| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Show surd expression equals value |
| Difficulty | Easy -1.2 This is a routine C1 surds question testing standard techniques: simplifying surds by factoring out perfect squares, and rationalizing denominators. Part (a) requires basic surd simplification (√50 = 5√2, √18 = 3√2, √8 = 2√2), while part (b) uses the standard method of multiplying by the conjugate. Both are textbook exercises with no problem-solving or insight required, making this easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{(x+2)3e^{3x} - e^{3x}(1)}{(x+2)^2}\) | B1, M1, A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 0\), \(\frac{dy}{dx} = \frac{6e^0 - e^0}{2^2} = \frac{5}{4}\) | M1, A1F | Attempt to find \(dy/dx\) at \(x = 0\) |
| \(A\left(0, \frac{1}{2}\right)\) | B1 | |
| Equation of tangent at \(A\): \(y - \frac{1}{2} = \frac{5}{4}(x - 0)\) | A1 | 4 |
# Question 4(a) [XMCA2]:
$\frac{dy}{dx} = \frac{(x+2)3e^{3x} - e^{3x}(1)}{(x+2)^2}$ | B1, M1, A1 | **3** | $(e^{3x})' = 3e^{3x}$; Quotient rule OE
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# Question 4(b) [XMCA2]:
When $x = 0$, $\frac{dy}{dx} = \frac{6e^0 - e^0}{2^2} = \frac{5}{4}$ | M1, A1F | Attempt to find $dy/dx$ at $x = 0$
$A\left(0, \frac{1}{2}\right)$ | B1 |
Equation of tangent at $A$: $y - \frac{1}{2} = \frac{5}{4}(x - 0)$ | A1 | **4** | ACF
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\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \sqrt { 50 } + \sqrt { 18 } } { \sqrt { 8 } }$ is an integer and find its value.\\
(3 marks)
\item Express $\frac { 2 \sqrt { 7 } - 1 } { 2 \sqrt { 7 } + 5 }$ in the form $m + n \sqrt { 7 }$, where $m$ and $n$ are integers.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q4 [7]}}