1.07j Differentiate exponentials: e^(kx) and a^(kx)

6 A curve has equation \(y = \frac { 9 \mathrm { e } ^ { 2 x } + 16 } { \mathrm { e } ^ { x } - 1 }\).

1. Show that the \(x\)-coordinate of any stationary point on the curve satisfies the equation $$\mathrm { e } ^ { x } \left( 3 \mathrm { e } ^ { x } - 8 \right) \left( 3 \mathrm { e } ^ { x } + 2 \right) = 0$$
2. Hence show that the curve has only one stationary point and find its exact coordinates.

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 \(C\).

5 A curve has equation \(\mathrm { y } = \frac { 1 + \mathrm { e } ^ { 2 \mathrm { x } } } { 1 + 3 \mathrm { x } }\). The curve has exactly one stationary point \(P\).

1. Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 1 } { 6 } + \frac { 1 } { 2 } \mathrm { e } ^ { - 2 x }\).
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.35 and 0.45 .
3. Use an iterative formula based on the equation in part (a) to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{971a1d8d-a82e-4a3a-b72d-3c147e4f30bb-10_451_647_258_699} The diagram shows the curve with equation \(\mathrm { y } = \sqrt { \sin 2 \mathrm { x } + \sin ^ { 2 } 2 \mathrm { x } }\) for \(0 \leqslant x \leqslant \frac { 1 } { 6 } \pi\). The shaded region is bounded by the curve and the straight lines \(x = \frac { 1 } { 6 } \pi\) and \(y = 0\).

3 \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-04_776_483_310_769} The diagram shows the curve with equation \(y = 8 \mathrm { e } ^ { - x } - \mathrm { e } ^ { 2 x }\). The curve crosses the \(y\)-axis at the point \(A\) and the \(x\)-axis at the point \(B\). The shaded region is bounded by the curve and the two axes.

1. Find the gradient of the curve at \(A\). \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-04_2715_35_141_2011}
2. Show that the \(x\)-coordinate of \(B\) is \(\ln 2\) and hence find the area of the shaded region.

6 A curve has equation \(y = x ^ { 3 } \mathrm { e } ^ { 0.2 x }\) where \(x \geqslant 0\). At the point \(P\) on the curve, the gradient of the curve is 15 .

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \sqrt { \frac { 75 \mathrm { e } ^ { - 0.2 x } } { 15 + x } }\).
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(P\) lies between 1.7 and 1.8.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4 The curve with equation \(y = x \mathrm { e } ^ { 2 x } + 5 \mathrm { e } ^ { - x }\) has a minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { 1 } { 3 } \ln 5 - \frac { 1 } { 3 } \ln ( 1 + 2 x )\).
2. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(M\) correct to 3 significant figures. Use an initial value of 0.35 and give the result of each iteration to 5 significant figures.

3 \includegraphics[max width=\textwidth, alt={}, center]{b104e2a7-06c8-4e2e-a4f9-5095ad56897a-04_652_392_274_872} The diagram shows the curve with equation \(y = 6 \mathrm { e } ^ { - \frac { 1 } { 2 } x }\). The points on the curve with \(x\)-coordinates 0 and 2 are denoted by \(A\) and \(B\) respectively. The shaded region is enclosed by the curve, the line through \(A\) parallel to the \(x\)-axis and the line through \(B\) parallel to the \(y\)-axis.

1. Find the exact gradient of the curve at \(B\).
2. Find the exact area of the shaded region.

5 \includegraphics[max width=\textwidth, alt={}, center]{beb8df77-e091-4248-812b-20e885c42e37-3_528_757_251_694} The diagram shows the curve \(y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) can be written in the form \(\ln a\), where the value of \(a\) is to be stated.
2. Find the exact value of the area of the region enclosed by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).

5 \includegraphics[max width=\textwidth, alt={}, center]{0a45a806-007f-4840-85e7-16d4c1a2c599-3_528_757_251_694} The diagram shows the curve \(y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) can be written in the form \(\ln a\), where the value of \(a\) is to be stated.
2. Find the exact value of the area of the region enclosed by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).

7

1. Find the exact area of the region bounded by the curve \(y = 1 + \mathrm { e } ^ { 2 x - 1 }\), the \(x\)-axis and the lines \(x = \frac { 1 } { 2 }\) and \(x = 2\).
2. \includegraphics[max width=\textwidth, alt={}, center]{e3ee4932-8219-4332-9cd2-e7f835522469-3_469_719_397_753} The diagram shows the curve \(y = \frac { \mathrm { e } ^ { 2 x } } { \sin 2 x }\) for \(0 < x < \frac { 1 } { 2 } \pi\), and its minimum point \(M\). Find the exact \(x\)-coordinate of \(M\).

2 Find the gradient of each of the following curves at the point for which \(x = 0\).

1. \(y = 3 \sin x + \tan 2 x\)
2. \(y = \frac { 6 } { 1 + \mathrm { e } ^ { 2 x } }\)

1 Find the gradient of the curve $$y = 3 e ^ { 4 x } - 6 \ln ( 2 x + 3 )$$ at the point for which \(x = 0\).

4 Find the equation of the tangent to the curve \(y = \frac { \mathrm { e } ^ { 4 x } } { 2 x + 3 }\) at the point on the curve for which \(x = 0\). Give your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers.

9 \includegraphics[max width=\textwidth, alt={}, center]{b2136f5d-0d66-4524-bb76-fcc4cb59150c-3_639_387_1749_879} The diagram shows the curve \(y = \mathrm { e } ^ { 2 \sin x } \cos x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Using the substitution \(u = \sin x\), find the exact value of the area of the shaded region bounded by the curve and the axes.
2. Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.

4 The curve \(y = \mathrm { e } ^ { x } + 4 \mathrm { e } ^ { - 2 x }\) has one stationary point.

1. Find the \(x\)-coordinate of this point.
2. Determine whether the stationary point is a maximum or a minimum point.

6 \includegraphics[max width=\textwidth, alt={}, center]{79efa364-da5a-4888-85a9-dc4de1e0908e-3_543_825_287_660} The diagram shows the curve \(y = ( 3 - x ) \mathrm { e } ^ { - 2 x }\) and its minimum point \(M\). The curve intersects the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).

1. Calculate the \(x\)-coordinate of \(M\).
2. Find the area of the region bounded by \(O A , O B\) and the curve, giving your answer in terms of e.

3 The curve with equation \(y = 6 \mathrm { e } ^ { x } - \mathrm { e } ^ { 3 x }\) has one stationary point.

1. Find the \(x\)-coordinate of this point.
2. Determine whether this point is a maximum or a minimum point.

3 Use integration by parts to show that $$\int _ { 2 } ^ { 4 } \ln x \mathrm {~d} x = 6 \ln 2 - 2$$

3 The curve \(y = \frac { \mathrm { e } ^ { x } } { \cos x }\), for \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), has one stationary point. Find the \(x\)-coordinate of this point.

4 A curve has equation \(y = \mathrm { e } ^ { - 3 x } \tan x\). Find the \(x\)-coordinates of the stationary points on the curve in the interval \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\). Give your answers correct to 3 decimal places.

2 The equation of a curve is \(y = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). Show that the gradient of the curve at the point for which \(x = \ln 3\) is \(\frac { 9 } { 50 }\).

2 Prove by mathematical induction that, for all positive integers \(n\), $$\frac { d ^ { n } } { d x ^ { n } } \left( x ^ { 2 } e ^ { x } \right) = \left( x ^ { 2 } + 2 n x + n ( n - 1 ) \right) e ^ { x }$$

7 \includegraphics[max width=\textwidth, alt={}, center]{25dffd43-9456-449b-be77-8402109ee603-3_608_672_283_733} The diagram shows the curve \(y = 2 \mathrm { e } ^ { x } + 3 \mathrm { e } ^ { - 2 x }\). The curve cuts the \(y\)-axis at \(A\).

1. Write down the coordinates of \(A\).
2. Find the equation of the tangent to the curve at \(A\), and state the coordinates of the point where this tangent meets the \(x\)-axis.
3. Calculate the area of the region bounded by the curve and by the lines \(x = 0 , y = 0\) and \(x = 1\), giving your answer correct to 2 significant figures.

6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { - x }\). The point \(( 0,1 )\) lies on the curve.

1. Find the equation of the curve.
2. The curve has one stationary point. Find the \(x\)-coordinate of this point and determine whether it is a maximum or a minimum point.

8 \includegraphics[max width=\textwidth, alt={}, center]{8f815127-61b2-4a7f-8687-747950ea6597-3_693_1061_262_541} The diagram shows the curve \(y = x ^ { 2 } \mathrm { e } ^ { - x }\) and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Show that the tangent to the curve at the point where \(x = 1\) passes through the origin.
3. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { 1 } ^ { 3 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.