| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 15 |
| Topic | Areas Between Curves |
| Type | Two Curves Intersection Area |
| Difficulty | Challenging +1.3 This is a multi-part question requiring curve intersection analysis, integration with parameter cases, inequality solving, and inverse function work. While it demands careful case analysis (k<1 vs k>1) and involves several connected concepts, the individual techniques are standard A-level fare: solving simultaneous equations, definite integration, and function theory. The proof elements are straightforward rather than requiring deep insight. More demanding than typical textbook exercises but accessible to strong A-level students. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Intersection satisfies \(x^k = x^2 \Rightarrow x^{k^2} = x\) | M1 | Rearrange and factorise: \(x(x^{k^2-1} - 1) = 0\) |
| (ii) Consider \(\int_0^1\left(x^k - x^{\frac{k}{k}}\right)dx\) | M1 | Obtain \(\left[\frac{1}{k+1}x^{k+1} - \frac{1}{\frac{k}{k}+1}x^{\frac{k}{k}+1}\right]_0^1\) |
| (iii) Attempt to solve \(A = 0.5\) | M1 | Obtain \(k = \frac{1}{3}\) if \(k < 1\) and \(k = 3\) if \(k > 1\) |
| (iv) Attempt to process \(\frac{k_1 - 1}{k_1 + 1} = \frac{k_2 - 1}{k_2 + 1}\) | M1 | Convincingly obtain \(k_1 = k_2 \Rightarrow\) 1-1 function |
(i) Intersection satisfies $x^k = x^2 \Rightarrow x^{k^2} = x$ | M1 | Rearrange and factorise: $x(x^{k^2-1} - 1) = 0$ | M1 | Obtain $x = 0$ and $1$ | A1 | Unless $k^2 - 1 = 0 \Rightarrow k = 1 \Rightarrow x$ can be any value | A1 | 4 marks
(ii) Consider $\int_0^1\left(x^k - x^{\frac{k}{k}}\right)dx$ | M1 | Obtain $\left[\frac{1}{k+1}x^{k+1} - \frac{1}{\frac{k}{k}+1}x^{\frac{k}{k}+1}\right]_0^1$ | A1 | $= \frac{1-k}{1+k}$ | A1 | $A = \begin{cases}\frac{1-k}{1+k} & \text{if } k < 1 \\ \frac{k-1}{1+k} & \text{if } k > 1\end{cases}$ | A1 | [Note: the function could alternatively be given in modulus form] | 4 marks
(iii) Attempt to solve $A = 0.5$ | M1 | Obtain $k = \frac{1}{3}$ if $k < 1$ and $k = 3$ if $k > 1$ | A1 | State $\frac{1}{3} \leq k \leq 3$ | A1 | 3 marks
(iv) Attempt to process $\frac{k_1 - 1}{k_1 + 1} = \frac{k_2 - 1}{k_2 + 1}$ | M1 | Convincingly obtain $k_1 = k_2 \Rightarrow$ 1-1 function | A1 | Attempt to solve $A = \frac{k-1}{k+1}$ | M1 | Obtain $k = \frac{A+1}{1-A}$ | A1 | 4 marksTwo curves are defined by $y = x^k$ and $y = x^{\frac{1}{k}}$, for $x \geqslant 0$, where $k > 0$.
\begin{enumerate}[label=(\roman*)]
\item Prove that, except for one value of $k$, the curves intersect in exactly two points. [4]
\end{enumerate}
The two curves enclose a finite region $R$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the area, $A$, of $R$, giving your answer in the form $A = f(k)$ and distinguishing clearly between the cases $k < 1$ and $k > 1$. [4]
\item Determine the set of values of $k$ for which $A \leqslant 0.5$. [3]
\item The function $f$ is given by $f : x \mapsto A$ with $k > 1$. Prove that $f$ is one-one and determine its inverse. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q9 [15]}}