Finding stationary points after integration

10 A function f with domain \(x > 0\) is such that \(\mathrm { f } ^ { \prime } ( x ) = 8 ( 2 x - 3 ) ^ { \frac { 1 } { 3 } } - 10 x ^ { \frac { 2 } { 3 } }\). It is given that the curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(( 1,0 )\).

1. Find the equation of the normal to the curve at the point \(( 1,0 )\).
2. Find \(\mathrm { f } ( x )\).
   \includegraphics[max width=\textwidth, alt={}, center]{39393c20-34df-4167-a8bf-e4067371de81-16_2715_41_110_2008}
   \includegraphics[max width=\textwidth, alt={}, center]{39393c20-34df-4167-a8bf-e4067371de81-17_2723_35_101_20} It is given that the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\) can be expressed in the form $$125 x ^ { 2 } - 128 x + 192 = 0$$
3. Determine, making your reasoning clear, whether f is an increasing function, a decreasing function or neither.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.
   \includegraphics[max width=\textwidth, alt={}, center]{39393c20-34df-4167-a8bf-e4067371de81-18_2714_38_109_2010}

9 A curve for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 5 x - 1 ) ^ { \frac { 1 } { 2 } } - 2\) passes through the point ( 2,3 ).

1. Find the equation of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
3. Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point.

8 A function f is defined for \(x > \frac { 1 } { 2 }\) and is such that \(\mathrm { f } ^ { \prime } ( x ) = 3 ( 2 x - 1 ) ^ { \frac { 1 } { 2 } } - 6\).

1. Find the set of values of \(x\) for which f is decreasing.
2. It is now given that \(\mathrm { f } ( 1 ) = - 3\). Find \(\mathrm { f } ( x )\).

7 A new design for a company logo is to be made from two sectors of a circle, ORP and OQS, and a rhombus OSTR, as shown in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-08_509_584_408_817} The points \(P , O\) and \(Q\) lie on a straight line and the angle \(R O S\) is \(\theta\) radians.
A large copy of the logo, with \(P Q = 5\) metres, is to be put on a wall.
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1. Show that the area of the logo, \(A\) square metres, is given by $$A = \frac { 25 } { 8 } ( \pi - \theta + 2 \sin \theta )$$ \section*{-
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2. 1. Show that the maximum value of \(A\) occurs when \(\theta = \frac { \pi } { 3 }\)
      Fully justify your answer.} 7
3. (ii) Find the exact maximum value of \(A\)
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4. Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if \(P Q\) were increased to 10 metres.
   \includegraphics[max width=\textwidth, alt={}, center]{6fba7e53-de46-460b-9bef-f1a6962f2e7d-11_2488_1716_219_153} Use the substitution \(u = x ^ { 5 } + 2\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 9 } } { \left( x ^ { 5 } + 2 \right) ^ { 3 } } \mathrm {~d} x = \frac { 1 } { 180 }$$