AQA C2 2006 January — Question 1 5 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2006
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeFind stationary points coordinates
DifficultyModerate -0.5 This is a straightforward stationary points question requiring differentiation of a simple function (power rule only), setting the derivative to zero, and solving a quadratic equation. It's slightly easier than average because it involves standard techniques with no complications, though it does require multiple steps and solving x² = 1/16 correctly.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives
1 Given that \(y = 16 x + x ^ { - 1 }\), find the two values of \(x\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).
(5 marks)
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y'(x) = 16 - x^{-2}\) (one term correct)M1 One term correct
Both terms correctA1 Both correct
\(y'(x) = 16 - \frac{1}{x^2}\)B1 \(x^{-2} = \frac{1}{x^2}\) OE PI
\(y'(x) = 0 \Rightarrow 16x^2 = 1\)M1 c's \(y'(x)=0\) and one relevant further step
\(\Rightarrow x = \pm\frac{1}{4}\)A1 Both answers required
Total5 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y'(x) = 16 - x^{-2}$ (one term correct) | M1 | One term correct |
| Both terms correct | A1 | Both correct |
| $y'(x) = 16 - \frac{1}{x^2}$ | B1 | $x^{-2} = \frac{1}{x^2}$ OE PI |
| $y'(x) = 0 \Rightarrow 16x^2 = 1$ | M1 | c's $y'(x)=0$ and one relevant further step |
| $\Rightarrow x = \pm\frac{1}{4}$ | A1 | Both answers required |
| **Total** | **5** | |

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1 Given that $y = 16 x + x ^ { - 1 }$, find the two values of $x$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.\\
(5 marks)

\hfill \mbox{\textit{AQA C2 2006 Q1 [5]}}
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Q1 5 Q2 5 Q3 9 Q4 11 Q5 9 Q6 12 Q7 5 Q8 18