Graphs & Transformations: Logarithmic transformations — JPJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
(a) Show that \(\displaystyle y=\ln\!\left(\frac{\mathrm{e}^{2}}{3x}\right)\) can be written in the form \(y=a+b\ln(cx)\), where \(a\), \(b\) and \(c\) are integers to be found. Hence, state a sequence of transformations which transform the graph of \(y=\ln x\) onto the graph of \(\displaystyle y=\ln\!\left(\frac{\mathrm{e}^{2}}{3x}\right)\).
(b) The curve \(y=\mathrm{f}(x)\) passes through the point \(P\) with coordinates \((a,b)\), where \(b\neq 0\). The tangent to the curve at \(P\) has gradient \(5\). When \(y=\mathrm{f}(x)\) is transformed onto the curve \(y=\mathrm{g}(x)\), \(P\) corresponds to the point \(R\) on \(y=\mathrm{g}(x)\). For each of the following curves, state the coordinates of \(R\) and find the gradient of the curve at \(R\).
- \(\mathrm{g}(x)=2\mathrm{f}(x-1)\)
- \(\mathrm{g}(x)=\dfrac{1}{\mathrm{f}(x)}\)
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(a) [3 marks] \[ y = \ln\!\left(\frac{\mathrm{e}^{2}}{3x}\right) = \ln\mathrm{e}^{2}-\ln 3x = 2-\ln 3x. \] \[ y = \ln x \;\xrightarrow{<strong>A</strong>}\; y = \ln 3x \;\xrightarrow{<strong>B</strong>}\; y = -\ln 3x \;\xrightarrow{<strong>C</strong>}\; y = 2-\ln 3x \]
A: A scaling parallel to the \(x\)-axis by a factor of \(\tfrac{1}{3}\).
B: Reflection in the \(x\)-axis.
C: A translation of \(2\) units in the positive direction of the \(y\)-axis.
(b)(i) [3 marks]
\(\mathrm{g}(x) = 2\mathrm{f}(x-1)\). \[ (a,b) \;\to\; (a+1,b) \;\to\; (a+1,2b). \] Hence the corresponding point \(R\) is \((a+1,2b)\).
\(\mathrm{g}'(x) = 2\mathrm{f}'(x-1)\).
Given \(\mathrm{f}'(a) = 5\): \(\mathrm{f}'(a+1-1) = \mathrm{f}'(a) = 5\) (gradient remains the same after translation).
At \((a+1,2b)\): \(\mathrm{g}'(a+1) = 2\mathrm{f}'(a) = 2(5) = 10\).
(b)(ii) [3 marks]
\(\mathrm{g}(x) = \dfrac{1}{\mathrm{f}(x)}\).
The corresponding point \(R\) is \(\left(a,\dfrac{1}{b}\right)\). \[ \mathrm{g}'(x) = -\frac{\mathrm{f}'(x)}{[\mathrm{f}(x)]^{2}}. \]
Given \(\mathrm{f}'(a) = 5\), at \(\left(a,\dfrac{1}{b}\right)\): \[ \mathrm{g}'(a) = -\frac{\mathrm{f}'(a)}{[\mathrm{f}(a)]^{2}} = -\frac{5}{b^{2}}. \]