AQA C2 2005 June — Question 7 9 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2005
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeEquation of normal line
DifficultyModerate -0.8 This is a straightforward multi-part question requiring basic algebraic manipulation (simplifying indices), routine differentiation using power rule, and finding a normal gradient. All techniques are standard C2 level with no problem-solving insight needed—easier than average A-level questions.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

7 A curve is defined, for \(x > 0\), by the equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \frac { x ^ { 8 } - 1 } { x ^ { 3 } }$$
  1. Express \(\frac { x ^ { 8 } - 1 } { x ^ { 3 } }\) in the form \(x ^ { p } - x ^ { q }\), where \(p\) and \(q\) are integers.
    1. Hence differentiate \(\mathrm { f } ( x )\) to find \(\mathrm { f } ^ { \prime } ( x )\).
    2. Hence show that f is an increasing function.
  2. Find the gradient of the normal to the curve at the point \(( 1,0 )\).

AnswerMarks Guidance
7(a)........ = \(x^2 - x^{-3}\) M1, A1
7(b)(i)\(f'(x) = \frac{5x^4 + 3x^{-4}}{}\) B1 ft, B1 ft
7(b)(ii)\(f'(x) = \{5x^4 + \frac{3}{x^4}\} > 0\) ⟹ f is increasing {function} M1, A1
7(c)At (1,0), f'(1) = 5 + 3 = 8 ⟹ grad. of normal = \(-\frac{1}{8}\) M1, m1, A1 ft
7(a) | ........ = $x^2 - x^{-3}$ | M1, A1 | One power correct; Accept p = 5, q = −3 |

7(b)(i) | $f'(x) = \frac{5x^4 + 3x^{-4}}{}$ | B1 ft, B1 ft | ft on $px^{p-1}$; ft on $-qx^{q-1}$ provided q < 0 |

7(b)(ii) | $f'(x) = \{5x^4 + \frac{3}{x^4}\} > 0$ ⟹ f is increasing {function} | M1, A1 | M1 Considers sign of f'(x); a statement "f'(x) > 0" OE " with 'f increasing'; A1 needs f'(x) of the form $ax^c + \frac{b}{x^d}$, where a and b both ≥ 0 and no incorrect statements based on f'(x) at different values of x |

7(c) | At (1,0), f'(1) = 5 + 3 = 8 ⟹ grad. of normal = $-\frac{1}{8}$ | M1, m1, A1 ft | Attempts to find f'(1); Use of $m \times m' = -1$ PI; ft on wrong f'(x) |
7 A curve is defined, for $x > 0$, by the equation $y = \mathrm { f } ( x )$, where

$$\mathrm { f } ( x ) = \frac { x ^ { 8 } - 1 } { x ^ { 3 } }$$
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { x ^ { 8 } - 1 } { x ^ { 3 } }$ in the form $x ^ { p } - x ^ { q }$, where $p$ and $q$ are integers.
\item \begin{enumerate}[label=(\roman*)]
\item Hence differentiate $\mathrm { f } ( x )$ to find $\mathrm { f } ^ { \prime } ( x )$.
\item Hence show that f is an increasing function.
\end{enumerate}\item Find the gradient of the normal to the curve at the point $( 1,0 )$.
\end{enumerate}

\hfill \mbox{\textit{AQA C2 2005 Q7 [9]}}