SPS SPS SM Pure (SPS SM Pure) 2021 May

1.

1. For a small angle \(\theta\), where \(\theta\) is in radians, show that \(2 \cos \theta + ( 1 - \tan \theta ) ^ { 2 } \approx 3 - 2 \theta\).
2. Hence determine an approximate solution to \(2 \cos \theta + ( 1 - \tan \theta ) ^ { 2 } = 28 \sin \theta\).
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2. Solve the equation \(| 2 x - 1 | = | x + 3 |\).
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3. Solve the equation \(2 ^ { 4 x - 1 } = 3 ^ { 5 - 2 x }\), giving your answer in the form \(x = \frac { \log _ { 10 } a } { \log _ { 10 } b }\).
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4. A sequence of transformations maps the curve \(y = \mathrm { e } ^ { x }\) to the curve \(y = \mathrm { e } ^ { 2 x + 3 }\). Give details of these transformations.
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5. A curve has equation \(x ^ { 3 } - 3 x ^ { 2 } y + y ^ { 2 } + 1 = 0\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x y - 3 x ^ { 2 } } { 2 y - 3 x ^ { 2 } }\).
2. Find the equation of the normal to the curve at the point \(( 1,2 )\).
   [0pt] [BLANK PAGE] \section*{6. In this question you must show detailed reasoning.} A circle touches the lines \(y = \frac { 1 } { 2 } x\) and \(y = 2 x\) at \(( 6,3 )\) and \(( 3,6 )\) respectively.
   \includegraphics[max width=\textwidth, alt={}, center]{f9e0bca6-c2a3-4868-b38b-942ceabd4992-14_515_524_338_790} Find the equation of the circle.
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7. It is given that there is exactly one value of \(x\), where \(0 < x < \pi\), that satisfies the equation $$3 \tan 2 x - 8 \tan x = 4$$

1. Show that \(t = \sqrt [ 3 ] { \frac { 1 } { 2 } + \frac { 1 } { 4 } t - \frac { 1 } { 2 } t ^ { 2 } }\), where \(t = \tan x\).
2. Show by calculation that the value of \(t\) satisfying the equation in part (i) lies between 0.7 and 0.8 .
3. Use an iterative process based on the equation in part (i) to find the value of \(t\) correct to 4 significant figures. Use a starting value of 0.75 and show the result of each iteration.
4. Solve the equation \(3 \tan 4 y - 8 \tan 2 y = 4\) for \(0 < y < \frac { 1 } { 2 } \pi\).
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8. Find the general solution of the differential equation $$\left( 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 20 x - 35 )$$ Give your answer in the form \(y = \mathrm { f } ( x )\).
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9. (i) Show that the two non-stationary points of inflection on the curve \(y = \ln \left( 1 + 4 x ^ { 2 } \right)\) are at \(x = \pm \frac { 1 } { 2 }\).
\includegraphics[max width=\textwidth, alt={}, center]{f9e0bca6-c2a3-4868-b38b-942ceabd4992-20_492_1064_237_513} The diagram shows the curve \(y = \ln \left( 1 + 4 x ^ { 2 } \right)\). The shaded region is bounded by the curve and a line parallel to the \(x\)-axis which meets the curve where \(x = \frac { 1 } { 2 }\) and \(x = - \frac { 1 } { 2 }\).
(ii) Show that the area of the shaded region is given by $$\int _ { 0 } ^ { \ln 2 } \sqrt { \mathrm { e } ^ { y } - 1 } \mathrm {~d} y$$ (iii) Show that the substitution \(\mathrm { e } ^ { y } = \sec ^ { 2 } \theta\) transforms the integral in part (ii) to \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } 2 \tan ^ { 2 } \theta \mathrm {~d} \theta\).
(iv) Hence find the exact area of the shaded region.
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