Moderate -0.3 This is a straightforward application of the quotient rule to find dy/dx, evaluate at x=1 to get the gradient, find the perpendicular gradient for the normal, then use point-slope form. It's slightly easier than average because it's a standard procedure with no conceptual challenges, though the quotient rule and algebraic manipulation add minor complexity beyond the simplest differentiation questions.
3 Find the equation of the normal to the curve \(y = \frac { x ^ { 2 } + 4 } { x + 2 }\) at the point \(\left( 1 , \frac { 5 } { 3 } \right)\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Condone \(u/v\) muddles but needs \((x+2)^2\) in denominator; or product rule to produce terms involving \((x+2)^{-1}\) and \((x+2)^{-2}\)
Obtain \(\frac{2x(x+2)-(x^2+4)}{(x+2)^2}\)
A1
Or equiv; brackets may be implied by subsequent recovery
Substitute 1 into attempt at first derivative
M1
Also allow if sign slip leads to derivative cancelling to 1
Obtain \(\frac{1}{9}\)
A1
Use \(-9\) as gradient of normal
A1ft
Following their value of first derivative
Attempt to find equation of normal
M1
Not equation of tangent; needs use of negative reciprocal of their derivative value
Obtain \(27x + 3y - 32 = 0\)
A1
Or equiv of requested form
# Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of quotient rule | M1 | Condone $u/v$ muddles but needs $(x+2)^2$ in denominator; or product rule to produce terms involving $(x+2)^{-1}$ and $(x+2)^{-2}$ |
| Obtain $\frac{2x(x+2)-(x^2+4)}{(x+2)^2}$ | A1 | Or equiv; brackets may be implied by subsequent recovery |
| Substitute 1 into attempt at first derivative | M1 | Also allow if sign slip leads to derivative cancelling to 1 |
| Obtain $\frac{1}{9}$ | A1 | |
| Use $-9$ as gradient of normal | A1ft | Following their value of first derivative |
| Attempt to find equation of normal | M1 | Not equation of tangent; needs use of negative reciprocal of their derivative value |
| Obtain $27x + 3y - 32 = 0$ | A1 | Or equiv of requested form |
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3 Find the equation of the normal to the curve $y = \frac { x ^ { 2 } + 4 } { x + 2 }$ at the point $\left( 1 , \frac { 5 } { 3 } \right)$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C3 2012 Q3 [7]}}