Question
Solve \({\log _2}x + {\log _2}(x – 2) = 3\) , for \(x > 2\) .
Answer/Explanation
Markscheme
recognizing \(\log a + \log b = \log ab\) (seen anywhere) (A1)
e.g. \({\log _2}(x(x – 2))\) , \({x^2} – 2x\)
recognizing \({\log _a}b = x \Leftrightarrow {a^x} = b\) (A1)
e.g. \({2^3} = 8\)
correct simplification A1
e.g. \(x(x – 2) = {2^3}\) , \({x^2} – 2x – 8\)
evidence of correct approach to solve (M1)
e.g. factorizing, quadratic formula
correct working A1
e.g. \((x – 4)(x + 2)\) , \(\frac{{2 \pm \sqrt {36} }}{2}\)
\(x = 4\) A2 N3
[7 marks]
Question
Let \(f(x) = {x^2} + x – 6\).
Write down the \(y\)-intercept of the graph of \(f\).
Solve \(f(x) = 0\).
On the following grid, sketch the graph of \(f\), for \( – 4 \le x \le 3\).
Answer/Explanation
Markscheme
\(y\)-intercept is \( – 6,{\text{ }}(0,{\text{ }} – 6),{\text{ }}y = – 6\) A1
[1 mark]
valid attempt to solve (M1)
eg\(\;\;\;(x – 2)(x + 3) = 0,{\text{ }}x = \frac{{ – 1 \pm \sqrt {1 + 24} }}{2}\), one correct answer
\(x = 2,{\text{ }}x = – 3\) A1A1 N3
[3 marks]
A1A1A1
Note: The shape must be an approximately correct concave up parabola. Only if the shape is correct, award the following:
A1 for the \(y\)-intercept in circle and the vertex approximately on \(x = – \frac{1}{2}\), below \(y = – 6\),
A1 for both the \(x\)-intercepts in circles,
A1 for both end points in ovals.
[3 marks]
Total [7 marks]
Examiners report
Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).
Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).
In part (c), although most candidates were familiar with the general parabolic shape of the graph, many placed the vertex at the \(y\)-intercept \((0,{\text{ }} – 6)\), and very few candidates considered the endpoints of the function with the given domain.