OCR MEI AS Paper 1 2021 November — Question 7 6 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2021
SessionNovember
Marks6
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Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (straightforward integration + point)
DifficultyModerate -0.3 This is a straightforward integration question requiring students to integrate a polynomial and power term, then use a given point to find the constant of integration, and finally check if another point lies on the curve. While it involves multiple steps (integrate, find C, substitute), each step uses standard AS-level techniques with no conceptual challenges or problem-solving insight required. It's slightly easier than average due to the routine nature of the integration and the clear structure of the question.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)
7 The diagram shows part of a curve which passes through the point \(( 1,0 )\). \includegraphics[max width=\textwidth, alt={}, center]{5428eabf-431d-4db1-8c25-1f2b9570d9aa-4_711_704_1722_258} The gradient of the curve is given by \(\frac { d y } { d x } = 6 x + \frac { 8 } { x ^ { 3 } }\).
Determine whether the curve passes through the point \(( 2,12 )\).
Question 7:
AnswerMarks Guidance
AnswerMarks Guidance
\(y = \int\left(6x + \frac{8}{x^3}\right)dx = \int(6x + 8x^{-3})\,dx\)M1 Attempt to integrate with a negative power. Condone missing \(+c\)
\(= 6 \times \frac{x^2}{2} + 8 \times \frac{x^{-2}}{-2} + c\)A1 First two terms correct. Condone missing \(+c\)
When \(x = 1\), \(y = 0\): \(0 = 3 \times 1^2 - 4 \times 1^{-2} + c\) giving \(c = 1\)M1 Attempt to evaluate \(c\)
So \(y = 3x^2 - 4x^{-2} + 1\)A1 Correct equation for the curve
When \(x = 2\), \(y = 3 \times 2^2 - 4 \times 2^{-2} + 1 = 12 - 1 + 1 = 12\)M1 Using \(x = 2\) in their non-linear equation
So the point \((2, 12)\) lies on the curve.E1 [6] Conclusion from their \(y\)-value FT 
# Question 7:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \int\left(6x + \frac{8}{x^3}\right)dx = \int(6x + 8x^{-3})\,dx$ | M1 | Attempt to integrate with a negative power. Condone missing $+c$ |
| $= 6 \times \frac{x^2}{2} + 8 \times \frac{x^{-2}}{-2} + c$ | A1 | First two terms correct. Condone missing $+c$ |
| When $x = 1$, $y = 0$: $0 = 3 \times 1^2 - 4 \times 1^{-2} + c$ giving $c = 1$ | M1 | Attempt to evaluate $c$ |
| So $y = 3x^2 - 4x^{-2} + 1$ | A1 | Correct equation for the curve |
| When $x = 2$, $y = 3 \times 2^2 - 4 \times 2^{-2} + 1 = 12 - 1 + 1 = 12$ | M1 | Using $x = 2$ in their non-linear equation |
| So the point $(2, 12)$ lies on the curve. | E1 [6] | Conclusion from their $y$-value FT |

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7 The diagram shows part of a curve which passes through the point $( 1,0 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{5428eabf-431d-4db1-8c25-1f2b9570d9aa-4_711_704_1722_258}

The gradient of the curve is given by $\frac { d y } { d x } = 6 x + \frac { 8 } { x ^ { 3 } }$.\\
Determine whether the curve passes through the point $( 2,12 )$.

\hfill \mbox{\textit{OCR MEI AS Paper 1 2021 Q7 [6]}}
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Q1 2 Q2 2 Q3 2 Q4 4 Q5 5 Q6 8 Q7 6 Q8 12 Q9 9 Q10 10 Q11 10