OCR MEI AS Paper 2 2023 June — Question 14 7 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind tangent at given point (polynomial/algebraic)
DifficultyModerate -0.8 This is a straightforward differentiation and tangent line question requiring standard techniques: rewrite terms as powers, differentiate using power rule, substitute x=4 to find gradient and y-coordinate, then use y-y₁=m(x-x₁). All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

14 In this question you must show detailed reasoning. The equation of a curve is \(y = 16 \sqrt { x } + \frac { 8 } { x }\).
Determine the equation of the tangent to the curve at the point where \(x = 4\).

Question 14:
AnswerMarks Guidance
\(y = 16x^{\frac{1}{2}} + 8x^{-1}\)B1 AO 3.1a
\(\frac{dy}{dx} = 8x^{-\frac{1}{2}} - 8x^{-2}\)M1 AO 1.1
A1 AO 1.1
\(x = 4, \frac{dy}{dx} = \frac{7}{2}\)B1FT AO 1.1
\(x = 4, y = 34\)B1 AO 1.1
\(y - \text{their } 34 = \left(\text{their } \frac{7}{2}\right)(x - 4)\) oeM1FT AO 1.1
\(y = \frac{7}{2}x + 20\) oeA1 AO 3.2a
## Question 14:
$y = 16x^{\frac{1}{2}} + 8x^{-1}$ | **B1** | AO 3.1a | May be implied by correct derivative

$\frac{dy}{dx} = 8x^{-\frac{1}{2}} - 8x^{-2}$ | **M1** | AO 1.1 | At least one term of the form $\alpha x^{-\frac{1}{2}}$ or $\beta x^{-2}$ obtained

— | **A1** | AO 1.1 | All correct

$x = 4, \frac{dy}{dx} = \frac{7}{2}$ | **B1FT** | AO 1.1 | FT their $\frac{dy}{dx}$, dep on **M1**

$x = 4, y = 34$ | **B1** | AO 1.1 |

$y - \text{their } 34 = \left(\text{their } \frac{7}{2}\right)(x - 4)$ oe | **M1FT** | AO 1.1 | Their 7/2 must come from substituting $x = 4$ into their derivative

$y = \frac{7}{2}x + 20$ oe | **A1** | AO 3.2a | All correct. Accept any form: $7x - 2y + 40 = 0$ or $y - 34 = \frac{7}{2}(x-4)$ oe. NOTE: Final answer can be obtained from incorrect working — check their derivative.

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14 In this question you must show detailed reasoning.
The equation of a curve is $y = 16 \sqrt { x } + \frac { 8 } { x }$.\\
Determine the equation of the tangent to the curve at the point where $x = 4$.

\hfill \mbox{\textit{OCR MEI AS Paper 2 2023 Q14 [7]}}