3 Integrate with respect to \(x\)
- \(\quad \int \frac { 3 } { 2 } ( 2 x - 7 ) ^ { 5 } d x\)
- \(\quad \int \frac { 3 x } { 2 } \left( 2 x ^ { 2 } - 7 \right) ^ { 5 } d x\)
- Express \(\frac { 5 - x } { 1 - x - 2 x ^ { 2 } }\) in partial fractions.
- Hence find the exact value of \(\int _ { 1 } ^ { 2 } \frac { 5 - x } { 1 - x - 2 x ^ { 2 } } d x\)
- Using your answer to part a, find a quadratic approximation for the expression \(\frac { 5 - x } { 1 - x - 2 x ^ { 2 } }\),
giving your answer in the form \(p + q x + r x ^ { 2 }\), where \(p , q\) and \(r\) are constants to be found.
The function f is defined by
$$\mathrm { f } ( x ) = x ^ { 2 } + 2 \cos x \text { for } - \pi \leq x \leq \pi$$
Determine whether the curve with equation \(y = \mathrm { f } ( x )\) has a point of inflection at the point where \(x = 0\)
Fully justify your answer.
a) Show that the expression
$$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta$$
can be written as
$$4 \cos \theta - \sec \theta$$
where \(\sin \theta \neq 0\) and \(\cos \theta \neq 0\)
b) A student is attempting to solve the equation
$$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta = 3 \text { for } 0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }$$
They use the result from part (a), and write the following incorrect solution:
$$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta = 3$$
Step \(1 \quad 4 \cos \theta - \sec \theta = 3\)
Step \(24 \cos \theta - \frac { 1 } { \cos \theta } - 3 = 0\)
Step \(34 \cos ^ { 2 } \theta - 3 \cos \theta - 1 = 0\)
Step \(4 \cos \theta = 1\) or \(\cos \theta = - 0.25\)
Step \(5 \quad \theta = 0 ^ { \circ } , 104.5 ^ { \circ } , 255.5 ^ { \circ } , 360 ^ { \circ }\) - Explain why the student should reject one of their values for \(\cos \theta\) in Step 4 .
- State the correct solutions to the equation
$$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta = 3 \text { for } 0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }$$
A particle moves in the \(x - y\) plane so that at time \(t\) seconds, where \(t \geqslant 0\), its coordinates are given by \(x = \mathrm { e } ^ { 2 t } - 4 \mathrm { e } ^ { t } + 3 , y = 2 \mathrm { e } ^ { - 3 t }\).
(a) Explain why the path of the particle never crosses the \(x\)-axis.
(b) Determine the exact values of \(t\) when the path of the particle intersects the \(y\)-axis.
(c) Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 \mathrm { e } ^ { 4 t } - \mathrm { e } ^ { 5 t } }\).
(d) Hence find the coordinates of the particle when its path is parallel to the \(y\)-axis.
\section*{In this question you must show detailed reasoning.}
(a) Express \(\cos x + \sqrt { 3 } \sin x\) in the form \(R \sin ( x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Give the values of \(R\) and \(\alpha\) in exact form.
(b) Hence solve the equation \(\cos x = \sqrt { 3 } ( 1 - \sin x )\) for values of \(x\) in the interval \(- \pi \leqslant x \leqslant \pi\). Give the roots of this equation in exact form.
A curve has equation \(y = \mathrm { f } ( x )\), where
$$\mathrm { f } ( x ) = \frac { 7 x \mathrm { e } ^ { x } } { \sqrt { \mathrm { e } ^ { 3 x } - 2 } } \quad x > \ln \sqrt [ 3 ] { 2 }$$
a) Show that
$$\mathrm { f } ^ { \prime } ( x ) = \frac { 7 \mathrm { e } ^ { x } \left( \mathrm { e } ^ { 3 x } ( 2 - x ) + A x + B \right) } { 2 \left( \mathrm { e } ^ { 3 x } - 2 \right) ^ { \frac { 3 } { 2 } } }$$
where \(A\) and \(B\) are constants to be found.
b) Hence show that the \(x\) coordinates of the turning points of the curve are solutions of the equation
$$x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }$$
In this question you must show detailed reasoning.
\includegraphics[max width=\textwidth, alt={}, center]{2af0aec0-4e21-4050-b26f-52f2c9c68b51-17_732_472_233_778}
The diagram shows the curve \(y = \frac { 4 \cos 2 x } { 3 - \sin 2 x }\), for \(x \geqslant 0\), and the normal to the curve at the point \(\left( \frac { 1 } { 4 } \pi , 0 \right)\). Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the \(y\)-axis is \(\ln \frac { 9 } { 4 } + \frac { 1 } { 128 } \pi ^ { 2 }\).