18 marks Standard +0.8
2 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]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-2_754_870_478_609} \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 Q3
18 marks Challenging +1.2
3 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]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-3_559_644_622_745} \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 \(\ln \frac { 32 } { 27 }\). [Hint: consider its reflection in \(y = x\).]
OCR MEI C3 Q1
5 marks Moderate -0.3
1 Show that \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 3 x - 2 } } \mathrm {~d} x = \frac { 2 } { 3 }\).
OCR MEI C3 Q2
23 marks Standard +0.3
2 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]{f7049002-f97a-4c83-a7d6-eba28e3b589a-1_904_937_785_604} \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 Q3
3 marks Moderate -0.8
3 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \sin 3 x \mathrm {~d} x\).
[0pt] [3]
OCR MEI C3 Q4
18 marks Standard +0.8
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]{f7049002-f97a-4c83-a7d6-eba28e3b589a-2_824_816_885_699} \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.