1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

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]

The function f(x) is defined by $$f(x) = 1 + 2\sin 3x, \quad -\frac{\pi}{6} \leqslant x \leqslant \frac{\pi}{6}.$$ You are given that this function has an inverse, \(f^{-1}(x)\). Find \(f^{-1}(x)\) and its domain. [6]

1. Use integration by parts to show that $$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx = \frac{1}{4}\pi - \frac{1}{2} \ln 2.$$ [6]

\includegraphics{figure_1} The finite region \(R\), bounded by the equation \(y = x^{\frac{1}{2}} \sec x\), the line \(x = \frac{\pi}{4}\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Find the volume of the solid of revolution generated. [2]
2. Find the gradient of the curve with equation \(y = x^{\frac{1}{2}} \sec x\) at the point where \(x = \frac{\pi}{4}\). [3]

Given that \(\cos\text{ec}^2\theta - \cot\theta = 3\), show that \(\cot^2\theta - \cot\theta - 2 = 0\). Hence solve the equation \(\cos\text{ec}^2\theta - \cot\theta = 3\) for \(0° \leq \theta \leq 180°\). [6]

\includegraphics{figure_1} Figure 1 shows the curve with equation \(y = 2\sin x + \cosec x\), \(0 < x < \pi\). The shaded region bounded by the curve, the \(x\)-axis and the lines \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\) is rotated through \(360°\) about the \(x\)-axis. Show that the volume of the solid formed is \(\frac{1}{2}\pi(4\pi + 3\sqrt{3})\). [8]

Given that \(\cosec^2 \theta - \cot \theta = 3\), show that \(\cot^2 \theta - \cot \theta - 2 = 0\). Hence solve the equation \(\cosec^2 \theta - \cot \theta = 3\) for \(0° \leqslant \theta \leqslant 180°\). [6]

\includegraphics{figure_1} Figure 1 shows a sketch of part of the curve with equation $$y = \sin (\cos x).$$ The curve cuts the \(x\)-axis at the points \(A\) and \(C\) and the \(y\)-axis at the point \(B\).

1. Find the coordinates of the points \(A\), \(B\) and \(C\). [3]
2. Prove that \(B\) is a stationary point. [2]

Given that the region \(OCB\) is convex,

1. show that, for \(0 \leq x \leq \frac{\pi}{2}\), $$\sin (\cos x) \leq \cos x$$ and $$(1 - \frac{2}{\pi} x) \sin 1 \leq \sin (\cos x)$$ and state in each case the value or values of \(x\) for which equality is achieved. [6]
2. Hence show that $$\frac{\pi}{4} \sin 1 < \int_0^{\frac{\pi}{2}} \sin(\cos x) \, dx < 1.$$ [4]

\includegraphics{figure_4} The diagram shows the part of the curve \(y = 3x \sin 2x\) for which \(0 \leqslant x \leqslant \frac{1}{2}\pi\). The maximum point on the curve is denoted by \(P\).

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(\tan 2x + 2x = 0\). [3]
2. Use the Newton-Raphson method, with a suitable initial value, to find the \(x\)-coordinate of \(P\), giving your answer correct to 4 decimal places. Show the result of each iteration. [4]
3. The trapezium rule, with four strips of equal width, is used to find an approximation to $$\int_0^{\frac{1}{2}\pi} 3x \sin 2x \, dx.$$ Show that the result can be expressed as \(k\pi^2(\sqrt{2} + 1)\), where \(k\) is a rational number to be determined. [4]
4. 1. Evaluate \(\int_0^{\frac{1}{2}\pi} 3x \sin 2x \, dx\). [1]
   2. Hence determine whether using the trapezium rule with four strips of equal width gives an under- or over-estimate for the area of the region enclosed by the curve \(y = 3x \sin 2x\) and the \(x\)-axis for \(0 \leqslant x \leqslant \frac{1}{2}\pi\). [1]
   3. Explain briefly why it is not easy to tell from the diagram alone whether the trapezium rule with four strips of equal width gives an under- or over-estimate for the area of the region in this case. [1]

In this question you must show detailed reasoning.

1. Show that the equation \(m \sec \theta + 3 \cos \theta = 4 \sin \theta\) can be expressed in the form $$m \tan^2 \theta - 4 \tan \theta + (m + 3) = 0.$$ [3]
2. It is given that there is only one value of \(\theta\), for \(0 < \theta < \pi\), satisfying the equation \(m \sec \theta + 3 \cos \theta = 4 \sin \theta\). Given also that \(m\) is a negative integer, find this value of \(\theta\), correct to 3 significant figures. [5]

Prove the identity \(\cot^2 \theta - \cos^2 \theta = \cot^2 \theta \cos^2 \theta\) [3 marks]

The height \(x\) metres, of a column of water in a fountain display satisfies the differential equation \(\frac{dx}{dt} = \frac{8\sin 2t}{3\sqrt{x}}\), where \(t\) is the time in seconds after the display begins.

1. Solve the differential equation, given that initially the column of water has zero height. Express your answer in the form \(x = f(t)\) [7 marks]
2. Find the maximum height of the column of water, giving your answer to the nearest cm. [1 mark]

Show that, for small values of \(x\), the graph of \(y = 5 + 4\sin\frac{x}{2} + 12\tan\frac{x}{3}\) can be approximated by a straight line. [3 marks]

Solve the equation $$2\tan^2\theta + 2\tan\theta - \sec^2\theta = 2,$$ for values of \(\theta\) between \(0°\) and \(360°\). [5]

Find a small positive value of \(x\) which is an approximate solution of the equation. $$\cos x - 4\sin x = x^2.$$ [4]

solve, for \(0° < \theta < 360°\), the equation $$2 \tan^2 \theta - \frac{1}{\cos \theta} = 4.$$ [4]

The curve \(C\) has equation $$r = a(p + 2\cos\theta) \quad 0 \leqslant \theta < 2\pi$$ where \(a\) and \(p\) are positive constants and \(p > 2\) There are exactly four points on \(C\) where the tangent is perpendicular to the initial line.

1. Show that the range of possible values for \(p\) is $$2 < p < 4$$ [5]
2. Sketch the curve with equation $$r = a(3 + 2\cos\theta) \quad 0 \leqslant \theta < 2\pi \quad \text{where } a > 0$$ [1]

John digs a hole in his garden in order to make a pond. The pond has a uniform horizontal cross section that is modelled by the curve with equation $$r = 20(3 + 2\cos\theta) \quad 0 \leqslant \theta < 2\pi$$ where \(r\) is measured in centimetres. The depth of the pond is 90 centimetres. Water flows through a hosepipe into the pond at a rate of 50 litres per minute. Given that the pond is initially empty,

1. determine how long it will take to completely fill the pond with water using the hosepipe, according to the model. Give your answer to the nearest minute. [7]

In this question you must show detailed reasoning. Fig. 4 shows the region bounded by the curve \(y = \sec \frac{1}{2}x\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{2}\pi\). \includegraphics{figure_4} This region is rotated through \(2\pi\) radians about the \(x\)-axis. Find, in exact form, the volume of the solid of revolution generated. [3]