Question
The graph of \(y = \ln (x)\) is transformed into the graph of \(y = \ln \left( {2x + 1} \right)\) .
a.Describe two transformations that are required to do this.[2]
b.Solve \(\ln \left( {2x + 1} \right) > 3\cos (x)\), \(x \in [0,10]\).[4]
▶️Answer/Explanation
Markscheme
EITHER
translation of \( – \frac{1}{2}\) parallel to the \(x\)-axis
stretch of a scale factor of \(\frac{1}{2}\) parallel to the \(x\)-axis A1A1
OR
stretch of a scale factor of \(\frac{1}{2}\) parallel to the \(x\)-axis
translation of \( – 1\) parallel to the \(x\)-axis A1A1
Note: Accept clear alternative terminologies for either transformation.
[2 marks]
EITHER
\(1.16 < x < 5.71 \cup 6.75 < x \leqslant 10\) A1A1A1A1
OR
]\(1.16\), \(5.71\)[ \(\cup\) ]\(6.75\), \(10\)] A1A1A1A1
Note: Award A1 for 1 intersection value, A1 for the other 2, A1A1 for the intervals.
[6 marks]
Question
Let \(f(x) = \ln x\) . The graph of f is transformed into the graph of the function g by a translation of \(\left( {\begin{array}{*{20}{c}}
3 \\
{ – 2}
\end{array}} \right)\), followed by a reflection in the x-axis. Find an expression for \(g(x)\), giving your answer as a single logarithm.
▶️Answer/Explanation
Markscheme
\(h(x) = f(x – 3) – 2 = \ln (x – 3) – 2\) (M1)(A1)
\(g(x) = -h(x) = 2 – \ln (x – 3)\) M1
Note: Award M1 only if it is clear the effect of the reflection in the x-axis:
the expression is correct OR
there is a change of signs of the previous expression OR
there’s a graph or an explanation making it explicit
\( = \ln {{\text{e}}^2} – \ln (x – 3)\) M1
\( = \ln \left( {\frac{{{{\text{e}}^2}}}{{x – 3}}} \right)\) A1
[5 marks]
Examiners report
This question was well attempted but many candidates could have scored better had they written down all the steps to obtain the final expression. In some cases, as the final expression was incorrect and the middle steps were missing, candidates scored just 1 mark. That could be a consequence of a small mistake, but the lack of working prevented them from scoring at least all method marks. Some candidates performed the transformations well but were not able to use logarithms properties to transform the answer and give it as a single logarithm.
Question
The graph of \(y = \ln (5x + 10)\) is obtained from the graph of \(y = \ln x\) by a translation of \(a\) units in the direction of the \(x\)-axis followed by a translation of \(b\) units in the direction of the \(y\)-axis.
a.Find the value of \(a\) and the value of \(b\).[4]
b.The region bounded by the graph of \(y = \ln (5x + 10)\), the \(x\)-axis and the lines \(x = {\text{e}}\) and \(x = 2{\text{e}}\), is rotated through \(2\pi \) radians about the \(x\)-axis. Find the volume generated.[2]
▶️Answer/Explanation
Markscheme
EITHER
\(y = \ln (x – a) + b = \ln (5x + 10)\) (M1)
\(y = \ln (x – a) + \ln c = \ln (5x + 10)\)
\(y = \ln \left( {c(x – a)} \right) = \ln (5x + 10)\) (M1)
OR
\(y = \ln (5x + 10) = \ln \left( {5(x + 2)} \right)\) (M1)
\(y = \ln (5) + \ln (x + 2)\) (M1)
THEN
\(a = – 2,{\text{ }}b = \ln 5\) A1A1
Note: Accept graphical approaches.
Note: Accept \(a = 2,{\text{ }}b = 1.61\)
[4 marks]
\(V = \pi {\int_e^{2e} {\left[ {\ln (5x + 10)} \right]} ^2}{\text{d}}x\) (M1)
\( = 99.2\) A1
[2 marks]
Total [6 marks]
Examiners report
[N/A]
[N/A]
Question
The graph of y = 2x2 + 4x +7 is translated using the vector \(\begin{pmatrix} 2\\-1\end{pmatrix}\). Find the equation of the translated graph, giving your answer in the form y = ax2 + bx c.
▶️Answer/Explanation
Ans
\(y=2(x-2)^2+4(x-2)+7-1\) \(=2(x^2-4x+4)+4x-8+6\) \(=2x^2-8x+8+4x-2\) \(\Rightarrow y = 2x^2-4x+6\)