CAIE P1 2016 June — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeSolve equation involving composites
DifficultyModerate -0.8 This question requires computing two function compositions and solving a linear equation, but all steps are straightforward applications of function notation. Finding ff(x) = f(f(x)) and gf(2) = g(f(2)) involves direct substitution with no conceptual challenges, making it easier than average for A-level.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence

1 Functions f and g are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto 10 - 3 x , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \mapsto \frac { 10 } { 3 - 2 x } , \quad x \in \mathbb { R } , x \neq \frac { 3 } { 2 } \end{aligned}$$ Solve the equation \(\mathrm { ff } ( x ) = \mathrm { gf } ( 2 )\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(ff(x) = 10 - 3(10 - 3x)\)B1 Correct unsimplified expression
\(gf(2) = \frac{10}{3-2(10-3(2))}\) (= -2)B1 Correct unsimplified expression with 2 in for \(x\)
\(x = 2\)B1
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $ff(x) = 10 - 3(10 - 3x)$ | B1 | Correct unsimplified expression |
| $gf(2) = \frac{10}{3-2(10-3(2))}$ (= -2) | B1 | Correct unsimplified expression with 2 in for $x$ |
| $x = 2$ | B1 | |
1 Functions f and g are defined by

$$\begin{aligned}
& \mathrm { f } : x \mapsto 10 - 3 x , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \mapsto \frac { 10 } { 3 - 2 x } , \quad x \in \mathbb { R } , x \neq \frac { 3 } { 2 }
\end{aligned}$$

Solve the equation $\mathrm { ff } ( x ) = \mathrm { gf } ( 2 )$.

\hfill \mbox{\textit{CAIE P1 2016 Q1 [3]}}