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Question
A particle moves in a straight line. Its velocity \(v{\text{ m}}\,{{\text{s}}^{ – 1}}\) after \(t\) seconds is given by
\[v = 6t – 6,{\text{ for }}0 \leqslant t \leqslant 2.\]
After \(p\) seconds, the particle is 2 m from its initial position. Find the possible values of \(p\).
Answer/Explanation
Markscheme
correct approach (A1)
eg\(\,\,\,\,\,\)\(s = \int {v,{\text{ }}\int_0^p {6t – 6{\text{d}}t} } \)
correct integration (A1)
eg\(\,\,\,\,\,\)\(\int {6t – 6{\text{d}}t = 3{t^2} – 6t + C,{\text{ }}\left[ {3{t^2} – 6t} \right]_0^p} \)
recognizing that there are two possibilities (M1)
eg\(\,\,\,\,\,\)2 correct answers, \(s = \pm 2,{\text{ }}c \pm 2\)
two correct equations in \(p\) A1A1
eg\(\,\,\,\,\,\)\(3{p^2} – 6p = 2,{\text{ }}3{p^2} – 6p = – 2\)
0.42265, 1.57735
\(p = 0.423{\text{ or }}p = 1.58\) A1A1 N3
[7 marks]
Question
Let \(f(x) = {x^2} + 2x + 1\) and \(g(x) = x – 5\), for \(x \in \mathbb{R}\).
Find \(f(8)\).
Find \((g \circ f)(x)\).
Solve \((g \circ f)(x) = 0\).
Answer/Explanation
Markscheme
attempt to substitute \(x = 8\) (M1)
eg\(\,\,\,\,\,\)\({8^2} + 2 \times 8 + 1\)
\(f(8) = 81\) A1 N2
[2 marks]
attempt to form composition (in any order) (M1)
eg\(\,\,\,\,\,\)\(f(x – 5),{\text{ }}g\left( {f(x)} \right),{\text{ }}\left( {{x^2} + 2x + 1} \right) – 5\)
\((g \circ f)(x) = {x^2} + 2x – 4\) A1 N2
[2 marks]
valid approach (M1)
eg \(x = \frac{{ – 2 \pm \sqrt {20} }}{2}\), 
\(1.23606,{\text{ }} – 3.23606\)
\(x = 1.24,{\text{ }}x = – 3.24\) A1A1 N3
[3 marks]