Edexcel C1 2010 January — Question 4 7 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2010
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (straightforward integration + point)
DifficultyModerate -0.8 This is a straightforward integration question requiring only basic power rule application and using an initial condition to find the constant. The algebraic manipulation (rewriting √x as x^{1/2} and x√x as x^{3/2}) is routine for C1 level, and finding the constant of integration from given coordinates is a standard textbook exercise with no problem-solving insight required.
Spec1.08b Integrate x^n: where n != -1 and sums

4. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 5 x ^ { - \frac { 1 } { 2 } } + x \sqrt { } x , \quad x > 0$$ Given that \(y = 35\) at \(x = 4\), find \(y\) in terms of \(x\), giving each term in its simplest form.

Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x\sqrt{x} = x^{3/2}\)B1 Seen or implied by correct integration; \(x^{3/2}\) from \(\sqrt{x}\) is B0
\(x^{-1/2} \to kx^{1/2}\) or \(x^{3/2} \to kx^{5/2}\) (\(k\) a non-zero constant)M1
\((y=)\ \frac{5x^{1/2}}{1/2} \ldots + \frac{x^{5/2}}{5/2}\ (+C)\)A1, A1 "y=" and "+C" not required for these marks; 1st A1: any unsimplified/simplified correct form e.g. \(\frac{5\sqrt{x}}{0.5}\); 2nd A1: e.g. \(\frac{x^2\sqrt{x}}{2.5}\)
\(35 = \frac{5\times 4^{1/2}}{1/2} + \frac{4^{5/2}}{5/2} + C\) (with terms simplified or unsimplified)M1 Attempting to use \(x=4\) and \(y=35\) in a changed function to form equation in \(C\)
\(C = \frac{11}{5}\) or equivalent \(2\frac{1}{5}\), \(2.2\)A1 Obtaining \(C=\frac{11}{5}\) with no earlier incorrect work
\(y = 10x^{1/2} + \frac{2x^{5/2}}{5} + \frac{11}{5}\) (or equivalent simplified)A1 ft Final A mark requires equation "\(y=\ldots\)" with correct \(x\) terms; follow-through on value of \(C\) only
Total: [7]
## Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\sqrt{x} = x^{3/2}$ | B1 | Seen or implied by correct integration; $x^{3/2}$ from $\sqrt{x}$ is B0 |
| $x^{-1/2} \to kx^{1/2}$ or $x^{3/2} \to kx^{5/2}$ ($k$ a non-zero constant) | M1 | |
| $(y=)\ \frac{5x^{1/2}}{1/2} \ldots + \frac{x^{5/2}}{5/2}\ (+C)$ | A1, A1 | "y=" and "+C" not required for these marks; 1st A1: any unsimplified/simplified correct form e.g. $\frac{5\sqrt{x}}{0.5}$; 2nd A1: e.g. $\frac{x^2\sqrt{x}}{2.5}$ |
| $35 = \frac{5\times 4^{1/2}}{1/2} + \frac{4^{5/2}}{5/2} + C$ (with terms simplified or unsimplified) | M1 | Attempting to use $x=4$ and $y=35$ in a changed function to form equation in $C$ |
| $C = \frac{11}{5}$ or equivalent $2\frac{1}{5}$, $2.2$ | A1 | Obtaining $C=\frac{11}{5}$ with no earlier incorrect work |
| $y = 10x^{1/2} + \frac{2x^{5/2}}{5} + \frac{11}{5}$ (or equivalent simplified) | A1 ft | Final A mark requires equation "$y=\ldots$" with correct $x$ terms; follow-through on value of $C$ only |

**Total: [7]**
4.

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 5 x ^ { - \frac { 1 } { 2 } } + x \sqrt { } x , \quad x > 0$$

Given that $y = 35$ at $x = 4$, find $y$ in terms of $x$, giving each term in its simplest form.\\

\hfill \mbox{\textit{Edexcel C1 2010 Q4 [7]}}