7
\(q\) 3 The line \(L\) has equation \(2 x + 3 y = 7\)
Which one of the following is perpendicular to \(L\) ?
Tick one box. $$\begin{aligned} & 2 x - 3 y = 7
& 3 x + 2 y = - 7
& 2 x + 3 y = - \frac { 1 } { 7 }
& 3 x - 2 y = 7 \end{aligned}$$ □
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□ 4 Sketch the graph of \(y = | 2 x + a |\), where \(a\) is a positive constant. Show clearly where the graph intersects the axes.
\includegraphics[max width=\textwidth, alt={}, center]{d9149857-5f94-4fa5-a6d8-550c0c07fefb-03_1001_1002_450_520} 5 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.
6 (b) Use the quotient rule to show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { x - 2 } { ( 2 x - 2 ) ^ { \frac { 3 } { 2 } } }\) 6 (a) State the maximum possible domain of f .
\(6 \quad\) A function f is defined by \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 x - 2 } }\) $$\begin{gathered} \text { Do not write }
\text { outside the }
\text { box } \end{gathered}$$ 6 (a)
6 (c) Show that the graph of \(y = \mathrm { f } ( x )\) has exactly one point of inflection.
6 (d) Write down the values of \(x\) for which the graph of \(y = \mathrm { f } ( x )\) is convex.
7 (a) Given that \(\log _ { a } y = 2 \log _ { a } 7 + \log _ { a } 4 + \frac { 1 } { 2 }\), find \(y\) in terms of \(a\).
7 (b) When asked to solve the equation $$2 \log _ { a } x = \log _ { a } 9 - \log _ { a } 4$$ a student gives the following solution: $$\begin{aligned} & 2 \log _ { a } x = \log _ { a } 9 - \log _ { a } 4
& \Rightarrow 2 \log _ { a } x = \log _ { a } \frac { 9 } { 4 }
& \Rightarrow \log _ { a } x ^ { 2 } = \log _ { a } \frac { 9 } { 4 }
& \Rightarrow x ^ { 2 } = \frac { 9 } { 4 }
& \therefore x = \frac { 3 } { 2 } \text { or } - \frac { 3 } { 2 } \end{aligned}$$ Explain what is wrong with the student's solution.