CAIE P1 2018 June — Question 9 11 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind intersection points
DifficultyModerate -0.3 This is a multi-part question covering standard function operations (intersection, inequalities, composition, inverse conditions) with straightforward algebraic manipulation. Part (i) requires solving a quadratic equation, (ii) uses the quadratic inequality, (iii) involves composition and completing the square to find range, and (iv) requires understanding that a function needs to be one-to-one (finding the vertex). All techniques are routine for P1 level with no novel problem-solving required, making it slightly easier than average.
Spec1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.02v Inverse and composite functions: graphs and conditions for existence

9 Functions f and g are defined for \(x \in \mathbb { R }\) by $$\begin{aligned} & \mathrm { f } : x \mapsto \frac { 1 } { 2 } x - 2 \\ & \mathrm {~g} : x \mapsto 4 + x - \frac { 1 } { 2 } x ^ { 2 } \end{aligned}$$
  1. Find the points of intersection of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\).
  2. Find the set of values of \(x\) for which \(\mathrm { f } ( x ) > \mathrm { g } ( x )\).
  3. Find an expression for \(\mathrm { fg } ( x )\) and deduce the range of fg .
    The function h is defined by \(\mathrm { h } : x \mapsto 4 + x - \frac { 1 } { 2 } x ^ { 2 }\) for \(x \geqslant k\).
  4. Find the smallest value of \(k\) for which h has an inverse.

Question 9(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(4 + x - \frac{x^2}{2} = \frac{x}{2} - 2 \rightarrow x^2 - x - 12 = 0\)M1 Equates and forms 3-term quadratic
\((4, 0)\) and \((-3, -3.5)\); Trial and improvement: B3 all correct or B0A1 A1 A1 for both \(x\) values or a correct pair; A1 all
Question 9(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(f(x) > g(x)\) for \(x > 4\), \(x < -3\)B1, B1 B1 for each part. Loses a mark for \(\leqslant\) or \(\geqslant\)
Question 9(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(fg(x) = 2 + \frac{x}{2} - \frac{x^2}{4} - 2\ \left(= \frac{x}{2} - \frac{x^2}{4}\right)\)B1 CAO, any correct form
\(-\frac{1}{4}((x-1)^2 - 1)\) or \(\frac{dy}{dx} = \frac{1}{2} - \frac{2x}{4} = 0 \rightarrow x = 1\)M1 A1 Completes the square or uses calculus. First A1 for \(x=1\) or completed square form
\(y = \frac{1}{4}\ \rightarrow\) Range of \(fg \leqslant \frac{1}{4}\)A1 CAO, OE e.g. \(y \leqslant \frac{1}{4}\), \(\left[-\infty, \frac{1}{4}\right)\) etc.
Question 9(iv):
AnswerMarks Guidance
AnswerMarks Guidance
Calculus or completing square on 'h' \(\rightarrow x = 1\)M1 May use a sketch or \(-\frac{b}{2a}\)
\(k = 1\) (accept \(k \geqslant 1\))A1 Complete method. CAO
## Question 9(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $4 + x - \frac{x^2}{2} = \frac{x}{2} - 2 \rightarrow x^2 - x - 12 = 0$ | **M1** | Equates and forms 3-term quadratic |
| $(4, 0)$ and $(-3, -3.5)$; Trial and improvement: B3 all correct or B0 | **A1 A1** | A1 for both $x$ values or a correct pair; A1 all |

## Question 9(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) > g(x)$ for $x > 4$, $x < -3$ | **B1, B1** | B1 for each part. Loses a mark for $\leqslant$ or $\geqslant$ |

## Question 9(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $fg(x) = 2 + \frac{x}{2} - \frac{x^2}{4} - 2\ \left(= \frac{x}{2} - \frac{x^2}{4}\right)$ | **B1** | CAO, any correct form |
| $-\frac{1}{4}((x-1)^2 - 1)$ or $\frac{dy}{dx} = \frac{1}{2} - \frac{2x}{4} = 0 \rightarrow x = 1$ | **M1 A1** | Completes the square or uses calculus. First A1 for $x=1$ or completed square form |
| $y = \frac{1}{4}\ \rightarrow$ Range of $fg \leqslant \frac{1}{4}$ | **A1** | CAO, OE e.g. $y \leqslant \frac{1}{4}$, $\left[-\infty, \frac{1}{4}\right)$ etc. |

## Question 9(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculus or completing square on 'h' $\rightarrow x = 1$ | **M1** | May use a sketch or $-\frac{b}{2a}$ |
| $k = 1$ (accept $k \geqslant 1$) | **A1** | Complete method. CAO |
9 Functions f and g are defined for $x \in \mathbb { R }$ by

$$\begin{aligned}
& \mathrm { f } : x \mapsto \frac { 1 } { 2 } x - 2 \\
& \mathrm {~g} : x \mapsto 4 + x - \frac { 1 } { 2 } x ^ { 2 }
\end{aligned}$$

(i) Find the points of intersection of the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$.\\

(ii) Find the set of values of $x$ for which $\mathrm { f } ( x ) > \mathrm { g } ( x )$.\\

(iii) Find an expression for $\mathrm { fg } ( x )$ and deduce the range of fg .\\

The function h is defined by $\mathrm { h } : x \mapsto 4 + x - \frac { 1 } { 2 } x ^ { 2 }$ for $x \geqslant k$.\\
(iv) Find the smallest value of $k$ for which h has an inverse.\\

\hfill \mbox{\textit{CAIE P1 2018 Q9 [11]}}