Find gradient at a point - at special curve features

Find the gradient at points defined by curve features (e.g., where curve crosses x-axis, at point P shown in diagram, at intersection points). Requires first finding the x-coordinate before evaluating the derivative.

10 questions · Standard +0.5

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CAIE P2 2021 June Q6
8 marks Standard +0.3
6
\includegraphics[max width=\textwidth, alt={}, center]{388d7076-636c-417d-84cb-e6e2a3e9a6a0-08_451_1086_260_525} The diagram shows the curve with equation $$y = ( \ln x ) ^ { 2 } - 2 \ln x$$ The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a minimum point \(M\).
  1. Find the exact value of the gradient of the curve at each of the points \(A\) and \(B\).
  2. Find the exact \(x\)-coordinate of \(M\).
CAIE P2 2021 June Q6
8 marks Moderate -0.3
6
\includegraphics[max width=\textwidth, alt={}, center]{61df367d-741f-4906-8ab9-2f32e8711aa6-08_451_1086_260_525} The diagram shows the curve with equation $$y = ( \ln x ) ^ { 2 } - 2 \ln x$$ The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a minimum point \(M\).
  1. Find the exact value of the gradient of the curve at each of the points \(A\) and \(B\).
  2. Find the exact \(x\)-coordinate of \(M\).
CAIE P2 2023 June Q5
9 marks Standard +0.3
5
\includegraphics[max width=\textwidth, alt={}, center]{4ce3208e-8ceb-4848-a9c7-fcda166319f4-06_526_947_276_591} The diagram shows the curve with equation \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \left( x ^ { 2 } - 5 x + 4 \right)\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a maximum at the point \(C\).
  1. Find the exact gradient of the curve at \(B\).
  2. Find the exact coordinates of \(C\).
CAIE P2 2023 June Q5
9 marks Standard +0.3
5
\includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-06_526_947_276_591} The diagram shows the curve with equation \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \left( x ^ { 2 } - 5 x + 4 \right)\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a maximum at the point \(C\).
  1. Find the exact gradient of the curve at \(B\).
  2. Find the exact coordinates of \(C\).
CAIE P2 2022 November Q7
9 marks Standard +0.3
7
\includegraphics[max width=\textwidth, alt={}, center]{389df578-e7a7-4d19-9416-5e580d107717-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).
  1. Find the gradient of the curve at \(A\).
  2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
  3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.
CAIE P2 2022 November Q7
9 marks Standard +0.3
7
\includegraphics[max width=\textwidth, alt={}, center]{1cd04df5-3fe3-4573-b880-d49262afd16a-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).
  1. Find the gradient of the curve at \(A\).
  2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
  3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.
CAIE P2 2017 June Q8
10 marks Standard +0.3
8
\includegraphics[max width=\textwidth, alt={}, center]{6295873e-7db4-4e7e-8dcd-912ad9c41675-10_643_414_260_863} The diagram shows the curve with equation $$y = 3 x ^ { 2 } \ln \left( \frac { 1 } { 6 } x \right) .$$ The curve crosses the \(x\)-axis at the point \(P\) and has a minimum point \(M\).
  1. Find the gradient of the curve at the point \(P\).
  2. Find the exact coordinates of the point \(M\).
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
18 marks Challenging +1.2
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 2012 January Q9
18 marks Challenging +1.2
9 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]{8b8958be-0ebc-4f72-ac3f-c16a8ec9e4ab-5_552_636_470_715} \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 \(y = x\).]