| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find intersection points |
| Difficulty | Standard +0.8 This is a multi-part question requiring finding an inverse function (routine), determining intercepts (straightforward), finding a perpendicular bisector (standard), and most challengingly, proving four points lie on a circle and finding its equation. The final part requires geometric insight about the symmetry of functions and their inverses about y=x, plus solving simultaneous equations or using the general circle equation—this elevates it above typical textbook exercises but remains within standard A-level techniques. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x^3 - 8\), \(x^3 = y + 8\), \(\sqrt[3]{(x+8)}\) oe isw | M1, A1 | Attempt to re-arrange; ignore labelling e.g. \(fh(x) = \sqrt[3]{(x+8)}\) scores 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A(0,-8)\) | B1 | |
| \(B(2,0)\) | B2 | |
| \(C(0,2)\) | B1 | FT their B |
| \(D(-8,0)\) | B1 | FT their A |
| Answer | Marks | Guidance |
|---|---|---|
| Midpoint is \((1,-4)\) and gradient of \(AB\) is \(4\) | B1 | FT their A and B; may be implied by later work |
| Gradient of perpendicular bisector is \(-\frac{1}{4}\) | B1 | FT negative reciprocal of their 4 |
| Equation \(y + 4 = -\frac{1}{4}(x-1)\) | M1 | Or using \(y = mx + c\) and attempting to evaluate \(c\); must be using their midpoint and their \(-\frac{1}{4}\) |
| \(y = -\frac{1}{4}x - 3\frac{3}{4}\) | A1 | Final answer |
| Answer | Marks |
|---|---|
| Setting up any 2 of: \((0-a)^2+(2-b)^2=r^2\), \((0-a)^2+(-8-b)^2=r^2\), \((2-a)^2+(0-b)^2=r^2\), \((-8-a)^2+(0-b)^2=r^2\) | M1 |
| Attempting to solve to find \(a\) or \(b\) | M1 |
| Centre \((-3,-3)\) oe | A1 |
| Using their centre and another point \((A, B, C\) or \(D)\) to find radius | M1 |
| \((x+3)^2 + (y+3)^2 = 34\) cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Intersection of any 2 of \(y=x\), \(x=-3\), \(y=-3\), \(y=-\frac{1}{4}x - 3\frac{3}{4}\), \(y=-4x-15\) | M1 | Identifying 2 perp bisectors of 2 chords e.g. \(BD\) and \(AC\), or \(AB\) and \(CD\) |
| Attempting to solve to find intersection point | M1 | |
| \((-3,-3)\) | A1 | |
| Using their centre and another point to find radius | M1 | |
| \((x+3)^2 + (y+3)^2 = 34\) | A1 |
## Question 8:
### Part (a):
$y = x^3 - 8$, $x^3 = y + 8$, $\sqrt[3]{(x+8)}$ oe isw | **M1, A1** | Attempt to re-arrange; ignore labelling e.g. $fh(x) = \sqrt[3]{(x+8)}$ scores 2
### Part (b):
$A(0,-8)$ | **B1** |
$B(2,0)$ | **B2** |
$C(0,2)$ | **B1** | FT their B
$D(-8,0)$ | **B1** | FT their A
### Part (c):
Midpoint is $(1,-4)$ and gradient of $AB$ is $4$ | **B1** | FT their A and B; may be implied by later work
Gradient of perpendicular bisector is $-\frac{1}{4}$ | **B1** | FT negative reciprocal of their 4
Equation $y + 4 = -\frac{1}{4}(x-1)$ | **M1** | Or using $y = mx + c$ and attempting to evaluate $c$; must be using their midpoint and their $-\frac{1}{4}$
$y = -\frac{1}{4}x - 3\frac{3}{4}$ | **A1** | Final answer
### Part (d):
**Either method:**
Setting up any 2 of: $(0-a)^2+(2-b)^2=r^2$, $(0-a)^2+(-8-b)^2=r^2$, $(2-a)^2+(0-b)^2=r^2$, $(-8-a)^2+(0-b)^2=r^2$ | **M1** |
Attempting to solve to find $a$ or $b$ | **M1** |
Centre $(-3,-3)$ oe | **A1** |
Using their centre and another point $(A, B, C$ or $D)$ to find radius | **M1** |
$(x+3)^2 + (y+3)^2 = 34$ cao | **A1** |
**Or:**
Intersection of any 2 of $y=x$, $x=-3$, $y=-3$, $y=-\frac{1}{4}x - 3\frac{3}{4}$, $y=-4x-15$ | **M1** | Identifying 2 perp bisectors of 2 chords e.g. $BD$ and $AC$, or $AB$ and $CD$
Attempting to solve to find intersection point | **M1** |
$(-3,-3)$ | **A1** |
Using their centre and another point to find radius | **M1** |
$(x+3)^2 + (y+3)^2 = 34$ | **A1** |8 The curves $\mathrm { y } = \mathrm { h } ( \mathrm { x } )$ and $\mathrm { y } = \mathrm { h } ^ { - 1 } ( \mathrm { x } )$, where $\mathrm { h } ( x ) = x ^ { 3 } - 8$, are shown below.\\
The curve $\mathrm { y } = \mathrm { h } ( \mathrm { x } )$ crosses the $x$-axis at B and the $y$-axis at A.\\
The curve $\mathrm { y } = \mathrm { h } ^ { - 1 } ( \mathrm { x } )$ crosses the $x$-axis at D and the $y$-axis at C .\\
\includegraphics[max width=\textwidth, alt={}, center]{c30a926b-d832-46f5-aa65-0066ef482c3d-7_826_819_520_255}
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\mathrm { h } ^ { - 1 } ( x )$.
\item Determine the coordinates of A, B, C and D.
\item Determine the equation of the perpendicular bisector of AB . Give your answer in the form $\mathrm { y } = \mathrm { mx } + c$, where $m$ and $c$ are constants to be determined.
\item Points A , B , C and D lie on a circle.
Determine the equation of the circle. Give your answer in the form $( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }$, where $a$, $b$ and $r ^ { 2 }$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2022 Q8 [16]}}