Question 1 - A-Level Maths

Edexcel C1 2007 January — Question 1 4 marks

Exam Board	Edexcel
Module	C1 (Core Mathematics 1)
Year	2007
Session	January
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Tangents, normals and gradients
Type	Find derivative after algebraic simplification (fractional/mixed powers)
Difficulty	Easy -1.8 This is a straightforward application of the power rule for differentiation requiring only direct recall of standard formulas. Each term differentiates independently with no algebraic manipulation, problem-solving, or conceptual insight needed—purely mechanical execution of a basic technique taught early in C1.
Spec	1.07i Differentiate x^n: for rational n and sums

Given that

$$y = 4 x ^ { 3 } - 1 + 2 x ^ { \frac { 1 } { 2 } } , \quad x > 0 ,$$ find $\frac { \mathrm { d } y } { \mathrm {~d} x }$. \includegraphics[max width=\textwidth, alt={}, center]{fff086fd-f5d8-45b7-8db1-8b22ba5aab31-02_29_45_2690_1852}

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
$4x^3 \to kx^2$ or $2x^2 \to kx^{-\frac{1}{2}}$	M1	(k a non-zero constant)
$12x^2 + x^{-\frac{1}{2}} \ldots$	A1, A1, B1	(−1 → 0)

Total: 4 marks

Guidance:

- Accept equivalent alternatives to $x^{-\frac{1}{2}}$, e.g. $\frac{1}{x^2}$, $\frac{1}{\sqrt{x}}$, $x^{-0.5}$

- M1: $4x^3$ 'differentiated' to give $kx^2$, or $2x^2$ 'differentiated' to give $kx^{-\frac{1}{2}}$ (but not for just −1 → 0)

- 1st A1: $12x^2$ (Do not allow just $3 \times 4x^2$)

- 2nd A1: $x^{-\frac{1}{2}}$ or equivalent (Do not allow just $\frac{1}{2} \times 2x^{-\frac{1}{2}}$, but allow $1x^{-\frac{1}{2}}$ or $\frac{2}{2} \times x^{-\frac{1}{2}}$)

- B1: −1 differentiated to give zero (or 'disappearing'). Can be given provided that at least one of the other terms has been changed. Adding an extra term, e.g. + C, is B0.

$4x^3 \to kx^2$ or $2x^2 \to kx^{-\frac{1}{2}}$ | M1 | (k a non-zero constant)

$12x^2 + x^{-\frac{1}{2}} \ldots$ | A1, A1, B1 | (−1 → 0)

**Total: 4 marks**

**Guidance:**
- Accept equivalent alternatives to $x^{-\frac{1}{2}}$, e.g. $\frac{1}{x^2}$, $\frac{1}{\sqrt{x}}$, $x^{-0.5}$
- M1: $4x^3$ 'differentiated' to give $kx^2$, or $2x^2$ 'differentiated' to give $kx^{-\frac{1}{2}}$ (but not for just −1 → 0)
- 1st A1: $12x^2$ (Do not allow just $3 \times 4x^2$)
- 2nd A1: $x^{-\frac{1}{2}}$ or equivalent (Do not allow just $\frac{1}{2} \times 2x^{-\frac{1}{2}}$, but allow $1x^{-\frac{1}{2}}$ or $\frac{2}{2} \times x^{-\frac{1}{2}}$)
- B1: −1 differentiated to give zero (or 'disappearing'). Can be given provided that at least one of the other terms has been changed. Adding an extra term, e.g. + C, is B0.

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Show LaTeX source

\begin{enumerate}
  \item Given that
\end{enumerate}

$$y = 4 x ^ { 3 } - 1 + 2 x ^ { \frac { 1 } { 2 } } , \quad x > 0 ,$$

find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\

\includegraphics[max width=\textwidth, alt={}, center]{fff086fd-f5d8-45b7-8db1-8b22ba5aab31-02_29_45_2690_1852}\\

\hfill \mbox{\textit{Edexcel C1 2007 Q1 [4]}}

This paper (10 questions)

View full paper

Q1 4 Q2 4 Q3 6 Q4 7 Q5 4 Q6 5 Q7 9 Q8 11 Q9 12 Q10 13

Edexcel C1 2007 January — Question 1 4 marks

Total: 4 marks

Guidance:

- Accept equivalent alternatives to \(x^{-\frac{1}{2}}\), e.g. \(\frac{1}{x^2}\), \(\frac{1}{\sqrt{x}}\), \(x^{-0.5}\)

- M1: \(4x^3\) 'differentiated' to give \(kx^2\), or \(2x^2\) 'differentiated' to give \(kx^{-\frac{1}{2}}\) (but not for just −1 → 0)

- 1st A1: \(12x^2\) (Do not allow just \(3 \times 4x^2\))

- 2nd A1: \(x^{-\frac{1}{2}}\) or equivalent (Do not allow just \(\frac{1}{2} \times 2x^{-\frac{1}{2}}\), but allow \(1x^{-\frac{1}{2}}\) or \(\frac{2}{2} \times x^{-\frac{1}{2}}\))

- B1: −1 differentiated to give zero (or 'disappearing'). Can be given provided that at least one of the other terms has been changed. Adding an extra term, e.g. + C, is B0.

This paper (10 questions)

Answer	Marks	Guidance
\(4x^3 \to kx^2\) or \(2x^2 \to kx^{-\frac{1}{2}}\)	M1	(k a non-zero constant)
\(12x^2 + x^{-\frac{1}{2}} \ldots\)	A1, A1, B1	(−1 → 0)