| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Linear function from conditions |
| Difficulty | Easy -1.2 This is a straightforward question requiring substitution into a linear function to form two simultaneous equations (2a+b=1, 5a+b=7), solving to get a=2, b=-3, then composing the function and solving a simple linear equation. All steps are routine with no problem-solving insight needed, making it 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) \(2a + b = 1\) and \(5a + b = 7\) \(\rightarrow a = 2\) and \(b = -3\) | M1, A1 [2] | Realising how one of these is formed. Co |
| (ii) \(f(x) = 2x - 3\); \(ff(x) = f(2x-3) \cdot 3 = 4x - 9\) | M1, DM1, A1 [3] | Replacing "x" by "his \(ax + b\)" and "+b". For his a and b and solved = 0. Co |
| \(= 0\) when \(x = 2.25\) | [Co] |
**(i)** $2a + b = 1$ and $5a + b = 7$ $\rightarrow a = 2$ and $b = -3$ | M1, A1 [2] | Realising how one of these is formed. Co
**(ii)** $f(x) = 2x - 3$; $ff(x) = f(2x-3) \cdot 3 = 4x - 9$ | M1, DM1, A1 [3] | Replacing "x" by "his $ax + b$" and "+b". For his a and b and solved = 0. Co
$= 0$ when $x = 2.25$ | [Co]5 The function f is defined by $\mathrm { f } : x \mapsto a x + b$, for $x \in \mathbb { R }$, where $a$ and $b$ are constants. It is given that $f ( 2 ) = 1$ and $f ( 5 ) = 7$.\\
(i) Find the values of $a$ and $b$.\\
(ii) Solve the equation $\operatorname { ff } ( x ) = 0$.
\hfill \mbox{\textit{CAIE P1 2003 Q5 [5]}}