| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | State domain or range |
| Difficulty | Standard +0.3 This question involves standard techniques: finding the vertex of a quadratic to determine the range and smallest value of c for a one-one function, then solving simultaneous equations from composite function values. All steps are routine applications of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Range is \((y) \geq c^2 + 4c\) | B1 | Allow \(>\) |
| \(x^2 + 4x = (x + 2)^2 - 4\) | M1 | OR \(\frac{dy}{dx} = 2x + 4 = 0\) |
| (Smallest value of \(c\) is) \(-2\) | A1 | \(-2\) with no (wrong) working gets B2 |
| [3] | ||
| (ii) \(5a + b = 11\) | B1 | |
| \((a + b)^2 + 4(a + b) = 21\) | B1 | |
| \((11 - 5a + a)^2 + 4(11 - 5a + a) = 21\) | M1 | OR corresponding equation in \(b\) |
| \((8)(2a^2 - 13a + 18) = (8)(2a - 9)(a - 2) = 0\) | M1 | OR \((8)(2b + 23)(b - 1) = 0\) |
| \(a = \frac{9}{2}, 2\) OR \(b = \left(-\frac{23}{2}\right), 1\) | A1 | A1 for either \(a\) or \(b\) correct. Condone 2nd |
| value. Spotted solution scores only B marks. | ||
| [6] | ||
| Alt. (ii) Last 5 marks | ||
| \(f^{-1}(x) = \sqrt{x + 4} - 2\) | B1 | Alt. (ii) Last 4 marks |
| \(g(1) = f^{-1}(21)\) used | M1 | \((a + b + 7)(a + b - 3) = 0\) |
| \(a + b = \sqrt{25} - 2 = 3\) | A1 | (Ignore solution involving \(a + b = -7\)) |
| Solve \(a + b = 3, 5a + b = 11\) | M1 | Solve \(a + b = 3, 5a + b = 11\) |
| \(a = 2, b = 1\) | A1 | \(a = 2, b = 1\) |
**(i)** Range is $(y) \geq c^2 + 4c$ | B1 | Allow $>$
$x^2 + 4x = (x + 2)^2 - 4$ | M1 | OR $\frac{dy}{dx} = 2x + 4 = 0$
(Smallest value of $c$ is) $-2$ | A1 | $-2$ with no (wrong) working gets B2
| | [3]
**(ii)** $5a + b = 11$ | B1 |
$(a + b)^2 + 4(a + b) = 21$ | B1 |
$(11 - 5a + a)^2 + 4(11 - 5a + a) = 21$ | M1 | OR corresponding equation in $b$
$(8)(2a^2 - 13a + 18) = (8)(2a - 9)(a - 2) = 0$ | M1 | OR $(8)(2b + 23)(b - 1) = 0$
$a = \frac{9}{2}, 2$ OR $b = \left(-\frac{23}{2}\right), 1$ | A1 | A1 for either $a$ or $b$ correct. Condone 2nd
| | | value. Spotted solution scores only B marks.
| | [6]
**Alt. (ii)** Last 5 marks | |
$f^{-1}(x) = \sqrt{x + 4} - 2$ | B1 | Alt. (ii) Last 4 marks
$g(1) = f^{-1}(21)$ used | M1 | $(a + b + 7)(a + b - 3) = 0$
$a + b = \sqrt{25} - 2 = 3$ | A1 | (Ignore solution involving $a + b = -7$)
Solve $a + b = 3, 5a + b = 11$ | M1 | Solve $a + b = 3, 5a + b = 11$
$a = 2, b = 1$ | A1 | $a = 2, b = 1$10 The function f is defined by $\mathrm { f } : x \mapsto x ^ { 2 } + 4 x$ for $x \geqslant c$, where $c$ is a constant. It is given that f is a one-one function.\\
(i) State the range of f in terms of $c$ and find the smallest possible value of $c$.
The function g is defined by $\mathrm { g } : x \mapsto a x + b$ for $x \geqslant 0$, where $a$ and $b$ are positive constants. It is given that, when $c = 0 , \operatorname { gf } ( 1 ) = 11$ and $\operatorname { fg } ( 1 ) = 21$.\\
(ii) Write down two equations in $a$ and $b$ and solve them to find the values of $a$ and $b$.
\hfill \mbox{\textit{CAIE P1 2013 Q10 [9]}}