| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.8 This is a straightforward composite and inverse functions question requiring only routine algebraic manipulation. All parts involve standard textbook procedures: composing linear and quadratic functions, solving simple equations, and finding inverses with restricted domains. No problem-solving insight or novel techniques are 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) fg(x) = 2x² − 3, gf(x) = 4x² + 4x − 1 | B1, B1 [2] | fg & gf clearly transposed gets B0B0 |
| (ii) \(2a^2 - 3 = 4a^2 + 4a - 1 \Rightarrow 2a^2 + 4a + 2 = 0\) | M1 | |
| \((a + 1)^2 = 0\) | M1 | |
| \(a = -1\) | A1 [3] | Dep. quadratic. Allow x for all 3 marks Allow marks in (ii) if transposed in (i) |
| (iii) \(b^2 - b - 2 = 0 \Rightarrow (b+1)(b-2) = 0\) | M1 | Allow in terms of x for M1 only |
| \(b = 2\) Allow \(b = -1\) in addition | A1 [2] | Correct answer without working B2 |
| (iv) \(f^{-1}(x) = \frac{1}{2}(x - 1)\) | B1 | |
| \(f^{-1} g(x) = \frac{1}{2}(x^2 - 3)\) | B1 ✓ [2] | Must be simplified. Ft from their f⁻¹ |
| (v) \(x = (±) \sqrt{y + 2}\) | M1 | |
| \(h^{-1}(x) = -\sqrt{x + 2}\) | A1 [2] |
**(i)** fg(x) = 2x² − 3, gf(x) = 4x² + 4x − 1 | B1, B1 [2] | fg & gf clearly transposed gets B0B0
**(ii)** $2a^2 - 3 = 4a^2 + 4a - 1 \Rightarrow 2a^2 + 4a + 2 = 0$ | M1 |
$(a + 1)^2 = 0$ | M1 |
$a = -1$ | A1 [3] | Dep. quadratic. Allow x for all 3 marks Allow marks in (ii) if transposed in (i)
**(iii)** $b^2 - b - 2 = 0 \Rightarrow (b+1)(b-2) = 0$ | M1 | Allow in terms of x for M1 only
$b = 2$ Allow $b = -1$ in addition | A1 [2] | Correct answer without working B2
**(iv)** $f^{-1}(x) = \frac{1}{2}(x - 1)$ | B1 |
$f^{-1} g(x) = \frac{1}{2}(x^2 - 3)$ | B1 ✓ [2] | Must be simplified. Ft from their f⁻¹
**(v)** $x = (±) \sqrt{y + 2}$ | M1 |
$h^{-1}(x) = -\sqrt{x + 2}$ | A1 [2] |11 Functions f and g are defined for $x \in \mathbb { R }$ by
$$\begin{aligned}
& \mathrm { f } : x \mapsto 2 x + 1 \\
& \mathrm {~g} : x \mapsto x ^ { 2 } - 2
\end{aligned}$$
(i) Find and simplify expressions for $\mathrm { fg } ( x )$ and $\mathrm { gf } ( x )$.\\
(ii) Hence find the value of $a$ for which $\mathrm { fg } ( a ) = \mathrm { gf } ( a )$.\\
(iii) Find the value of $b ( b \neq a )$ for which $\mathrm { g } ( b ) = b$.\\
(iv) Find and simplify an expression for $\mathrm { f } ^ { - 1 } \mathrm {~g} ( x )$.
The function h is defined by
$$\mathrm { h } : x \mapsto x ^ { 2 } - 2 , \quad \text { for } x \leqslant 0$$
(v) Find an expression for $\mathrm { h } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{CAIE P1 2011 Q11 [11]}}