7
7
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1.
$$f ( x ) \equiv a x ^ { 3 } + b x ^ { 2 } - 7 x + 14 , \text { where } a \text { and } b \text { are constants. }$$
Given that when \(\mathrm { f } ( x )\) is divided by ( \(x - 1\) ) the remainder is 9 ,
- write down an equation connecting \(a\) and \(b\).
Given also that \(( x + 2 )\) is a factor of \(\mathrm { f } ( x )\),
- find the values of \(a\) and \(b\).
2. (i) Differentiate with respect to \(x\)
$$2 x ^ { 3 } + \sqrt { } x + \frac { x ^ { 2 } + 2 x } { x ^ { 2 } }$$
(ii) Evaluate
$$\int _ { 1 } ^ { 4 } \left( \frac { x } { 2 } + \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x$$
- (a) An arithmetic series has first term \(a\) and common difference \(d\). Prove that the sum of the first \(n\) terms of the series is
$$\frac { 1 } { 2 } n [ 2 a + ( n - 1 ) d ] .$$
A company made a profit of \(\pounds 54000\) in the year 2001. A model for future performance assumes that yearly profits will increase in an arithmetic sequence with common difference \(\pounds d\). This model predicts total profits of \(\pounds 619200\) for the 9 years 2001 to 2009 inclusive. - Find the value of \(d\).
Using your value of \(d\),
- find the predicted profit for the year 2011.
An alternative model assumes that the company's yearly profits will increase in a geometric sequence with common ratio 1.06 . Using this alternative model and again taking the profit in 2001 to be \(\pounds 54000\),
- find the predicted profit for the year 2011.
(3 marks)
4. (a) Write down formulae for \(\sin ( \mathrm { A } + \mathrm { B } )\) and \(\sin ( \mathrm { A } - \mathrm { B } )\).
Using \(\mathrm { X } = \mathrm { A } + \mathrm { B }\) and \(\mathrm { Y } = \mathrm { A } - \mathrm { B }\), prove that
$$\operatorname { Sin } X + \sin Y = 2 \sin \frac { X + Y } { 2 } \cos \frac { X - Y } { 2 }$$ - Hence, or otherwise, solve, for \(0 \leq \theta < 360\),
$$\sin 40 ^ { \circ } + \sin 20 ^ { \circ } = 0$$
\section*{5.}
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{2561e267-6495-453d-ac50-0b1542215e0a-4_833_965_303_466}
\end{figure}
Figure 1 shows a gardener's design for the shape of a flower bed with perimeter \(A B C D\).
\(A D\) is an arc of a circle with centre \(O\) and radius 5 m .
\(B C\) is an arc of a circle with centre \(O\) and radius 7 m .
\(O A B\) and \(O D C\) are straight lines and the size of \(\angle A O D\) is \(\theta\) radians. - Find, in terms of \(\theta\), an expression for the area of the flower bed.
Given that the area of the flower bed is \(15 \mathrm {~m} ^ { 2 }\),
- show that \(\theta = 1.25\),
- calculate, in m , the perimeter of the flower bed.
The gardener now decides to replace arc \(A D\) with the straight line \(A D\).
- Find, to the nearest cm, the reduction in the perimeter of the flower bed.
6. (a) Given that
$$( 2 + x ) ^ { 5 } + ( 2 - x ) ^ { 5 } \equiv A + B x ^ { 2 } + C x ^ { 4 }$$
Find the values of the constants \(A , B\) and \(C\). - Using the substitution \(y = x ^ { 2 }\) and your answers to part (a), solve,
$$( 2 + x ) ^ { 5 } + ( 2 - x ) ^ { 5 } = 349$$
7.
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{2561e267-6495-453d-ac50-0b1542215e0a-5_780_974_922_630}
\end{figure}
Figure 2 shows part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where
$$f ( x ) = x ^ { 3 } - 6 x ^ { 2 } + 5 x$$
The curve crosses the \(x\)-axis at the origin \(O\) and at the points \(A\) and \(B\). - Factorise \(\mathrm { f } ( x )\) completely
- Write down the \(x\)-coordinates of the points \(A\) and \(B\).
- Find the gradient of \(C\) at \(A\).
The region \(R\) is bounded by \(C\) and the line \(O A\), and the region \(S\) is bounded by \(C\) and the line \(A B\).
- Use integration to find the area of the combined regions \(R\) and \(S\), shown shaded in Fig. 2.