Edexcel PURE 2024 October — Question 6

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicComposite & Inverse Functions
TypeEvaluate composite at point
DifficultyStandard +0.3 This is a straightforward composite and inverse functions question with standard techniques: (a) requires simple substitution, (b) involves routine algebraic manipulation to find an inverse (swap x and y, rearrange), and (c) is a nested composition that reduces to a simple quadratic equation. All parts are textbook exercises requiring no novel insight, making it slightly easier than average.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence
  1. The functions f and g are defined by
$$\begin{array} { l l } \mathrm { f } ( x ) = 6 - \frac { 21 } { 2 x + 3 } & x \geqslant 0 \\ \mathrm {~g} ( x ) = x ^ { 2 } + 5 & x \in \mathbb { R } \end{array}$$
  1. Find \(\mathrm { gf } ( 2 )\)
  2. Find \(f ^ { - 1 }\)
  3. Solve the equation $$\operatorname { gg } ( x ) = 126$$
Question 6:
AnswerMarks Guidance
6 (a)gf ( 2 )=g ( 3 )=3 2 +5 M1
= 14A1
(2)
AnswerMarks
(b)Correct attempt at inverse
2 1 2 1  2 1 
y = 6 −  2 x + 3 =  x = − 3  2
AnswerMarks
2 x + 3 6 − y 6 − yM1
3( x + 3 21 3
f − 1 ( x ) = or f−1(x)= −
AnswerMarks
2 6 − x ) 2(6−x) 2A1
− 1 „ x  6B1
(3)
AnswerMarks
(c)( ) 2
g g ( x ) = 1 2 6  x 2 + 5 + 5 = 1 2 6  x 2 + 5 = 1 2 1M1
 x 2 = '' 6 ''  x = (  ) 6dM1
x= 6A1
(3)

Total 8

(c)
M1: For an attempt to ''undo'' g once in an attempt to find the value of g ( x ) , x2or x2 +5.
This may be scored for any of the following methods (condoning slips)
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x 2 + 5 = 1 2 1
( )2 ( )( )
x2 +5 +5=126 x4 +10x 2 −96=0 x 2 −6 x 2 +16 =0 x 2 =...
• Condone ( x 2 + 5 ) 2 + 5 = 1 2 6  x 4 + 1 0 x 2 − 9 6 = 0  x = a w r t 2 . 4 5 OR x 2 = 6
• g ( x ) = 1 2 6 − 5
Do not award this mark if the candidate believes that to find x you only need to undo g once.
So, answers like x = 1 1 or x = 6 will score 0,0,0 unless further work is seen
dM1: For a complete attempt to find the exact solution to the equation. Examples of this would include
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x 2 + 5 = 1 2 1  x = 1 2 1  5
( )2 ( )( )
• x2 +5 +5=126 x4 +10x 2 −96=0 x 2 −6 x 2 +16 =0 x 2 =6 x = 6
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x = 6 or
(
x 2 + 5
) 2
+ 5 = 1 2 6  x = − 6
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x 4 + 1 0 x 2 − 9 6 = 0  x = 6
• x = '1 1 −' 5
A1: x =  6 only. Ignore any reference to  4 i if included.
Sight of  6
PMT
without incorrect working scores all 3 marks
Question 6:
--- 6 (a) ---
6 (a) | gf ( 2 )=g ( 3 )=3 2 +5 | M1
= 14 | A1
(2)
(b) | Correct attempt at inverse
2 1 2 1  2 1 
y = 6 −  2 x + 3 =  x = − 3  2
2 x + 3 6 − y 6 − y | M1
3( x + 3 21 3
f − 1 ( x ) = or f−1(x)= −
2 6 − x ) 2(6−x) 2 | A1
− 1 „ x  6 | B1
(3)
(c) | ( ) 2
g g ( x ) = 1 2 6  x 2 + 5 + 5 = 1 2 6  x 2 + 5 = 1 2 1 | M1
 x 2 = '' 6 ''  x = (  ) 6 | dM1
x= 6 | A1
(3)
Total 8
(c)
M1: For an attempt to ''undo'' g once in an attempt to find the value of g ( x ) , x2or x2 +5.
This may be scored for any of the following methods (condoning slips)
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x 2 + 5 = 1 2 1
( )2 ( )( )
x2 +5 +5=126 x4 +10x 2 −96=0 x 2 −6 x 2 +16 =0 x 2 =...
• Condone ( x 2 + 5 ) 2 + 5 = 1 2 6  x 4 + 1 0 x 2 − 9 6 = 0  x = a w r t 2 . 4 5 OR x 2 = 6
• g ( x ) = 1 2 6 − 5
Do not award this mark if the candidate believes that to find x you only need to undo g once.
So, answers like x = 1 1 or x = 6 will score 0,0,0 unless further work is seen
dM1: For a complete attempt to find the exact solution to the equation. Examples of this would include
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x 2 + 5 = 1 2 1  x = 1 2 1  5
( )2 ( )( )
• x2 +5 +5=126 x4 +10x 2 −96=0 x 2 −6 x 2 +16 =0 x 2 =6 x = 6
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x = 6 or
(
x 2 + 5
) 2
+ 5 = 1 2 6  x = − 6
•
(
x 2 + 5
) 2
+ 5 = 1 2 6  x 4 + 1 0 x 2 − 9 6 = 0  x = 6
• x = '1 1 −' 5
A1: x =  6 only. Ignore any reference to  4 i if included.
Sight of  6
PMT
without incorrect working scores all 3 marks
\begin{enumerate}
  \item The functions f and g are defined by
\end{enumerate}

$$\begin{array} { l l } 
\mathrm { f } ( x ) = 6 - \frac { 21 } { 2 x + 3 } & x \geqslant 0 \\
\mathrm {~g} ( x ) = x ^ { 2 } + 5 & x \in \mathbb { R }
\end{array}$$

(a) Find $\mathrm { gf } ( 2 )$\\
(b) Find $f ^ { - 1 }$\\
(c) Solve the equation

$$\operatorname { gg } ( x ) = 126$$

\hfill \mbox{\textit{Edexcel PURE 2024 Q6}}
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