Show gradient condition leads to equation

4 The equation of a curve is \(y = \frac { 1 + \mathrm { e } ^ { - x } } { 1 - \mathrm { e } ^ { - x } }\), for \(x > 0\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is always negative.
2. The gradient of the curve is equal to - 1 when \(x = a\). Show that \(a\) satisfies the equation \(\mathrm { e } ^ { 2 a } - 4 \mathrm { e } ^ { a } + 1 = 0\). Hence find the exact value of \(a\).

8. A scientist is studying a population of fish in a lake. The number of fish, \(N\), in the population, \(t\) years after the start of the study, is modelled by the equation $$N = \frac { 600 \mathrm { e } ^ { 0.3 t } } { 2 + \mathrm { e } ^ { 0.3 t } } \quad t \geqslant 0$$ Use the equation of the model to answer parts (a), (b), (c), (d) and (e).

1. Find the number of fish in the lake at the start of the study.
2. Find the upper limit to the number of fish in the lake.
3. Find the time, after the start of the study, when there are predicted to be 500 fish in the lake. Give your answer in years and months to the nearest month.
4. Show that $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { A \mathrm { e } ^ { 0.3 t } } { \left( 2 + \mathrm { e } ^ { 0.3 t } \right) ^ { 2 } }$$ where \(A\) is a constant to be found. Given that when \(t = T , \frac { \mathrm {~d} N } { \mathrm {~d} t } = 8\)
5. find the value of \(T\) to one decimal place.
   (Solutions relying entirely on calculator technology are not acceptable.) \includegraphics[max width=\textwidth, alt={}, center]{76205772-5395-4ab2-96f9-ad9803b8388c-27_2644_1840_118_111}

6. (i) The curve \(C _ { 1 }\) has equation $$y = 3 \ln \left( x ^ { 2 } - 5 \right) - 4 x ^ { 2 } + 15 \quad x > \sqrt { 5 }$$ Show that \(C _ { 1 }\) has a stationary point at \(x = \frac { \sqrt { p } } { 2 }\) where \(p\) is a constant to be found.
(ii) A different curve \(C _ { 2 }\) has equation $$y = 4 x - 12 \sin ^ { 2 } x$$

1. Show that, for this curve, $$\frac { \mathrm { d } y } { \mathrm {~d} x } = A + B \sin 2 x$$ where \(A\) and \(B\) are constants to be found.
2. Hence, state the maximum gradient of this curve.

\includegraphics{figure_5} Figure 5 shows a sketch of the curve with equation \(y = f(x)\), where $$f(x) = \frac{4\sin 2x}{e^{\sqrt{2}x-1}}, \quad 0 \leq x \leq \pi$$ The curve has a maximum turning point at \(P\) and a minimum turning point at \(Q\) as shown in Figure 5.

1. Show that the \(x\) coordinates of point \(P\) and point \(Q\) are solutions of the equation $$\tan 2x = \sqrt{2}$$ [4]
2. Using your answer to part (a), find the \(x\)-coordinate of the minimum turning point on the curve with equation $$y = 3 - 2f(x)$$ [2]