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

A curve has equation \(y = (x + 2)^2(2x - 3)\).

1. Sketch the curve, giving the coordinates of all points of intersection with the axes. [3]
2. Find an equation of the tangent to the curve at the point where \(x = -1\). Give your answer in the form \(ax + by + c = 0\). [9]

The curve \(C\) with equation \(y = p + qe^x\), where \(p\) and \(q\) are constants, passes through the point \((0, 2)\). At the point \(P\) (ln 2, \(p + 2q\)) on \(C\), the gradient is 5.

1. Find the value of \(p\) and the value of \(q\). [5]

The normal to \(C\) at \(P\) crosses the \(x\)-axis at \(L\) and the \(y\)-axis at \(M\).

1. Show that the area of \(\triangle OLM\), where \(O\) is the origin, is approximately 53.8 [5]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = \text{f}(x)\), where $$\text{f}(x) = 10 + \ln(3x) - \frac{1}{2}e^x, \quad 0.1 \leq x \leq 3.3.$$ Given that f(k) = 0,

1. show, by calculation, that \(3.1 < k < 3.2\). [2]
2. Find f'(x). [3]

The tangent to the graph at \(x = 1\) intersects the \(y\)-axis at the point \(P\).

1. 1. Find an equation of this tangent.
   2. Find the exact \(y\)-coordinate of \(P\), giving your answer in the form \(a + \ln b\). [5]

\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation \(y = \text{f}(x)\), where $$\text{f}(x) = 0.5e^x - x^2.$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).

1. Find an equation of the tangent to \(C\) at \(A\). [4]

The \(x\)-coordinate of \(B\) is approximately 2.15. A more exact estimate is to be made of this coordinate using iterations \(x_{n+1} = \ln g(x_n)\).

1. Show that a possible form for \(g(x)\) is \(g(x) = 4x\). [3]
2. Using \(x_{n+1} = \ln 4x_n\), with \(x_0 = 2.15\), calculate \(x_1\), \(x_2\) and \(x_3\). Give the value of \(x_3\) to 4 decimal places. [2]

The curve \(C\) has equation \(y = 2e^x + 3x^2 + 2\). The point \(A\) with coordinates \((0, 4)\) lies on \(C\). Find the equation of the tangent to \(C\) at \(A\). [5]

f(x) = \(x + \frac{e^x}{5}\), \(x \in \mathbb{R}\).

1. Find f'(x). [2]

The curve \(C\), with equation \(y = \)f(x), crosses the \(y\)-axis at the point \(A\).

1. Find an equation for the tangent to \(C\) at \(A\). [3]
2. Complete the table, giving the values of \(\sqrt{x + \frac{e^x}{5}}\) to 2 decimal places.

\(x\)	0	0.5	1	1.5	2
\(\sqrt{x + \frac{e^x}{5}}\)	0.45	0.91

[2]

The mass, \(m\) grams, of a substance at time \(t\) years is given by the formula $$m = 180e^{-0.017t}.$$

1. Find the value of \(t\) for which the mass is 25 grams. [3]
2. Find the rate at which the mass is decreasing when \(t = 55\). [3]

The mass, \(m\) grams, of a substance is increasing exponentially so that the mass at time \(t\) hours is given by $$m = 250e^{0.02t}.$$

1. Find the time taken for the mass to increase to twice its initial value, and deduce the time taken for the mass to increase to 8 times its initial value. [3]
2. Find the rate at which the mass is increasing at the instant when the mass is 400 grams. [3]

\includegraphics{figure_8} Fig. 8 shows the curve \(y = f(x)\), where \(f(x) = \frac{1}{e^x + e^{-x} + 2}\).

1. Show algebraically that \(f(x)\) is an even function, and state how this property relates to the curve \(y = f(x)\). [3]
2. Find \(f'(x)\). [3]
3. Show that \(f(x) = \frac{e^x}{(e^x + 1)^2}\). [2]
4. Hence, using the substitution \(u = e^x + 1\), or otherwise, find the exact area enclosed by the curve \(y = f(x)\), the \(x\)-axis, and the lines \(x = 0\) and \(x = 1\). [5]
5. Show that there is only one point of intersection of the curves \(y = f(x)\) and \(y = \frac{1}{4}e^x\), and find its coordinates. [5]

Fig. 9 shows the curve \(y = xe^{-2x}\) together with the straight line \(y = mx\), 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. \includegraphics{figure_9}

1. Show that the \(x\)-coordinate of P is \(-\frac{1}{2}\ln m\). [3]
2. Find, in terms of \(m\), the gradient of the tangent to the curve at P. [4]

You are given that OP and this tangent are equally inclined to the \(x\)-axis.

1. Show that \(m = e^{-2}\), and find the exact coordinates of P. [4]
2. Find the exact area of the shaded region between the line OP and the curve. [7]

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = e^{2x} + k e^{-2x}\) and \(k\) is a constant greater than 1. The curve crosses the \(y\)-axis at P and has a turning point Q. \includegraphics{figure_9}

1. Find the \(y\)-coordinate of P in terms of \(k\). [1]
2. Show that the \(x\)-coordinate of Q is \(\frac{1}{4}\ln k\), and find the \(y\)-coordinate in its simplest form. [5]
3. Find, in terms of \(k\), the area of the region enclosed by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\ln k\). Give your answer in the form \(ak + b\). [4]

The function \(g(x)\) is defined by \(g(x) = f(x + \frac{1}{4}\ln k)\).

1. 1. Show that \(g(x) = \sqrt{k}(e^{2x} + e^{-2x})\). [3]
   2. Hence show that \(g(x)\) is an even function. [2]
   3. Deduce, with reasons, a geometrical property of the curve \(y = f(x)\). [3]

Fig. 8 shows part of the curve \(y = \text{f}(x)\), where \(\text{f}(x) = e^{-\frac{1}{5}x} \sin x\), for all \(x\). \includegraphics{figure_8}

1. Sketch the graphs of (A) \(y = \text{f}(2x)\), (B) \(y = \text{f}(x + \pi)\). [4]
2. Show that the \(x\)-coordinate of the turning point P satisfies the equation \(\tan x = 5\). Hence find the coordinates of P. [6]
3. Show that \(\text{f}(x + \pi) = -e^{-\frac{1}{5}\pi}\text{f}(x)\). Hence, using the substitution \(u = x - \pi\), show that $$\int_{\pi}^{2\pi} \text{f}(x)\,dx = -e^{-\frac{1}{5}\pi} \int_{0}^{\pi} \text{f}(u)\,du.$$ Interpret this result graphically. [You should not attempt to integrate f(x).] [8]

Fig. 9 shows the curve \(y = f(x)\), where $$f(x) = (e^x - 2)^2 - 1, x \in \mathbb{R}.$$ The curve crosses the x-axis at O and P, and has a turning point at Q. \includegraphics{figure_9}

1. Find the exact x-coordinate of P. [2]
2. Show that the x-coordinate of Q is \(\ln 2\) and find its y-coordinate. [4]
3. Find the exact area of the region enclosed by the curve and the x-axis. [5]

The domain of f(x) is now restricted to \(x \geqslant \ln 2\).

1. Find the inverse function \(f^{-1}(x)\). Write down its domain and range, and sketch its graph on the copy of Fig. 9. [7]

A curve has the equation \(y = e^{-2x}\) At point \(P\) on the curve the tangent is parallel to the line \(x + 8y = 5\) Find the coordinates of \(P\) stating your answer in the form \((\ln p, q)\), where \(p\) and \(q\) are rational. [7 marks]

A curve has equation \(y = e^{2x}\) 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. [5 marks]

Given \(y = e^{kx}\), where \(k\) is a constant, find \(\frac{dy}{dx}\) Circle your answer. [1 mark] $$\frac{dy}{dx} = e^{kx} \quad \frac{dy}{dx} = ke^{kx} \quad \frac{dy}{dx} = kxe^{x-1} \quad \frac{dy}{dx} = \frac{e^{kx}}{k}$$

A curve, C, has equation $$y = \frac{e^{3x-5}}{x^2}$$ Show that C has exactly one stationary point. Fully justify your answer. [7 marks]

A function f has domain \(\mathbb{R}\) and range \(\{y \in \mathbb{R} : y \geq c\}\) The graph of \(y = f(x)\) is shown. \includegraphics{figure_2} The gradient of the curve at the point \((x, y)\) is given by \(\frac{dy}{dx} = (x - 1)e^x\) Find an expression for f(x). Fully justify your answer. [8 marks]

A particle is moving in a straight line through the origin \(O\) The displacement of the particle, \(r\) metres, from \(O\), at time \(t\) seconds is given by $$r = p + 2t - qe^{-0.2t}$$ where \(p\) and \(q\) are constants. When \(t = 3\), the acceleration of the particle is \(-1.8\) m s\(^{-2}\)

1. Show that \(q \approx 82\) [5 marks]
2. The particle has an initial displacement of 5 metres. Find the value of \(p\) Give your answer to two significant figures. [2 marks]

At time \(t = 0\), a parachutist jumps out of an airplane that is travelling horizontally. The velocity, \(\mathbf{v}\) m s\(^{-1}\), of the parachutist at time \(t\) seconds is given by: $$\mathbf{v} = (40e^{-0.2t})\mathbf{i} + 50(e^{-0.2t} - 1)\mathbf{j}$$ The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. Assume that the parachutist is at the origin when \(t = 0\) Model the parachutist as a particle.

1. Find an expression for the position vector of the parachutist at time \(t\). [4 marks]
2. The parachutist opens her parachute when she has travelled 100 metres horizontally. Find the vertical displacement of the parachutist from the origin when she opens her parachute. [4 marks]
3. Carefully, explaining the steps that you take, deduce the value of \(g\) used in the formulation of this model. [3 marks]

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