| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on the Factor and Remainder Theorem. Part (a) involves setting up two simultaneous equations using p(-2)=0 and p(2)=24, which is routine. Part (b) requires factorising a cubic and checking discriminant of the quadratic factor - standard technique. Part (c) involves solving a trigonometric equation after finding the root, which is mechanical substitution. All parts follow predictable patterns with no novel insight required, making this slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05o Trigonometric equations: solve in given intervals4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(x = -2\) and equate to zero | M1 | \(-8a + 4b + 2a + 8 = 0\) |
| Substitute \(x = 2\) and equate to \(24\) | M1 | \(8a + 4b - 2a + 8 = 24\) |
| Obtain \(-6a + 4b + 8 = 0\) and \(6a + 4b - 16 = 0\) | A1 | OE |
| Obtain \(a = 2\) and \(b = 1\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Divide by \(x + 2\) at least as far as the \(x\) term | M1 | OE |
| Obtain \((x+2)(2x^2 - 3x + 4)\) | A1 | SOI |
| Conclude with reference to root \(-2\), discriminant is \(-23\) and no further root | A1 | Or complete equivalent. |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(\cosec\theta = -4\) | B1 | May be implied by \(\sin\theta = -\frac{1}{4}\) |
| Attempt to find at least one value of \(\theta\) from \(\sin\theta = \pm\frac{1}{4}\) | M1 | Allow for \(14.5\), \(14.4\). |
| Obtain \(-14.5\) only and no others in the range | A1 | Or greater accuracy (\(14.4775\ldots\)). |
| Total | 3 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = -2$ and equate to zero | M1 | $-8a + 4b + 2a + 8 = 0$ |
| Substitute $x = 2$ and equate to $24$ | M1 | $8a + 4b - 2a + 8 = 24$ |
| Obtain $-6a + 4b + 8 = 0$ and $6a + 4b - 16 = 0$ | A1 | OE |
| Obtain $a = 2$ and $b = 1$ | A1 | |
| **Total** | **4** | |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Divide by $x + 2$ at least as far as the $x$ term | M1 | OE |
| Obtain $(x+2)(2x^2 - 3x + 4)$ | A1 | SOI |
| Conclude with reference to root $-2$, discriminant is $-23$ and no further root | A1 | Or complete equivalent. |
| **Total** | **3** | |
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $\cosec\theta = -4$ | B1 | May be implied by $\sin\theta = -\frac{1}{4}$ |
| Attempt to find at least one value of $\theta$ from $\sin\theta = \pm\frac{1}{4}$ | M1 | Allow for $14.5$, $14.4$. |
| Obtain $-14.5$ only and no others in the range | A1 | Or greater accuracy ($14.4775\ldots$). |
| **Total** | **3** | |
---5 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = a x ^ { 3 } + b x ^ { 2 } - a x + 8$$
where $a$ and $b$ are constants.It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$ ,and that the remainder is 24 when $\mathrm { p } ( x )$ is divided by $( x - 2 )$ .
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$ .\\
\includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-09_2723_35_101_20}
\item Factorise $\mathrm { p } ( x )$ and hence show that the equation $\mathrm { p } ( x ) = 0$ has exactly one real root.
\item Solve the equation $\mathrm { p } \left( \frac { 1 } { 2 } \operatorname { cosec } \theta \right) = 0$ for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-10_499_696_264_680}
The diagram shows the curves with equations $y = \sqrt [ 3 ] { 5 x ^ { 2 } + 7 }$ and $y = \frac { 27 } { 2 x + 5 }$ for $x \geqslant 0$.\\
The curves meet at the point $( 2,3 )$.\\
Region $A$ is bounded by the curve $y = \sqrt [ 3 ] { 5 x ^ { 2 } + 7 }$ and the straight lines $x = 0 , x = 2$ and $y = 0$.\\
Region $B$ is bounded by the two curves and the straight line $x = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2024 Q5 [10]}}