| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Quadratic in exponential form |
| Difficulty | Moderate -0.3 This is a slightly easier than average A-level question. Part (a) requires identifying errors in index law manipulation (incorrectly splitting 2^{2x+4} and wrong quadratic coefficient), which tests understanding rather than just recall. Part (b) requires correct application of index laws and solving a quadratic, which are standard AS techniques. The 4-mark total and straightforward solution path place it just below average difficulty. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| 12(a) | 22x (cid:14)24 is wrong in line 2 - it should be 22x(cid:117)24 | B1 |
| In line 4, 24 has been replaced by 8 instead of by 16 | B1 | 2.3 |
| Answer | Marks |
|---|---|
| (b) | Way 1: |
| Answer | Marks | Guidance |
|---|---|---|
| 16y2 (cid:16)9y (cid:32)0 | Way 2: | |
| (2x(cid:14)4)log2(cid:16)log9(cid:16)xlog2(cid:32)0 | M1 | 2.1 |
| Answer | Marks |
|---|---|
| o.e. with no second answer | log9 |
| Answer | Marks | Guidance |
|---|---|---|
| log2 | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 12:
--- 12(a) ---
12(a) | 22x (cid:14)24 is wrong in line 2 - it should be 22x(cid:117)24 | B1 | 2.3
In line 4, 24 has been replaced by 8 instead of by 16 | B1 | 2.3
(2)
(b) | Way 1:
22x(cid:14)4 (cid:16)9(2x)(cid:32)0
22x (cid:117)24 (cid:16)9(2x)(cid:32)0
Let 2x (cid:32) y
16y2 (cid:16)9y (cid:32)0 | Way 2:
(2x(cid:14)4)log2(cid:16)log9(cid:16)xlog2(cid:32)0 | M1 | 2.1
y (cid:32) 9 or y (cid:32)0
16
log(cid:11) 9 (cid:12)
So x (cid:32)log (cid:11) 9 (cid:12) or 16
2 16 log2
o.e. with no second answer | log9
x = (cid:16)4 o.e.
log2 | A1 | 1.1b
(2)
(4 marks)
Notes:
(a)
B1: Lists error in line 2 (as above)
B1: Lists error in line 4 (as above)
(b)
M1: Correct work with powers reaching this equation
A1: Correct answer here – there are many exact equivalents
Question | Scheme | Marks | AOsA student was asked to give the exact solution to the equation
$$2^{2x+4} - 9(2^x) = 0$$
The student's attempt is shown below:
$$2^{2x+4} - 9(2^x) = 0$$
$$2^{2x} + 2^4 - 9(2^x) = 0$$
Let $2^x = y$
$$y^2 - 9y + 8 = 0$$
$$(y - 8)(y - 1) = 0$$
$$y = 8 \text{ or } y = 1$$
$$\text{So } x = 3 \text{ or } x = 0$$
\begin{enumerate}[label=(\alph*)]
\item Identify the two errors made by the student.
[2]
\item Find the exact solution to the equation.
[2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q12 [4]}}