| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constants using remainder theorem |
| Difficulty | Moderate -0.3 This is a standard multi-part C4 question testing routine application of the Remainder Theorem, polynomial factorization, and differentiation. Part (a) is direct substitution, part (b)(i) requires algebraic division but with a given root, and part (b)(ii) involves simplification and showing a function is decreasing—all textbook techniques with no novel insight required. Slightly easier than average due to the scaffolded structure and given information. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f\!\left(-\frac{1}{2}\right) = -2\): \(8\left(-\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) - 2\left(-\frac{1}{2}\right) + d = -2\) | M1 | Remainder theorem applied correctly |
| \(-1 - 3 + 1 + d = -2\), so \(d = 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2x+1)\) is a factor of \(g(x)\) | B1 | Stated or implied |
| \(g(x) = (2x+1)(4x^2 - 8x + 3)\) | M1 | Attempting to find quadratic factor |
| \(= (2x+1)(2x-1)(2x-3)\) | A1 | All three linear factors correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h(x) = \frac{(2x-1)(2x+1)}{(2x+1)(2x-1)(2x-3)} = \frac{1}{2x-3}\) | M1 A1 | Correct simplification |
| \(h'(x) = \frac{-2}{(2x-3)^2}\) | M1 | Differentiating |
| Since \((2x-3)^2 > 0\) for \(x>2\), \(h'(x) < 0\), so h is decreasing | A1 | Complete argument |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f\!\left(-\frac{1}{2}\right) = -2$: $8\left(-\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) - 2\left(-\frac{1}{2}\right) + d = -2$ | M1 | Remainder theorem applied correctly |
| $-1 - 3 + 1 + d = -2$, so $d = 1$ | A1 | |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2x+1)$ is a factor of $g(x)$ | B1 | Stated or implied |
| $g(x) = (2x+1)(4x^2 - 8x + 3)$ | M1 | Attempting to find quadratic factor |
| $= (2x+1)(2x-1)(2x-3)$ | A1 | All three linear factors correct |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h(x) = \frac{(2x-1)(2x+1)}{(2x+1)(2x-1)(2x-3)} = \frac{1}{2x-3}$ | M1 A1 | Correct simplification |
| $h'(x) = \frac{-2}{(2x-3)^2}$ | M1 | Differentiating |
| Since $(2x-3)^2 > 0$ for $x>2$, $h'(x) < 0$, so h is decreasing | A1 | Complete argument |
I can see these are exam answer space pages (P/Jun15/MPC4) but they don't contain mark scheme content - they are blank answer spaces for students to write in. The actual mark scheme content is not shown in these images.
To get the mark scheme for these questions, I can however work out the answers from the questions shown:
---3
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 8 x ^ { 3 } - 12 x ^ { 2 } - 2 x + d$, where $d$ is a constant. When $\mathrm { f } ( x )$ is divided by ( $2 x + 1$ ), the remainder is - 2 . Use the Remainder Theorem to find the value of $d$.
\item The polynomial $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 8 x ^ { 3 } - 12 x ^ { 2 } - 2 x + 3$.
\begin{enumerate}[label=(\roman*)]
\item Given that $x = - \frac { 1 } { 2 }$ is a solution of the equation $\mathrm { g } ( x ) = 0$, write $\mathrm { g } ( x )$ as a product of three linear factors.
\item The function h is defined by $\mathrm { h } ( x ) = \frac { 4 x ^ { 2 } - 1 } { \mathrm {~g} ( x ) }$ for $x > 2$.
Simplify $\mathrm { h } ( x )$, and hence show that h is a decreasing function.\\[0pt]
[4 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2015 Q3 [9]}}