| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Constant in parametric equations, find its value |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring dy/dx = (dy/dt)/(dx/dt), setting equal to 1, and algebraic manipulation. Part (b) uses the factor theorem (routine A-level technique), and part (c) requires showing uniqueness via discriminant or factorization. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt use of quotient rule (or equivalent) to find \(\frac{dx}{dt}\) | M1* | |
| Obtain \(\frac{1}{(t+2)^2}\) | A1 | Or (unsimplified) equivalent |
| Equate \(\frac{dy}{dx}\) to 1 | DM1 | Must be using *their* \(\frac{dy}{dx}\) |
| Obtain \((2t+a)(t+2)^2 = 1\) and expand to confirm given result | A1 | AG necessary detail required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(t = -1\), equate to zero and attempt solution for \(a\) | M1 | Allow a complete method using algebraic long division or synthetic division |
| Obtain \(a = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Divide their cubic by \(t+1\) at least as far as the \(x\) term | M1 | Or equivalent (inspection, identity, …) |
| Obtain \(2t^2 + 9t + 11\) | A1 | |
| Calculate discriminant, obtain \(-7\) and conclude no further value of \(t\) | A1 | OE |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of quotient rule (or equivalent) to find $\frac{dx}{dt}$ | M1* | |
| Obtain $\frac{1}{(t+2)^2}$ | A1 | Or (unsimplified) equivalent |
| Equate $\frac{dy}{dx}$ to 1 | DM1 | Must be using *their* $\frac{dy}{dx}$ |
| Obtain $(2t+a)(t+2)^2 = 1$ and expand to confirm given result | A1 | AG necessary detail required |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $t = -1$, equate to zero and attempt solution for $a$ | M1 | Allow a complete method using algebraic long division or synthetic division |
| Obtain $a = 3$ | A1 | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Divide their cubic by $t+1$ at least as far as the $x$ term | M1 | Or equivalent (inspection, identity, …) |
| Obtain $2t^2 + 9t + 11$ | A1 | |
| Calculate discriminant, obtain $-7$ and conclude no further value of $t$ | A1 | OE |7 A curve has parametric equations
$$x = \frac { 2 t + 3 } { t + 2 } , \quad y = t ^ { 2 } + a t + 1$$
where $a$ is a constant. It is given that, at the point $P$ on the curve, the gradient is 1 .
\begin{enumerate}[label=(\alph*)]
\item Show that the value of $t$ at $P$ satisfies the equation
$$2 t ^ { 3 } + ( a + 8 ) t ^ { 2 } + ( 4 a + 8 ) t + 4 a - 1 = 0$$
\item It is given that $( t + 1 )$ is a factor of
$$2 t ^ { 3 } + ( a + 8 ) t ^ { 2 } + ( 4 a + 8 ) t + 4 a - 1$$
Find the value of $a$.
\item Hence show that $P$ is the only point on the curve at which the gradient is 1 .\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q7 [9]}}