Edexcel C1 — Question 8 5 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeDifferentiate after index conversion
DifficultyModerate -0.8 This is a straightforward C1 differentiation question requiring algebraic manipulation (expanding and simplifying with negative indices) followed by term-by-term differentiation using the power rule. Part (a) is routine algebra, and part (b) applies a standard technique with no problem-solving required. Easier than average for A-level.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07i Differentiate x^n: for rational n and sums

8. $$f ( x ) = \frac { \left( x ^ { 2 } - 3 \right) ^ { 2 } } { x ^ { 3 } } , x \neq 0$$
  1. Show that \(\mathrm { f } ( x ) \equiv x - 6 x ^ { - 1 } + 9 x ^ { - 3 }\).
  2. Hence, or otherwise, differentiate \(\mathrm { f } ( x )\) with respect to \(x\). END

\(f(x) = \frac{(x^2 – 3)^2}{x^3}\), \(x \neq 0\).
(a) Show that \(f(x) \equiv x – 6x^{-1} + 9x^{-3}\).
(2)
(b) Hence, or otherwise, differentiate \(f(x)\) with respect to \(x\).
(3)
END
$f(x) = \frac{(x^2 – 3)^2}{x^3}$, $x \neq 0$.

(a) Show that $f(x) \equiv x – 6x^{-1} + 9x^{-3}$.
(2)

(b) Hence, or otherwise, differentiate $f(x)$ with respect to $x$.
(3)

END
8.

$$f ( x ) = \frac { \left( x ^ { 2 } - 3 \right) ^ { 2 } } { x ^ { 3 } } , x \neq 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) \equiv x - 6 x ^ { - 1 } + 9 x ^ { - 3 }$.
\item Hence, or otherwise, differentiate $\mathrm { f } ( x )$ with respect to $x$.

END
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q8 [5]}}