| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Complete the square |
| Difficulty | Moderate -0.3 This is a standard multi-part question covering completing the square, stationary points, and inverse functions—all routine P1 techniques. While it has five parts, each part follows directly from the previous with no novel insight required. The completing the square is straightforward, and finding the inverse function for a restricted quadratic is a textbook exercise, making it slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02g Inequalities: linear and quadratic in single variable1.02u 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) \(8x - x^2 = a - x^2 - b^2 - 2bx +\) equating \(\rightarrow b = -4\); \(a = b^2 = 16\) (i.e. \(16 - (x - 4)^2\)) | M1, B1, A1 [3] | Knows what to do – some equating anywhere – may be independent. For \(16 - ( )^2\) |
| (ii) \(\frac{dy}{dx} = 8 - 2x = 0\) when \(\rightarrow (4, 16)\) (or from –b and a) | M1, A1 [2] | Any valid complete method. Needs both values |
| (iii) \(8x - x^2 = -20\); \(x^2 - 8x - 20 = (x - 10)(x + 2)\); End values –2 and 10; Interval \(-2 \le x \le 10\) | M1, A1, A1 [3] | Sets to 0 + correct method of solution. Co – independent of < or > or =. Co – including \(\le\) (< gets A0) |
| \(g: x \rightarrow 8x - x^2\) for \(x = 4\) | [Co] | |
| (iv) domain of \(g^{-1}\) is \(x \le 16\); range of \(g^{-1}\) is \(g^{-1} \ge 4\) | B1∨, B1 [2] | From answer to (i) or (ii). Accept <16. Not f.t since domain of g given |
| (v) \(y = 8x - x^2 \rightarrow x^2 - 8x + y = 0\) | M1 | Use of quadratic or completed square expression to make x subject |
| \(x = 8 \pm \sqrt{(64 - 4y)} \div 2\); \(g^{-1}(x) = 4 + \sqrt{(16 - x)}\) | DM1, A1 [3] | Replaces y by x. Co (inc. omission of –) |
| or \((x - 4)^2 = 16 - y \rightarrow x = 4 + \sqrt{(16 - y)}\) \(\rightarrow y = 4 + \sqrt{(16 - x)}\) | [Co] |
**(i)** $8x - x^2 = a - x^2 - b^2 - 2bx +$ equating $\rightarrow b = -4$; $a = b^2 = 16$ (i.e. $16 - (x - 4)^2$) | M1, B1, A1 [3] | Knows what to do – some equating anywhere – may be independent. For $16 - ( )^2$
**(ii)** $\frac{dy}{dx} = 8 - 2x = 0$ when $\rightarrow (4, 16)$ (or from –b and a) | M1, A1 [2] | Any valid complete method. Needs both values
**(iii)** $8x - x^2 = -20$; $x^2 - 8x - 20 = (x - 10)(x + 2)$; End values –2 and 10; Interval $-2 \le x \le 10$ | M1, A1, A1 [3] | Sets to 0 + correct method of solution. Co – independent of < or > or =. Co – including $\le$ (< gets A0)
$g: x \rightarrow 8x - x^2$ for $x = 4$ | [Co]
**(iv)** domain of $g^{-1}$ is $x \le 16$; range of $g^{-1}$ is $g^{-1} \ge 4$ | B1∨, B1 [2] | From answer to (i) or (ii). Accept <16. Not f.t since domain of g given
**(v)** $y = 8x - x^2 \rightarrow x^2 - 8x + y = 0$ | M1 | Use of quadratic or completed square expression to make x subject
$x = 8 \pm \sqrt{(64 - 4y)} \div 2$; $g^{-1}(x) = 4 + \sqrt{(16 - x)}$ | DM1, A1 [3] | Replaces y by x. Co (inc. omission of –)
or $(x - 4)^2 = 16 - y \rightarrow x = 4 + \sqrt{(16 - y)}$ $\rightarrow y = 4 + \sqrt{(16 - x)}$ | [Co]11 The equation of a curve is $y = 8 x - x ^ { 2 }$.\\
(i) Express $8 x - x ^ { 2 }$ in the form $a - ( x + b ) ^ { 2 }$, stating the numerical values of $a$ and $b$.\\
(ii) Hence, or otherwise, find the coordinates of the stationary point of the curve.\\
(iii) Find the set of values of $x$ for which $y \geqslant - 20$.
The function g is defined by $\mathrm { g } : x \mapsto 8 x - x ^ { 2 }$, for $x \geqslant 4$.\\
(iv) State the domain and range of $\mathrm { g } ^ { - 1 }$.\\
(v) Find an expression, in terms of $x$, for $\mathrm { g } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{CAIE P1 2003 Q11 [13]}}