| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Inverse function with exponentials |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard techniques: rearranging an exponential equation using logarithms (part i is a 'show that'), differentiating using the chain rule for inverse functions (part ii), and applying the relationship dy/dx = 1/(dx/dy) (part iii). All steps are routine C3 material with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\ln y = (x-1)\ln 5\) | B1 | whether following \(\ln y = \ln 5^{x-1}\) or not; brackets needed |
| Obtain \(x = 1 + \frac{\ln y}{\ln 5}\) | B1 2 | AG; correct working needed; missing brackets maybe now implied |
| (ii) Differentiate to obtain single term of form \(\frac{k}{y}\) | M1 | any constant \(k\) |
| Obtain \(\frac{1}{y \ln 5}\) | A1 2 | or equiv involving \(y\) |
| (iii) Substitute for \(y\) and attempt reciprocal | M1 | or equiv method for finding derivative without using part (ii) |
| Obtain \(25\ln 5\) | A1 2 | or exact equiv |
**(i)** State $\ln y = (x-1)\ln 5$ | B1 | whether following $\ln y = \ln 5^{x-1}$ or not; brackets needed
Obtain $x = 1 + \frac{\ln y}{\ln 5}$ | B1 2 | AG; correct working needed; missing brackets maybe now implied
**(ii)** Differentiate to obtain single term of form $\frac{k}{y}$ | M1 | any constant $k$
Obtain $\frac{1}{y \ln 5}$ | A1 2 | or equiv involving $y$
**(iii)** Substitute for $y$ and attempt reciprocal | M1 | or equiv method for finding derivative without using part (ii)
Obtain $25\ln 5$ | A1 2 | or exact equiv4 It is given that $y = 5 ^ { x - 1 }$.\\
(i) Show that $x = 1 + \frac { \ln y } { \ln 5 }$.\\
(ii) Find an expression for $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.\\
(iii) Hence find the exact value of the gradient of the curve $y = 5 ^ { x - 1 }$ at the point (3, 25).
\hfill \mbox{\textit{OCR C3 2006 Q4 [6]}}