CAIE P1 2020 June — Question 9 9 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2020
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeComplete the square
DifficultyModerate -0.3 This is a standard multi-part question on composite and inverse functions requiring completing the square, finding inverse functions, and composite functions. All techniques are routine A-level procedures with no novel problem-solving required, though the multiple parts and domain/range considerations make it slightly more substantial than the most basic exercises.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence

9 The functions f and g are defined by $$\begin{aligned} & \mathrm { f } ( x ) = x ^ { 2 } - 4 x + 3 \text { for } x > c , \text { where } c \text { is a constant, } \\ & \mathrm { g } ( x ) = \frac { 1 } { x + 1 } \quad \text { for } x > - 1 \end{aligned}$$
  1. Express \(\mathrm { f } ( x )\) in the form \(( x - a ) ^ { 2 } + b\).
    It is given that f is a one-one function.
  2. State the smallest possible value of \(c\).
    It is now given that \(c = 5\).
  3. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\).
  4. Find an expression for \(\mathrm { gf } ( x )\) and state the range of gf .

Question 9:
AnswerMarks
9(a): \([(x-2)^2]\ [-1]\)B1 B1
9(b): Smallest \(c = 2\) (FT on their part (a))B1FT
9(c): \(y = (x-2)^2 - 1 \rightarrow (x-2)^2 = y+1\)*M1
\(x = 2(\pm)\sqrt{y+1}\)DM1
\((f^{-1}(x)) = 2 + \sqrt{x+1}\) for \(x > 8\)A1
Question 9(d):
AnswerMarks Guidance
AnswerMark Guidance
\(\text{gf}(x) = \frac{1}{(x-2)^2 - 1 + 1} = \frac{1}{(x-2)^2}\)B1 OE
Range of gf is \(0 < \text{gf}(x) < \frac{1}{9}\)B1 B1
## Question 9:

**9(a):** $[(x-2)^2]\ [-1]$ | B1 B1 |

**9(b):** Smallest $c = 2$ (FT on their part (a)) | B1FT |

**9(c):** $y = (x-2)^2 - 1 \rightarrow (x-2)^2 = y+1$ | *M1 |

$x = 2(\pm)\sqrt{y+1}$ | DM1 |

$(f^{-1}(x)) = 2 + \sqrt{x+1}$ for $x > 8$ | A1 |

## Question 9(d):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{gf}(x) = \frac{1}{(x-2)^2 - 1 + 1} = \frac{1}{(x-2)^2}$ | B1 | OE |
| Range of gf is $0 < \text{gf}(x) < \frac{1}{9}$ | B1 B1 | |

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9 The functions f and g are defined by

$$\begin{aligned}
& \mathrm { f } ( x ) = x ^ { 2 } - 4 x + 3 \text { for } x > c , \text { where } c \text { is a constant, } \\
& \mathrm { g } ( x ) = \frac { 1 } { x + 1 } \quad \text { for } x > - 1
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ in the form $( x - a ) ^ { 2 } + b$.\\

It is given that f is a one-one function.
\item State the smallest possible value of $c$.\\

It is now given that $c = 5$.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state the domain of $\mathrm { f } ^ { - 1 }$.
\item Find an expression for $\mathrm { gf } ( x )$ and state the range of gf .
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2020 Q9 [9]}}