| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify factor then solve related equation |
| Difficulty | Moderate -0.3 This is a structured multi-part question that guides students through standard C1 techniques: Factor Theorem verification (substituting x=-3), polynomial division, differentiation of polynomials, finding stationary points, and using the second derivative test. While it has many parts (9 marks total), each step is routine and heavily scaffolded, requiring only direct application of learned procedures rather than problem-solving insight. Slightly easier than average due to the extensive guidance provided. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(-3) = (-3)^3 - 4 \times (-3) + 15\) | M1 | f(−3) attempted not long division |
| \(f(-3) = -27 + 12 + 15 = 0 \Rightarrow x + 3\) is a factor | A1 | shown = 0 plus statement; 2 marks total |
| (ii) Quadratic factor \(\left(x^2 - 3x + 5\right)\) | M1 | −3x or +5 term by inspection or full long division attempt |
| \(\left(f(x) = \right) (x + 3)(x^2 - 3x + 5)\) | A1 | must see correct product; 2 marks total |
| (b)(i) \(\left(\frac{dy}{dx} =\right) 4x^3 - 16x + 60\) | M1 | one of these terms correct |
| A1 | another term correct | |
| A1 | all correct (no +c etc); 3 marks total | |
| (ii) \(4x^3 - 16x + 60 = 0\) | must see this line OE | |
| \(\Rightarrow x^3 - 4x + 15 = 0\) | B1 | AG; 1 mark total |
| (iii) Discriminant of quadratic \(= (-3)^2 - 4 \times 5\) | M1 | discriminant of "their" quadratic or correct use of quad eqn "formula" |
| \(b^2 - 4ac = -11\) (or \(b^2 - 4ac < 0\)) therefore quadratic has no (real) roots | A1 | correct discriminant evaluated correctly (or shown to be \(< 0\)) with appropriate conclusion |
| Hence only stationary point is when \(x = -3\) | plus final statement; 2 marks total | |
| (iv) \(\left(\frac{d^2y}{dx^2}\right) = 12x^2 - 16\) | B1 | |
| \(= 12(-3)^2 - 16\) (or \(12 \times 9 - 16\) etc) \(= 92\) | M1 | sub \(x = -3\) into their \(\frac{d^2y}{dx^2}\) |
| A1 | 3 marks total | |
| (v) Minimum since \(\frac{d^2y}{dx^2} > 0\) (or 92 > 0 etc) | E1 | FT appropriate conclusion from their value from (iv) plus reason; 1 mark total |
| treat parts (iv) & (v) holistically |
**(a)** $f(-3) = (-3)^3 - 4 \times (-3) + 15$ | M1 | f(−3) attempted not long division
$f(-3) = -27 + 12 + 15 = 0 \Rightarrow x + 3$ is a factor | A1 | shown = 0 plus statement; 2 marks total
**(ii)** Quadratic factor $\left(x^2 - 3x + 5\right)$ | M1 | −3x or +5 term by inspection or full long division attempt
$\left(f(x) = \right) (x + 3)(x^2 - 3x + 5)$ | A1 | must see correct product; 2 marks total
**(b)(i)** $\left(\frac{dy}{dx} =\right) 4x^3 - 16x + 60$ | M1 | one of these terms correct
| | | A1 | another term correct
| | | A1 | all correct (no +c etc); 3 marks total
**(ii)** $4x^3 - 16x + 60 = 0$ | | must see this line OE
$\Rightarrow x^3 - 4x + 15 = 0$ | B1 | AG; 1 mark total
**(iii)** Discriminant of quadratic $= (-3)^2 - 4 \times 5$ | M1 | discriminant of "their" quadratic or correct use of quad eqn "formula"
$b^2 - 4ac = -11$ (or $b^2 - 4ac < 0$) therefore quadratic has no (real) roots | A1 | correct discriminant evaluated correctly (or shown to be $< 0$) with appropriate conclusion
| | | Hence only stationary point is when $x = -3$ | | plus final statement; 2 marks total
**(iv)** $\left(\frac{d^2y}{dx^2}\right) = 12x^2 - 16$ | B1 |
$= 12(-3)^2 - 16$ (or $12 \times 9 - 16$ etc) $= 92$ | M1 | sub $x = -3$ into their $\frac{d^2y}{dx^2}$
| | | A1 | 3 marks total
**(v)** Minimum since $\frac{d^2y}{dx^2} > 0$ (or 92 > 0 etc) | E1 | FT appropriate conclusion from their value from (iv) plus reason; 1 mark total
| | | | | treat parts (iv) & (v) holistically
**Total: 14 marks**
---4
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is given by $\mathrm { f } ( x ) = x ^ { 3 } - 4 x + 15$.
\begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $x + 3$ is a factor of $\mathrm { f } ( x )$.
\item Express $\mathrm { f } ( x )$ in the form $( x + 3 ) \left( x ^ { 2 } + p x + q \right)$, where $p$ and $q$ are integers.
\end{enumerate}\item A curve has equation $y = x ^ { 4 } - 8 x ^ { 2 } + 60 x + 7$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that the $x$-coordinates of any stationary points of the curve satisfy the equation
$$x ^ { 3 } - 4 x + 15 = 0$$
\item Use the results above to show that the only stationary point of the curve occurs when $x = - 3$.
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ when $x = - 3$.
\item Hence determine, with a reason, whether the curve has a maximum point or a minimum point when $x = - 3$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q4 [14]}}