Find gradient at a point - given gradient condition

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\includegraphics[max width=\textwidth, alt={}, center]{76371b0f-0145-4cc4-a147-27bcd749816a-3_451_933_1777_605} The diagram shows the curve \(y = x ^ { 2 } \mathrm { e } ^ { - x }\).

1. Show that the area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 3\) is equal to \(2 - \frac { 17 } { \mathrm { e } ^ { 3 } }\).
2. Find the \(x\)-coordinate of the maximum point \(M\) on the curve.
3. Find the \(x\)-coordinate of the point \(P\) at which the tangent to the curve passes through the origin.

5 A curve has equation \(y = \frac { 2 } { 3 } \ln \left( 1 + 3 \cos ^ { 2 } x \right)\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\tan x\).
2. Hence find the \(x\)-coordinate of the point on the curve where the gradient is - 1 . Give your answer correct to 3 significant figures.

4 Find the exact coordinates of the point on the curve \(y = \frac { x } { 1 + \ln x }\) at which the gradient of the tangent is equal to \(\frac { 1 } { 4 }\).

3 A curve has equation \(y = \frac { 3 + 2 \ln x } { 1 + \ln x }\). Find the exact gradient of the curve at the point for which \(y = 4\).

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\includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_476_608_287_756} The diagram shows the curve \(y = ( \ln x ) ^ { 2 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. The point \(P\) on the curve is the point at which the gradient takes its maximum value. Show that the tangent at \(P\) passes through the point \(( 0 , - 1 )\).

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] \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.

2 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] \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. END OF QUESTION PAPER

9 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] \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. \section*{END OF QUESTION PAPER} \section*{OCR \(^ { \text {N } }\)}

9 A curve has equation \(y = \mathrm { e } ^ { 2 x }\)
Find the coordinates of the point on the curve where the gradient of the curve is \(\frac { 1 } { 2 }\) Give your answer in an exact form.
[0pt] [5 marks]
David has been investigating the population of rabbits on an island during a three-year period. Based on data that he has collected, David decides to model the population of rabbits, \(R\), by the formula $$R = 50 \mathrm { e } ^ { 0.5 t }$$ where \(t\) is the time in years after 1 January 2016.

5 A curve has equation \(y = 3 \mathrm { e } ^ { 2 x }\) Find the gradient of the curve at the point where \(y = 10\)