| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find minimum domain for inverse |
| Difficulty | Moderate -0.3 This is a comprehensive composite/inverse functions question covering standard techniques (composition, range, inequalities, inverse domain restriction). Part (iv) requires discriminant reasoning and (v)-(vi) involve finding the vertex for domain restriction, but all are routine A-level procedures with no novel insight required. Slightly easier than average due to straightforward linear/quadratic functions. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e 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 existence |
| Answer | Marks | Guidance |
|---|---|---|
| (i) ff = 2(2x−3)−3 Solves = 11 → x = 5 (or 2x−3 = 11, x = 7. 2x−3−7 → x = 5) | M1 A1 [2] | Either forms ff correctly, or solves 2 equations co |
| (ii) min at x = −2 → Range ≥ −4 | M1 A1 [2] | Any valid method – could be guesswork. |
| (iii) \(x^2+4x-12\) (> 0) → x = 2 or − 6 → \(x < -6\), \(x > 2\). | M1 A1 A1 [3] | Makes quadratic = 0 + 2 solutions Correct limits – even if >, <, =, ≤ co unsimplified |
| (iv) \(g f(x) = (2x-3)^2+4(2x-3) = p\) → \(4x^2-4x-3-p=0\) Uses "\(b^2-4ac\)" \(16 = 16(−3−p)\) → \(p = −4\) | B1 M1 A1 [3] | co Use of discriminant co |
| (v) −2 | B1 [1] | co |
| (vi) \(y = (x+2)^2-4\) \(\sqrt{y+4} = x+2\) \(h^{-1}(x) = \sqrt{x+4}-2\) | B2,1 M1 A1 [4] | −1 for each error Correct order of operations co with x, not y. ± left A0. |
$f: x \mapsto 2x-3, x \in \mathbb{R},$
$g: x \mapsto x^2+4x, x \in \mathbb{R}.$
**(i)** ff = 2(2x−3)−3 Solves = 11 → x = 5 (or 2x−3 = 11, x = 7. 2x−3−7 → x = 5) | M1 A1 [2] | Either forms ff correctly, or solves 2 equations co
**(ii)** min at x = −2 → Range ≥ −4 | M1 A1 [2] | Any valid method – could be guesswork.
**(iii)** $x^2+4x-12$ (> 0) → x = 2 or − 6 → $x < -6$, $x > 2$. | M1 A1 A1 [3] | Makes quadratic = 0 + 2 solutions Correct limits – even if >, <, =, ≤ co unsimplified
**(iv)** $g f(x) = (2x-3)^2+4(2x-3) = p$ → $4x^2-4x-3-p=0$ Uses "$b^2-4ac$" $16 = 16(−3−p)$ → $p = −4$ | B1 M1 A1 [3] | co Use of discriminant co
**(v)** −2 | B1 [1] | co
**(vi)** $y = (x+2)^2-4$ $\sqrt{y+4} = x+2$ $h^{-1}(x) = \sqrt{x+4}-2$ | B2,1 M1 A1 [4] | −1 for each error Correct order of operations co with x, not y. ± left A0.10 Functions $f$ and $g$ are defined by
$$\begin{aligned}
& \mathrm { f } : x \mapsto 2 x - 3 , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \mapsto x ^ { 2 } + 4 x , \quad x \in \mathbb { R } .
\end{aligned}$$
(i) Solve the equation $\mathrm { ff } ( x ) = 11$.\\
(ii) Find the range of g .\\
(iii) Find the set of values of $x$ for which $\mathrm { g } ( x ) > 12$.\\
(iv) Find the value of the constant $p$ for which the equation $\mathrm { gf } ( x ) = p$ has two equal roots.
Function h is defined by $\mathrm { h } : x \mapsto x ^ { 2 } + 4 x$ for $x \geqslant k$, and it is given that h has an inverse.\\
(v) State the smallest possible value of $k$.\\
(vi) Find an expression for $\mathrm { h } ^ { - 1 } ( x )$.
\hfill \mbox{\textit{CAIE P1 2014 Q10 [15]}}