| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Topic | Differentiation from First Principles |
| Type | First principles: other functions |
| Difficulty | Standard +0.3 Part (a) is a standard first-principles differentiation of √x (routine but requires careful algebraic manipulation with rationalizing). Part (b)(i) involves finding two tangent equations and their intersection—straightforward coordinate geometry requiring algebraic manipulation but no novel insight. Part (b)(ii) is trivial pattern-spotting once (b)(i) is done. Overall slightly easier than average due to being mostly procedural with well-rehearsed techniques. |
| Spec | 1.07g Differentiation from first principles: for small positive integer powers of x1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\) | M1, A1 | Attempt \(\frac{1}{h}[f(x+h)-f(x)]\). Obtain correct expression (allow unsimplified denominator of \(x+h-x\)). |
| \(= \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \lim_{h \to 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{\sqrt{x}+\sqrt{x}} = \frac{1}{2\sqrt{x}}\) | M1, A1, A1 [5] | Multiply top and bottom by \(\sqrt{(x+h)}+\sqrt{x}\). Simplify expression as far as possible. Complete proof by considering \(\lim h \to 0\). Could also go via \(^{de}/_6x\), from \(x = y^2\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h \to 0} \frac{\sqrt{x}\left(1 + \frac{h}{2x} - \frac{h^2}{8x^2} + ...\right) - \sqrt{x}}{h} = \lim_{h \to 0} \frac{\sqrt{x} + \frac{1}{2}\frac{h}{\sqrt{x}} + h^2(...) - \sqrt{x}}{h} = \lim_{h \to 0} \frac{1}{2}\frac{1}{\sqrt{x}} + h(...)\) | M1, M1, A1 | Attempt binomial expansion with \(h/x\). Simplify expression as far as possible. |
| \(= \frac{1}{2\sqrt{x}}\) | M1 | Complete proof by considering \(\lim h \to 0\). |
| (b) (i) \(y - \sqrt{a} = \frac{1}{2\sqrt{a}}(x-a)\), \(y - \sqrt{b} = \frac{1}{2\sqrt{b}}(x-b)\) | M1, A1 | Attempt equations of both tangents. Obtain both correct equations. |
| \(\frac{1}{2\sqrt{a}}(x-a) + \sqrt{a} = \frac{1}{2\sqrt{b}}(x-b) + \sqrt{b}\) | M1 | Eliminate one variable and attempt to solve – as far as a correct equation in which \(x\) appears only once. Allow M1 if solving normals not tangents. |
| \(x = \sqrt{ab}\) AG; \(y = \frac{1}{2}\left(\sqrt{a}+\sqrt{b}\right)\) AG | A1, A1 [5] | Obtain \(x = \sqrt{ab}\), detail required. Obtain \(y = \frac{1}{2}\left(\sqrt{a}+\sqrt{b}\right)\), detail required. |
| (ii) Any valid solution e.g. \(a = 4\) and \(b = 16\) | M1, A1 [2] | State a pair of values that give one integer coord. State a pair of values that give both integer coords. |
**(a)** $f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$ | M1, A1 | Attempt $\frac{1}{h}[f(x+h)-f(x)]$. Obtain correct expression (allow unsimplified denominator of $x+h-x$).
$= \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \lim_{h \to 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{\sqrt{x}+\sqrt{x}} = \frac{1}{2\sqrt{x}}$ | M1, A1, A1 [5] | Multiply top and bottom by $\sqrt{(x+h)}+\sqrt{x}$. Simplify expression as far as possible. Complete proof by considering $\lim h \to 0$. Could also go via $^{de}/_6x$, from $x = y^2$.
**OR**
$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h \to 0} \frac{\sqrt{x}\left(1 + \frac{h}{2x} - \frac{h^2}{8x^2} + ...\right) - \sqrt{x}}{h} = \lim_{h \to 0} \frac{\sqrt{x} + \frac{1}{2}\frac{h}{\sqrt{x}} + h^2(...) - \sqrt{x}}{h} = \lim_{h \to 0} \frac{1}{2}\frac{1}{\sqrt{x}} + h(...)$ | M1, M1, A1 | Attempt binomial expansion with $h/x$. Simplify expression as far as possible.
$= \frac{1}{2\sqrt{x}}$ | M1 | Complete proof by considering $\lim h \to 0$.
**(b) (i)** $y - \sqrt{a} = \frac{1}{2\sqrt{a}}(x-a)$, $y - \sqrt{b} = \frac{1}{2\sqrt{b}}(x-b)$ | M1, A1 | Attempt equations of both tangents. Obtain both correct equations.
$\frac{1}{2\sqrt{a}}(x-a) + \sqrt{a} = \frac{1}{2\sqrt{b}}(x-b) + \sqrt{b}$ | M1 | Eliminate one variable and attempt to solve – as far as a correct equation in which $x$ appears only once. Allow M1 if solving normals not tangents.
$x = \sqrt{ab}$ **AG**; $y = \frac{1}{2}\left(\sqrt{a}+\sqrt{b}\right)$ **AG** | A1, A1 [5] | Obtain $x = \sqrt{ab}$, detail required. Obtain $y = \frac{1}{2}\left(\sqrt{a}+\sqrt{b}\right)$, detail required.
**(ii)** Any valid solution e.g. $a = 4$ and $b = 16$ | M1, A1 [2] | State a pair of values that give one integer coord. State a pair of values that give both integer coords.The function f is defined by $f(x) = \sqrt{x}, x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Use differentiation from first principles to find an expression for $f'(x)$. [5]
\end{enumerate}
The lines $l_1$ and $l_2$ are the tangents to the curve $y = f(x)$ at the points $A$ and $B$ where $x = a$ and $x = b$ respectively, $a \neq b$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item \begin{enumerate}[label=(\roman*)]
\item Show that the tangents intersect at the point $\left(\sqrt{ab}, \frac{1}{2}(\sqrt{a} + \sqrt{b})\right)$. [5]
\item Given that $l_1$ and $l_2$ intersect at a point with integer coordinates, write down a possible pair of values for $a$ and $b$. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2016 Q11 [12]}}