CAIE P1 2023 November — Question 8 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyStandard +0.3 This is a straightforward composite and inverse functions question requiring standard techniques: finding an inverse by swapping and rearranging (with domain restriction already given), stating domain/range from the original function's range/domain, and solving a composite equation by substitution. All steps are routine for P1 level with no novel insight required, making it slightly easier than average.
Spec1.02v Inverse and composite functions: graphs and conditions for existence
Functions f and g are defined by $$f(x) = (x + a)^2 - a \text{ for } x \leqslant -a,$$ $$g(x) = 2x - 1 \text{ for } x \in \mathbb{R},$$ where \(a\) is a positive constant.
  1. Find an expression for \(f^{-1}(x)\). [3]
    1. State the domain of the function \(f^{-1}\). [1]
    2. State the range of the function \(f^{-1}\). [1]
  3. Given that \(a = \frac{7}{2}\), solve the equation \(gf(x) = 0\). [3]
Question 8:
AnswerMarks Guidance
8(a)y=(x+a)2 −a leading to (x+a)2 = ya *M1
Allow  errors for these method marks.
AnswerMarks
x= yaaDM1
Alternative method for first 2 marks of Question 8(a)
AnswerMarks Guidance
x=(y+a)2 −a leading to y2 +2ay+a2−a−x =0 *M1 Allow  errors for this method mark.
−2a 4a2 −4 ( a2 −a−x )
y=
AnswerMarks Guidance
2DM1
[y or f−1(x) =]− x+a −aA1 OE
Must choose negative root.
3
AnswerMarks Guidance
8(b)(i)x −a B1
1
AnswerMarks Guidance
8(b)(ii)y or f−1  (x)  −a B1
1
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(c) 7    7 2 7  7 7 2 7
gf = 2  x+  −   −1 or 2x2 +4  x+2  −2 −1 =0 
AnswerMarks Guidance
 2    2 2  2 2 2B1 OE
1
Alternatively,  gf(x)=0  f(x)= .
2
33
−14 142 −42
x=− 7 2 or 2
2 4
−14 64
  or factorising
AnswerMarks Guidance
 4 M1 OE
Solving their three term quadratic equation as far as two
solutions or correctly selecting the negative root only.
1 7 7
Alternatively,  + − .
2 2 2
11
x=−
AnswerMarks Guidance
2A1 If B1M0 scored then award SCB1 for the correct final
answer.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(a) ---
8(a) | y=(x+a)2 −a leading to (x+a)2 = ya | *M1 | x and y may be interchanged initially.
Allow  errors for these method marks.
x= yaa | DM1
Alternative method for first 2 marks of Question 8(a)
x=(y+a)2 −a leading to y2 +2ay+a2−a−x =0  | *M1 | Allow  errors for this method mark.
−2a 4a2 −4 ( a2 −a−x )
y=
2 | DM1
[y or f−1(x) =]− x+a −a | A1 | OE
Must choose negative root.
3
--- 8(b)(i) ---
8(b)(i) | x −a | B1 | Ignore infinity limit if included.
1
--- 8(b)(ii) ---
8(b)(ii) | y or f−1  (x)  −a | B1 | Ignore negative infinity limit if included.
1
Question | Answer | Marks | Guidance
--- 8(c) ---
8(c) |  7    7 2 7  7 7 2 7
gf = 2  x+  −   −1 or 2x2 +4  x+2  −2 −1 =0 
 2    2 2  2 2 2 | B1 | OE
1
Alternatively,  gf(x)=0  f(x)= .
2
33
−14 142 −42
x=− 7 2 or 2
2 4
−14 64
  or factorising
 4  | M1 | OE
Solving their three term quadratic equation as far as two
solutions or correctly selecting the negative root only.
1 7 7
Alternatively,  + − .
2 2 2
11
x=−
2 | A1 | If B1M0 scored then award SCB1 for the correct final
answer.
3
Question | Answer | Marks | Guidance
Functions f and g are defined by
$$f(x) = (x + a)^2 - a \text{ for } x \leqslant -a,$$
$$g(x) = 2x - 1 \text{ for } x \in \mathbb{R},$$
where $a$ is a positive constant.

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $f^{-1}(x)$. [3]

\item \begin{enumerate}[label=(\roman*)]
\item State the domain of the function $f^{-1}$. [1]

\item State the range of the function $f^{-1}$. [1]
\end{enumerate}

\item Given that $a = \frac{7}{2}$, solve the equation $gf(x) = 0$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q8 [8]}}
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