OCR C1 2012 June — Question 6 7 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2012
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyModerate -0.3 This is a straightforward application of differentiation to find a normal line. It requires finding dy/dx using the power rule, evaluating at x=2, finding the negative reciprocal for the normal gradient, then using point-slope form. All steps are routine C1 techniques with no problem-solving insight needed, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations
6 Find the equation of the normal to the curve \(y = \frac { 6 } { x ^ { 2 } } - 5\) at the point on the curve where \(x = 2\). Give your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers.
Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = -12x^{-3}\)M1 Attempt to differentiate (i.e. \(kx^{-3}\) seen). "+ C" is A0
A1Correct derivative
When \(x = 2\), \(\frac{dy}{dx} = -\frac{3}{2}\)A1 Correct value of \(\frac{dy}{dx}\). Allow equivalent fractions.
Gradient of normal \(= \frac{2}{3}\)B1 FT Follow through their evaluated \(\frac{dy}{dx}\). Must be processed correctly
When \(x = 2\), \(y = -\frac{7}{2}\)B1 Correct \(y\) coordinate, accept equivalent forms
\(y + \frac{7}{2} = \frac{2}{3}(x - 2)\)M1 Correct equation of straight line through \((2,\) their evaluated \(y)\), any non-zero gradient
\(4x - 6y - 29 = 0\)A1 [7] Correct equation in required form i.e. \(k(4x - 6y - 29) = 0\) for integer \(k\). Must have "\(= 0\)" 
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -12x^{-3}$ | M1 | Attempt to differentiate (i.e. $kx^{-3}$ seen). "+ C" is A0 |
| | A1 | Correct derivative |
| When $x = 2$, $\frac{dy}{dx} = -\frac{3}{2}$ | A1 | Correct value of $\frac{dy}{dx}$. Allow equivalent fractions. |
| Gradient of normal $= \frac{2}{3}$ | B1 FT | Follow through their evaluated $\frac{dy}{dx}$. Must be processed correctly |
| When $x = 2$, $y = -\frac{7}{2}$ | B1 | Correct $y$ coordinate, accept equivalent forms |
| $y + \frac{7}{2} = \frac{2}{3}(x - 2)$ | M1 | Correct equation of straight line through $(2,$ their evaluated $y)$, any non-zero gradient |
| $4x - 6y - 29 = 0$ | A1 [7] | Correct equation in required form i.e. $k(4x - 6y - 29) = 0$ for integer $k$. **Must have** "$= 0$" |

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6 Find the equation of the normal to the curve $y = \frac { 6 } { x ^ { 2 } } - 5$ at the point on the curve where $x = 2$. Give your answer in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.

\hfill \mbox{\textit{OCR C1 2012 Q6 [7]}}
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