| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find minimum domain for inverse |
| Difficulty | Standard +0.3 This is a slightly above-average C3 question requiring understanding of one-to-one functions and finding turning points to determine domain restrictions. Part (i) requires differentiation to find the maximum, part (ii) needs recognition that k must be the x-coordinate of the turning point, and part (iii) involves straightforward differentiation and solving an equation. The concepts are standard for C3 but require careful application rather than just routine recall. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | Attempt use of quotient rule | *M1 |
| Obtain \(\frac{75-15x^2}{(x^2+5)^2}\) | A1 | or (unsimplified) equiv; this M1A1 available at any stage of question |
| Equate attempt at first derivative to zero and rearrange to solvable form | M1 | dep *M |
| Obtain \(x = \sqrt{5}\) or 2.24 | A1 | or greater accuracy |
| Recognise range as values less than \(y\)-coord of st pt | M1 | allowing < here |
| Obtain \(0 \leq y \leq \frac{2}{3}\sqrt{5}\) | A1 | any notation; with ≤ now; any exact equiv |
| (ii) | State \(\sqrt{5}\) | B1√ |
| (iii) | Equate attempt at first derivative to −1 and attempt simplification | *M1 |
| Obtain \(x^4 - 5x^2 + 100 = 0\) | A1 | or equiv involving 3 non-zero terms |
| Attempt evaluation of discriminant or equiv | M1 | dep *M |
| Obtain −375 or equiv and conclude appropriately | A1 |
(i) | Attempt use of quotient rule | *M1 | or equiv; allow $u/v$ muddles |
| Obtain $\frac{75-15x^2}{(x^2+5)^2}$ | A1 | or (unsimplified) equiv; this M1A1 available at any stage of question |
| Equate attempt at first derivative to zero and rearrange to solvable form | M1 | dep *M |
| Obtain $x = \sqrt{5}$ or 2.24 | A1 | or greater accuracy |
| Recognise range as values less than $y$-coord of st pt | M1 | allowing < here |
| Obtain $0 \leq y \leq \frac{2}{3}\sqrt{5}$ | A1 | any notation; with ≤ now; any exact equiv |
(ii) | State $\sqrt{5}$ | B1√ | following their $x$-coord of st pt; condone answer $x \geq \sqrt{5}$ but not inequality with $k$ |
(iii) | Equate attempt at first derivative to −1 and attempt simplification | *M1 | and dependent on first M in part (i) |
| Obtain $x^4 - 5x^2 + 100 = 0$ | A1 | or equiv involving 3 non-zero terms |
| Attempt evaluation of discriminant or equiv | M1 | dep *M |
| Obtain −375 or equiv and conclude appropriately | A1 | |9\\
\includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-4_534_935_264_605}
The function f is defined for the domain $x \geqslant 0$ by
$$f ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$
The diagram shows the curve with equation $y = \mathrm { f } ( x )$.\\
(i) Find the range of f .\\
(ii) The function g is defined for the domain $x \geqslant k$ by
$$\mathrm { g } ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$
Given that g is a one-one function, state the least possible value of $k$.\\
(iii) Show that there is no point on the curve $y = \mathrm { g } ( x )$ at which the gradient is - 1 .
\hfill \mbox{\textit{OCR C3 2008 Q9 [11]}}