OCR PURE — Question 3 7 marks

Exam BoardOCR
ModulePURE
Marks7
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyModerate -0.8 This is a straightforward application of differentiation to find a normal line. It requires finding dy/dx, evaluating at x=4, finding the normal gradient (negative reciprocal), then using point-slope form. All steps are routine with no conceptual challenges beyond standard C1/C2 material.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations
3 In this question you must show detailed reasoning. Find the equation of the normal to the curve \(y = 4 \sqrt { x } - 3 x + 1\) at the point on the curve where \(x = 4\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 3\)M1, A1 M1 for attempt to differentiate (answer of form \(\frac{dy}{dx} = \frac{k}{\sqrt{x}} - 3\)); A1 for correct derivative. A0 if "\(+c\)"
When \(x = 4\), \(\frac{dy}{dx} = -2\)A1 Correct value of \(\frac{dy}{dx}\)
Gradient of normal \(= \frac{1}{2}\)B1 FT Follow through their evaluated \(\frac{dy}{dx}\); must be processed correctly
When \(x = 4\), \(y = -3\)B1 Correct \(y\) coordinate, accept equivalent forms
\(y + 3 = \frac{1}{2}(x - 4)\)M1 Correct method for equation of straight line through \((4, \text{their } y)\), any non-zero gradient
\(x - 2y - 10 = 0\)A1 Correct equation in required form i.e. \(k(x - 2y - 10) = 0\) for integer \(k\). Must have \(= 0\)
[7 marks]
# Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 3$ | M1, A1 | M1 for attempt to differentiate (answer of form $\frac{dy}{dx} = \frac{k}{\sqrt{x}} - 3$); A1 for correct derivative. A0 if "$+c$" |
| When $x = 4$, $\frac{dy}{dx} = -2$ | A1 | Correct value of $\frac{dy}{dx}$ |
| Gradient of normal $= \frac{1}{2}$ | B1 FT | Follow through their evaluated $\frac{dy}{dx}$; must be processed correctly |
| When $x = 4$, $y = -3$ | B1 | Correct $y$ coordinate, accept equivalent forms |
| $y + 3 = \frac{1}{2}(x - 4)$ | M1 | Correct method for equation of straight line through $(4, \text{their } y)$, any non-zero gradient |
| $x - 2y - 10 = 0$ | A1 | Correct equation in required form i.e. $k(x - 2y - 10) = 0$ for integer $k$. Must have $= 0$ |

**[7 marks]**

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3 In this question you must show detailed reasoning.
Find the equation of the normal to the curve $y = 4 \sqrt { x } - 3 x + 1$ at the point on the curve where $x = 4$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.

\hfill \mbox{\textit{OCR PURE  Q3 [7]}}
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