| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Solve exponential equation by substitution |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C2 exponential content: sketching y=4^x (routine), describing a translation (basic transformation), and solving an exponential equation using substitution Y=2^x to form a quadratic. While it requires multiple techniques (substitution, quadratic formula, logarithms), each step follows a predictable pattern with no novel insight needed. The substitution is explicitly given, making this slightly easier than average but still requiring competent execution across several steps. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Graph only crossing \(y\)-axis at \((0,1)\) | B1 | Any graph only crossing the \(y\)-axis at \((0,1)\) stated/indicated… (accept 1 on \(y\)-axis as equivalent) and not drawn below \(x\)-axis |
| Correct shaped graph | B1 | Correct shaped graph, must clearly go below the intersection point and an indication of correct behaviour of curve for large positive and large negative values of \(x\). Ignore any scaling on axes |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Translation | B1 | Accept 'transl…' as equivalent. [T or Tr is NOT sufficient] |
| \(\begin{bmatrix} 0 \\ -5 \end{bmatrix}\) | B1 | If vector not given, accept full equivalent to vector in words provided linked to 'transl.../ move/shift'. (B0B0 if >1 transformation) |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4^x = (2^2)^x = 2^{2x} = (2^x)^2 = Y^2\) and \(2^{x+2} = 2^x \times 2^2 = 4Y\) | M1 | Justifying either \(4^x = Y^2\) or \(2^{x+2} = 4Y\) with no errors seen |
| \(4^x - 2^{x+2} - 5 = 0 \Rightarrow Y^2 - 4Y - 5 = 0\) | A1 | AG. Be convinced; must have justified both of the above |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((Y-5)(Y+1) = 0\) | M1 | Correct factorising or use of quadratic formula or completing sq. PI by both solns \(5\&-1\) seen |
| (Since) \(2^x > 0\) (for all real \(x\),) \(2^x = 5\) so only one (real) solution | E1 | Rejection of \(2^x\) (condone \(Y\)) negative, with justification, (condone "\(2^x\) not negative") followed by statement |
| \(\log 2^x = \log 5 \Rightarrow x\log 2 = \log 5\) | M1 | Eqn of form \(p^x = q \Rightarrow x\log p = \log q\) provided \(p > 0\) & \(q > 0\) OE eg \(x = \log_2 5\) |
| \(x = 2.3219\ldots = 2.322\) (to 3dp) | A1 | Condone > 3dp but must see explicit use of logs and must only be the one solution |
| Total: 4 marks |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Graph only crossing $y$-axis at $(0,1)$ | B1 | Any graph only crossing the $y$-axis at $(0,1)$ stated/indicated… (accept 1 on $y$-axis as equivalent) and not drawn **below** $x$-axis |
| Correct shaped graph | B1 | Correct shaped graph, must clearly go below the intersection point and an indication of correct behaviour of curve for large positive and large negative values of $x$. Ignore any scaling on axes |
| **Total: 2 marks** | | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Translation | B1 | Accept 'transl…' as equivalent. [T or Tr is NOT sufficient] |
| $\begin{bmatrix} 0 \\ -5 \end{bmatrix}$ | B1 | If vector not given, accept **full** equivalent to vector in words provided linked to 'transl.../ move/shift'. (B0B0 if >1 transformation) |
| **Total: 2 marks** | | |
## Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4^x = (2^2)^x = 2^{2x} = (2^x)^2 = Y^2$ and $2^{x+2} = 2^x \times 2^2 = 4Y$ | M1 | Justifying either $4^x = Y^2$ or $2^{x+2} = 4Y$ with no errors seen |
| $4^x - 2^{x+2} - 5 = 0 \Rightarrow Y^2 - 4Y - 5 = 0$ | A1 | AG. Be convinced; must have justified both of the above |
| **Total: 2 marks** | | |
## Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(Y-5)(Y+1) = 0$ | M1 | Correct factorising or use of quadratic formula or completing sq. PI by both solns $5\&-1$ seen |
| (Since) $2^x > 0$ (for all real $x$,) $2^x = 5$ so only one (real) solution | E1 | Rejection of $2^x$ (condone $Y$) negative, with justification, (condone "$2^x$ not negative") followed by statement |
| $\log 2^x = \log 5 \Rightarrow x\log 2 = \log 5$ | M1 | Eqn of form $p^x = q \Rightarrow x\log p = \log q$ provided $p > 0$ & $q > 0$ OE eg $x = \log_2 5$ |
| $x = 2.3219\ldots = 2.322$ (to 3dp) | A1 | Condone > 3dp but must see explicit use of logs and must only be the one solution |
| **Total: 4 marks** | | |4
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve with equation $y = 4 ^ { x }$, indicating the coordinates of any point where the curve intersects the coordinate axes.\\
(2 marks)
\item Describe the geometrical transformation that maps the graph of $y = 4 ^ { x }$ onto the graph of $y = 4 ^ { x } - 5$.
\item \begin{enumerate}[label=(\roman*)]
\item Use the substitution $Y = 2 ^ { x }$ to show that the equation $4 ^ { x } - 2 ^ { x + 2 } - 5 = 0$ can be written as $Y ^ { 2 } - 4 Y - 5 = 0$.
\item Hence show that the equation $4 ^ { x } - 2 ^ { x + 2 } - 5 = 0$ has only one real solution. Use logarithms to find this solution, giving your answer to three decimal places.\\
(4 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q4 [10]}}