Moderate -0.5 This is a standard logarithmic linearization problem requiring students to convert the power law to linear form (ln y = -n/2 ln x + ln C/2), find the gradient from two points, then use substitution to find C. While it involves multiple steps, the technique is routine for P3 students and follows a well-practiced procedure with no novel insight required.
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The variables \(x\) and \(y\) satisfy the equation \(x ^ { n } y ^ { 2 } = C\), where \(n\) and \(C\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points \(( 0.31,1.21 )\) and \(( 1.06,0.91 )\), as shown in the diagram.
Find the value of \(n\) and find the value of \(C\) correct to 2 decimal places.
\(\left(\frac{2.886}{1.363}\right)^n \times \left(\frac{2.484}{3.353}\right)^2 = 1\) leading to \(n = 0.7995\)
Solve for \(C\)
M1
Using \(x\) and \(y\) values
Obtain \(C = 14.41\)
A1
Must be 2 d.p.
Total: 5
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $n\ln x + 2\ln y = \ln C$ | B1 | |
| Substitute values of $\ln y$ and $\ln x$, or equate gradient of line to $\pm\frac{1}{2}n$, but not $\pm n$, and solve for $n$ | M1 | Using $\ln x$ and $\ln y$ values |
| Obtain $n = 0.8[0]$ or $0.8[00]$ or $\frac{4}{5}$ | A1 | |
| Solve for $C$ | M1 | Using $\ln x$ and $\ln y$ values in equation of correct form, that is $\ln C$ not $C$. Allow $C = e^{2.668}$ |
| Obtain $C = 14.41$ | A1 | Must be 2 d.p. |
| **Alternative method:** | | |
| Obtain two correct equations in $n$ and $C$ by substituting $x$ and $y$ values | B1 | $(2.886)^n \times (2.484)^2 = C$ and $(1.363)^n \times (3.353)^2 = C$ |
| Solve for $n$ | M1 | Using $x$ and $y$ values |
| Obtain $n = 0.8[0]$ or $0.8[00]$ or $4/5$ | A1 | $\left(\frac{2.886}{1.363}\right)^n \times \left(\frac{2.484}{3.353}\right)^2 = 1$ leading to $n = 0.7995$ |
| Solve for $C$ | M1 | Using $x$ and $y$ values |
| Obtain $C = 14.41$ | A1 | Must be 2 d.p. |
| **Total: 5** | | |
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The variables $x$ and $y$ satisfy the equation $x ^ { n } y ^ { 2 } = C$, where $n$ and $C$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points $( 0.31,1.21 )$ and $( 1.06,0.91 )$, as shown in the diagram.
Find the value of $n$ and find the value of $C$ correct to 2 decimal places.\\
\hfill \mbox{\textit{CAIE P3 2022 Q3 [5]}}