| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Solve exponential equation by substitution |
| Difficulty | Moderate -0.3 This is a straightforward C2 exponential equation requiring standard substitution technique. Part (a) guides students through the substitution process (3^(x+1) = 3y, 3^(2x-1) = y²/3), making part (b) a routine quadratic equation. The scaffolding and standard method make this slightly easier than average for A-level. |
| Spec | 1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| (i) \(= 3^1 \times 3^x = 3y\) | M1 A1 | |
| (ii) \(= 3^{-1} \times (3^x)^2 = \frac{1}{3}y^2\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(3y - \frac{1}{3}y^2 = 6\) | ||
| \(y^2 - 9y + 18 = 0\) | ||
| \((y-3)(y-6) = 0\) | M1 | |
| \(y = 3, 6\) | A1 | |
| \(3^x = 3, 6\) | ||
| \(x = 1, \frac{\lg 6}{\lg 3}\) | B1 M1 | |
| \(x = 1, 1.63\) (3sf) | A1 | (9) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| (i) $= 3^1 \times 3^x = 3y$ | M1 A1 | |
| (ii) $= 3^{-1} \times (3^x)^2 = \frac{1}{3}y^2$ | M1 A1 | |
## Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $3y - \frac{1}{3}y^2 = 6$ | | |
| $y^2 - 9y + 18 = 0$ | | |
| $(y-3)(y-6) = 0$ | M1 | |
| $y = 3, 6$ | A1 | |
| $3^x = 3, 6$ | | |
| $x = 1, \frac{\lg 6}{\lg 3}$ | B1 M1 | |
| $x = 1, 1.63$ (3sf) | A1 | **(9)** |
---7. (a) Given that $y = 3 ^ { x }$, find expressions in terms of $y$ for
\begin{enumerate}[label=(\roman*)]
\item $3 ^ { x + 1 }$,
\item $3 ^ { 2 x - 1 }$.\\
(b) Hence, or otherwise, solve the equation
$$3 ^ { x + 1 } - 3 ^ { 2 x - 1 } = 6$$
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q7 [9]}}