CAIE P1 2019 June — Question 4 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind composite function expression
DifficultyModerate -0.3 This is a straightforward composite and inverse functions question requiring standard techniques: finding range of f to determine domain of g (simple substitution of endpoints), composing two simple functions, and inverting a rational function. All steps are routine for P1 level with no novel problem-solving required, making it slightly easier than average.
Spec1.02v Inverse and composite functions: graphs and conditions for existence

The function f is defined by \(\text{f}(x) = \frac{48}{x - 1}\) for \(3 \leqslant x \leqslant 7\). The function g is defined by \(\text{g}(x) = 2x - 4\) for \(a \leqslant x \leqslant b\), where \(a\) and \(b\) are constants.
  1. Find the greatest value of \(a\) and the least value of \(b\) which will permit the formation of the composite function gf. [2] It is now given that the conditions for the formation of gf are satisfied.
  2. Find an expression for \(\text{gf}(x)\). [1]
  3. Find an expression for \((\text{gf})^{-1}(x)\). [2]

Question 4:

AnswerMarks Guidance
4(i)Max(a) is 8 B1
Min(b) is 24B1 Allow b = 24 or b(cid:46)24
2SCB1 for 8 and 24 seen

AnswerMarks
4(ii)96 100−4x
gf(x) = −4 or gf(x) =
AnswerMarks Guidance
x−1 x−1B1  48 
2  −4 is insufficient
x−1
Apply ISW
1

AnswerMarks
4(iii)96 96 96
y= −4 → y+4= → x−1=
AnswerMarks Guidance
x−1 x−1 y+4M1 FT from their(ii) provided (ii) involves algebraic fraction.
Allow sign errors
( gf )−1( x )= 96 +1
AnswerMarks Guidance
x+4A1 100+x
OR . Must be a function of x. Apply ISW
x+4
2
AnswerMarks Guidance
QuestionAnswer Marks
Question 4:
--- 4(i) ---
4(i) | Max(a) is 8 | B1 | Allow a = 8 or a(cid:45)8
Min(b) is 24 | B1 | Allow b = 24 or b(cid:46)24
2 | SCB1 for 8 and 24 seen
--- 4(ii) ---
4(ii) | 96 100−4x
gf(x) = −4 or gf(x) =
x−1 x−1 | B1 |  48 
2  −4 is insufficient
x−1
Apply ISW
1
--- 4(iii) ---
4(iii) | 96 96 96
y= −4 → y+4= → x−1=
x−1 x−1 y+4 | M1 | FT from their(ii) provided (ii) involves algebraic fraction.
Allow sign errors
( gf )−1( x )= 96 +1
x+4 | A1 | 100+x
OR . Must be a function of x. Apply ISW
x+4
2
Question | Answer | Marks | Guidance
The function f is defined by $\text{f}(x) = \frac{48}{x - 1}$ for $3 \leqslant x \leqslant 7$. The function g is defined by $\text{g}(x) = 2x - 4$ for $a \leqslant x \leqslant b$, where $a$ and $b$ are constants.

\begin{enumerate}[label=(\roman*)]
\item Find the greatest value of $a$ and the least value of $b$ which will permit the formation of the composite function gf. [2]

It is now given that the conditions for the formation of gf are satisfied.

\item Find an expression for $\text{gf}(x)$. [1]

\item Find an expression for $(\text{gf})^{-1}(x)$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2019 Q4 [5]}}