Question 6 - A-Level Maths

AQA C2 2013 June — Question 6 12 marks

Exam Board	AQA
Module	C2 (Core Mathematics 2)
Year	2013
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Indices and Surds
Type	Differentiate after index conversion
Difficulty	Moderate -0.3 This is a straightforward C2 differentiation question requiring index law manipulation (converting to powers), basic differentiation rules, and finding a normal line. Part (a) is routine algebra, parts (b)(i)-(ii) are standard applications, and part (b)(iii) involves solving dy/dx=0 and expressing the answer in a given form—all well-practiced techniques with no novel insight required. Slightly easier than average due to its highly structured, step-by-step nature.
Spec	1.07i Differentiate x^n: for rational n and sums 1.07m Tangents and normals: gradient and equations 1.07n Stationary points: find maxima, minima using derivatives

6 A curve has the equation $$y = \frac { 12 + x ^ { 2 } \sqrt { x } } { x } , \quad x > 0$$

Express $\frac { 12 + x ^ { 2 } \sqrt { x } } { x }$ in the form $12 x ^ { p } + x ^ { q }$.
1. Hence find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
2. Find an equation of the normal to the curve at the point on the curve where $x = 4$.
3. The curve has a stationary point $P$. Show that the $x$-coordinate of $P$ can be written in the form $2 ^ { k }$, where $k$ is a rational number.

Show mark scheme Show mark scheme source

6(a)

Answer	Marks	Guidance
$\sqrt{x} = x^{0.5}$	B1	$\sqrt{x} = x^{0.5}$ or $\sqrt{x} = x^{\frac{1}{2}}$ seen or used
$\frac{12 + x^2\sqrt{x}}{x} = \frac{12 + x^{2.5}}{x}$	B1	$12x^{-1}$ or $p = -1$
$= 12x^{-1} + x^{1.5}$	B1	$x^{1.5}$ or $q = \frac{3}{2}$ (=1.5)

6(b)(i)

Answer	Marks	Guidance
$\frac{dy}{dx} = -12x^{-2} + 1.5x^{0.5}$	B1F	Ft on c's $p$ only if c's $p$ is a negative integer
B1F	Ft on c's $q$ only if c's $q$ is a pos non-integer	2

6(b)(ii)

Answer	Marks	Guidance
When $x = 4$, $y = 11$	B1
When $x = 4$, $\frac{dy}{dx} = \frac{-12}{16} + 3 = \frac{9}{4}$	M1	Attempt to find $\frac{dy}{dx}$ when $x = 4$ PI
Gradient of normal = $-\frac{4}{9}$	m1	$m \times m' = -1$ used
Eqn of normal: $y - 11 = -\frac{4}{9}(x - 4)$	A1	ACF eg 4x + 9y = 115

6(b)(iii)

Answer	Marks	Guidance
At St Pt $\frac{dy}{dx} = -12x^{-2} + 1.5x^{0.5} = 0$	M1	Equating c's $\frac{dy}{dx}$ to zero.
$\Rightarrow x^2 x^{0.5} = 8, \Rightarrow x^{\frac{5}{2}} = 8 \Rightarrow x = 8^{\frac{2}{5}}$	A1	A correct eqn in the form $x^n = c$ or $x = c^n$ correctly obtained.
$\Rightarrow x = (2^3)^{\frac{2}{5}} \Rightarrow x = 2^{\frac{6}{5}}$	A1	CSO $x = 2^{\frac{6}{5}}$. All working must be correct and in an exact form. If 'x=0' also appears then A0 CSO
Total		12

**6(a)**
$\sqrt{x} = x^{0.5}$ | B1 | $\sqrt{x} = x^{0.5}$ or $\sqrt{x} = x^{\frac{1}{2}}$ seen or used
$\frac{12 + x^2\sqrt{x}}{x} = \frac{12 + x^{2.5}}{x}$ | B1 | $12x^{-1}$ or $p = -1$
$= 12x^{-1} + x^{1.5}$ | B1 | $x^{1.5}$ or $q = \frac{3}{2}$ (=1.5) | 3

**6(b)(i)**
$\frac{dy}{dx} = -12x^{-2} + 1.5x^{0.5}$ | B1F | Ft on c's $p$ only if c's $p$ is a negative integer
 | B1F | Ft on c's $q$ only if c's $q$ is a pos non-integer | 2

**6(b)(ii)**
When $x = 4$, $y = 11$ | B1 | 
When $x = 4$, $\frac{dy}{dx} = \frac{-12}{16} + 3 = \frac{9}{4}$ | M1 | Attempt to find $\frac{dy}{dx}$ when $x = 4$ PI
Gradient of normal = $-\frac{4}{9}$ | m1 | $m \times m' = -1$ used
Eqn of normal: $y - 11 = -\frac{4}{9}(x - 4)$ | A1 | ACF eg 4x + 9y = 115 | 4

**6(b)(iii)**
At St Pt $\frac{dy}{dx} = -12x^{-2} + 1.5x^{0.5} = 0$ | M1 | Equating c's $\frac{dy}{dx}$ to zero.
$\Rightarrow x^2 x^{0.5} = 8, \Rightarrow x^{\frac{5}{2}} = 8 \Rightarrow x = 8^{\frac{2}{5}}$ | A1 | A correct eqn in the form $x^n = c$ or $x = c^n$ correctly obtained.
$\Rightarrow x = (2^3)^{\frac{2}{5}} \Rightarrow x = 2^{\frac{6}{5}}$ | A1 | CSO $x = 2^{\frac{6}{5}}$. All working must be correct and in an exact form. If 'x=0' also appears then A0 CSO | 3

Total | | 12

---

Show LaTeX source

6 A curve has the equation

$$y = \frac { 12 + x ^ { 2 } \sqrt { x } } { x } , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 12 + x ^ { 2 } \sqrt { x } } { x }$ in the form $12 x ^ { p } + x ^ { q }$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find an equation of the normal to the curve at the point on the curve where $x = 4$.
\item The curve has a stationary point $P$. Show that the $x$-coordinate of $P$ can be written in the form $2 ^ { k }$, where $k$ is a rational number.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C2 2013 Q6 [12]}}

This paper (9 questions)

View full paper

Q1 5 Q2 8 Q3 9 Q4 5 Q5 9 Q6 12 Q7 6 Q8 7 Q9 14

Answer	Marks	Guidance
\(\sqrt{x} = x^{0.5}\)	B1	\(\sqrt{x} = x^{0.5}\) or \(\sqrt{x} = x^{\frac{1}{2}}\) seen or used
\(\frac{12 + x^2\sqrt{x}}{x} = \frac{12 + x^{2.5}}{x}\)	B1	\(12x^{-1}\) or \(p = -1\)
\(= 12x^{-1} + x^{1.5}\)	B1	\(x^{1.5}\) or \(q = \frac{3}{2}\) (=1.5)

Answer	Marks	Guidance
\(\frac{dy}{dx} = -12x^{-2} + 1.5x^{0.5}\)	B1F	Ft on c's \(p\) only if c's \(p\) is a negative integer
B1F	Ft on c's \(q\) only if c's \(q\) is a pos non-integer	2

Answer	Marks	Guidance
When \(x = 4\), \(y = 11\)	B1
When \(x = 4\), \(\frac{dy}{dx} = \frac{-12}{16} + 3 = \frac{9}{4}\)	M1	Attempt to find \(\frac{dy}{dx}\) when \(x = 4\) PI
Gradient of normal = \(-\frac{4}{9}\)	m1	\(m \times m' = -1\) used
Eqn of normal: \(y - 11 = -\frac{4}{9}(x - 4)\)	A1	ACF eg 4x + 9y = 115

Answer	Marks	Guidance
At St Pt \(\frac{dy}{dx} = -12x^{-2} + 1.5x^{0.5} = 0\)	M1	Equating c's \(\frac{dy}{dx}\) to zero.
\(\Rightarrow x^2 x^{0.5} = 8, \Rightarrow x^{\frac{5}{2}} = 8 \Rightarrow x = 8^{\frac{2}{5}}\)	A1	A correct eqn in the form \(x^n = c\) or \(x = c^n\) correctly obtained.
\(\Rightarrow x = (2^3)^{\frac{2}{5}} \Rightarrow x = 2^{\frac{6}{5}}\)	A1	CSO \(x = 2^{\frac{6}{5}}\). All working must be correct and in an exact form. If 'x=0' also appears then A0 CSO
Total		12