Moderate -0.8 This is a standard logarithmic transformation question requiring students to recognize that ln y = ln A + px gives a straight line, then use two points to find the gradient p and y-intercept ln A. It's routine bookwork with straightforward arithmetic, making it easier than average but not trivial since it requires understanding the exponential-linear relationship.
\includegraphics{figure_2}
The variables \(x\) and \(y\) satisfy the equation \(y = Ae^{px}\), where \(A\) and \(p\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points \((5, 3.17)\) and \((10, 4.77)\), as shown in the diagram. Find the values of \(A\) and \(p\) correct to 2 decimal places. [5]
Question 2:
2 | State or imply lny=lnA+ px
Equate gradient of line to p
Obtain p=0.32
Substitute to find A
Obtain A=4.81
OR 1:
3.17=lnA+5p or 4.77=lnA+10p
Correct attempt to obtain lnA or p
Correct attempt to obtain the other unknown
Obtain A=4.81
Obtain p=0.32
OR 2:
e3.17 = Ae5p or e4.77 = Ae10p
Correct attempt to obtain p
Correct attempt to get A
Obtain A=4.81
Obtain p=0.32 | B1
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\includegraphics{figure_2}
The variables $x$ and $y$ satisfy the equation $y = Ae^{px}$, where $A$ and $p$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(5, 3.17)$ and $(10, 4.77)$, as shown in the diagram. Find the values of $A$ and $p$ correct to 2 decimal places. [5]
\hfill \mbox{\textit{CAIE P2 2016 Q2 [5]}}