Questions — OCR MEI C3 (386 questions)

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OCR MEI C3 Q1
6 marks Moderate -0.3
1 A curve has implicit equation \(y ^ { 2 } + 2 x \ln y = x ^ { 2 }\).
Verify that the point \(( 1,1 )\) lies on the curve, and find the gradient of the curve at this point.
OCR MEI C3 Q2
6 marks Standard +0.3
2 A curve has equation \(x ^ { 2 } + 2 y ^ { 2 } = 4 x\).
  1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
    [0pt]
  2. Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.]
OCR MEI C3 Q3
4 marks Moderate -0.3
3 Given that \(y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }\).
OCR MEI C3 Q4
8 marks Standard +0.8
4 Fig. 7 shows the curve \(x ^ { 3 } + y ^ { 3 } = 3 x y\). The point P is a turning point of the curve. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{09d318c7-27b9-43aa-b4a0-e32ea8bd53c5-1_593_531_1573_805} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }\).
  2. Hence find the exact \(x\)-coordinate of P .
OCR MEI C3 Q5
4 marks Standard +0.3
5 Find the gradient at the point \(( 0 , \ln 2 )\) on the curve with equation \(\mathrm { e } ^ { 2 y } = 5 - \mathrm { e } ^ { - x }\).
OCR MEI C3 Q6
4 marks Moderate -0.3
6 A curve is defined by the equation \(( x + y ) ^ { 2 } = 4 x\). The point \(( 1,1 )\) lies on this curve.
By differentiating implicitly, show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x + y } - 1\).
Hence verify that the curve has a stationary point at \(( 1,1 )\).
OCR MEI C3 Q7
6 marks Moderate -0.3
7 A curve is defined by the equation \(\sin 2 x + \cos y = \sqrt { 3 }\).
  1. Verify that the point \(\mathrm { P } \left( \frac { 1 } { 6 } \pi , \frac { 1 } { 6 } \pi \right)\) lies on the curve.
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point P .
OCR MEI C3 Q8
7 marks Moderate -0.3
8
  1. Given that \(y = \sqrt [ 3 ] { 1 + 3 x ^ { 2 } }\), use the chain rule to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
  2. Given that \(y ^ { 3 } = 1 + 3 x ^ { 2 }\), use implicit differentiation to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Show that this result is equivalent to the result in part (i).
OCR MEI C3 Q1
7 marks Moderate -0.3
1
  1. Differentiate \(\sqrt { 1 + 3 x ^ { 2 } }\).
  2. Hence show that the derivative of \(x \sqrt { 1 + 3 x ^ { 2 } }\) is \(\frac { 1 + 6 x ^ { 2 } } { \sqrt { 1 + 3 x ^ { 2 } } }\).
OCR MEI C3 Q2
7 marks Standard +0.3
2 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).
OCR MEI C3 Q3
18 marks Standard +0.3
3 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e3850377-bd1a-4e3c-8424-e3db7fd3c4db-2_1018_994_481_611} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the gradient of PR.
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
  3. Find the exact coordinates of the turning point Q .
  4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).
OCR MEI C3 Q4
7 marks Moderate -0.3
4 The equation of a curve is given by \(\mathrm { e } ^ { 2 y } = 1 + \sin x\).
  1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
  2. Find an expression for \(y\) in terms of \(x\), and differentiate it to verify the result in part (i).
OCR MEI C3 Q5
8 marks Standard +0.8
5 Fig. 6 shows the curve \(\mathrm { e } ^ { 2 y } = x ^ { 2 } + y\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e3850377-bd1a-4e3c-8424-e3db7fd3c4db-3_736_1331_893_459} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { 2 \mathrm { e } ^ { 2 y } - 1 }\).
  2. Hence find to 3 significant figures the coordinates of the point P , shown in Fig. 6, where the curve has infinite gradient.
OCR MEI C3 Q1
5 marks Standard +0.0
1 Given that \(x ^ { 2 } + x y + y ^ { 2 } = 12\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
OCR MEI C3 Q2
18 marks Standard +0.3
2 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).
  1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{ce82bfc4-90dd-4127-a11c-281cdcca70cf-1_621_934_1046_664} \captionsetup{labelformat=empty} \caption{Fig. 9}
    \end{figure}
  2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
    (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
    (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
  3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
  4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).
OCR MEI C3 Q3
8 marks Standard +0.3
3 Fig. 6 shows the triangle OAP , where O is the origin and A is the point \(( 0,3 )\). The point \(\mathrm { P } ( x , 0 )\) moves on the positive \(x\)-axis. The point \(\mathrm { Q } ( 0 , y )\) moves between O and A in such a way that \(\mathrm { AQ } + \mathrm { AP } = 6\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce82bfc4-90dd-4127-a11c-281cdcca70cf-2_488_848_514_640} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Write down the length AQ in terms of \(y\). Hence find AP in terms of \(y\), and show that $$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9$$
  2. Use this result to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }\).
  3. When \(x = 4\) and \(y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2\). Calculate \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) at this time.
OCR MEI C3 Q4
8 marks Standard +0.3
4 A curve has equation \(2 y ^ { 2 } + y = 9 x ^ { 2 } + 1\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point \(\mathrm { A } ( 1,2 )\).
  2. Find the coordinates of the points on the curve at which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).
OCR MEI C3 Q5
4 marks Moderate -0.8
5 Given that \(y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }\).
OCR MEI C3 Q6
5 marks Moderate -0.3
6 A curve is defined implicitly by the equation $$y ^ { 3 } = 2 x y + x ^ { 2 }$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( x + y ) } { 3 y ^ { 2 } - 2 x }\).
  2. Hence write down \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(x\) and \(y\).
OCR MEI C3 Q1
18 marks Standard +0.3
1 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{6555136d-0444-41f6-9063-21960352089d-1_705_864_525_635} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by $$\sqrt { } \text { for } x \geqslant 0 \text {. }$$
  2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
  3. Show that \(\int \left( \mathrm { e } ^ { x } 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } \quad 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } \quad 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
  4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).
OCR MEI C3 Q2
8 marks Standard +0.3
2 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]{6555136d-0444-41f6-9063-21960352089d-2_388_727_434_701} \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 Q3
19 marks Standard +0.3
3 The function \(f ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{6555136d-0444-41f6-9063-21960352089d-3_495_867_519_607} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure} (1) Show algebraically that the function is even. State how this property relates to the shape of the curve.
(ii) Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
(iii) Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\),
(iv) Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(g ( x )\).
Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
(v) Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the commection between this result and your answer to part (ii).
OCR MEI C3 Q1
18 marks Standard +0.3
1 Fig. 7 shows the curve \(y = _ { x - 1 }\). It has a minimum at the point P . The line \(l\) is an asymptote to the curve. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e0636807-d5bf-43c2-a484-68245e639cee-1_732_1049_467_547} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Write down the equation of the asymptote \(l\).
  2. Find the coordinates of P .
  3. Using the substitution \(u = x - 1\), show that the area of the region enclosed by the \(x\)-axis, the curve and the lines \(x = 2\) and \(x = 3\) is given by $$\int _ { 1 } ^ { 2 } \left( u + 2 + \frac { 4 } { u } \right) \mathrm { d } u$$ Evaluate this area exactly.
  4. Another curve is defined by the equation \(\mathrm { e } ^ { y } = \frac { x ^ { 2 } + 3 } { x - 1 }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\) by differentiating implicitly. Hence find the gradient of this curve at the point where \(x = 2\).
OCR MEI C3 Q2
8 marks Standard +0.3
2 Fig. 7 shows the curve defined implicitly by the equation $$y ^ { 2 } + y = x ^ { 9 } + 2 x$$ together with the line \(x = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e0636807-d5bf-43c2-a484-68245e639cee-2_462_385_657_858} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} Not to scale Find the coordinates of the points of intersection of the line and the curve.
Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.
OCR MEI C3 Q3
18 marks Standard +0.3
3 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]{e0636807-d5bf-43c2-a484-68245e639cee-3_476_674_498_708} \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.