Moderate -0.3 This is a straightforward application of the quotient rule to find dy/dx, evaluate at x=2, find the negative reciprocal for the normal gradient, then write the line equation. It's slightly easier than average because the point is given (no need to verify), the algebra is clean, and it's a standard textbook exercise with no conceptual challenges beyond routine differentiation and line equations.
5 A curve has the equation \(\mathrm { y } = \frac { 3 } { 2 \mathrm { x } ^ { 2 } - 5 }\).
Find the equation of the normal to the curve at the point \(( 2,1 )\), giving your answer in the form \(\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0\), where \(a , b\) and \(c\) are integers.
Substitute \((2,1)\) to obtain gradient \(-\frac{24}{9}\)
A1
OE e.g. \(-\frac{8}{3}\). Allow \(-2.67\).
Apply negative reciprocal to *their* numerical gradient to obtain gradient of normal
\*M1
Must have been some attempt at differentiation. Expect \(\frac{3}{8}\)
Attempt equation of normal using *their* gradient of the normal and \((2,1)\)
DM1
Expect \(y-1=\frac{3}{8}(x-2)\)
Obtain \(3x-8y+2=0\) (allow multiples)
A1
Or equivalent of requested form e.g. \(8y-3x-2=0\)
Total: 6
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain form $kx(2x^2-5)^{-2}$ | M1 | |
| Obtain correct $-12x(2x^2-5)^{-2}$ | A1 | OE |
| Substitute $(2,1)$ to obtain gradient $-\frac{24}{9}$ | A1 | OE e.g. $-\frac{8}{3}$. Allow $-2.67$. |
| Apply negative reciprocal to *their* numerical gradient to obtain gradient of normal | \*M1 | Must have been some attempt at differentiation. Expect $\frac{3}{8}$ |
| Attempt equation of normal using *their* gradient of the normal and $(2,1)$ | DM1 | Expect $y-1=\frac{3}{8}(x-2)$ |
| Obtain $3x-8y+2=0$ (allow multiples) | A1 | Or equivalent of requested form e.g. $8y-3x-2=0$ |
| **Total: 6** | | |
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5 A curve has the equation $\mathrm { y } = \frac { 3 } { 2 \mathrm { x } ^ { 2 } - 5 }$.\\
Find the equation of the normal to the curve at the point $( 2,1 )$, giving your answer in the form $\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0$, where $a , b$ and $c$ are integers.\\
\hfill \mbox{\textit{CAIE P1 2024 Q5 [6]}}