CAIE P1 2017 November — Question 9 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2017
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind composite function expression
DifficultyModerate -0.8 This is a straightforward composite and inverse functions question requiring routine techniques: gg(x) is simple substitution, f^{-1}(x) involves standard algebraic manipulation with a restricted domain, and solving fg(x)=1/7 requires substituting g into f then solving a quadratic. All steps are mechanical with no novel insight required, making it easier than average.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence
9 Functions f and g are defined for \(x > 3\) by $$\begin{aligned} & \mathrm { f } : x \mapsto \frac { 1 } { x ^ { 2 } - 9 } \\ & \mathrm {~g} : x \mapsto 2 x - 3 \end{aligned}$$
  1. Find and simplify an expression for \(\operatorname { gg } ( x )\).
  2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\).
  3. Solve the equation \(\operatorname { fg } ( x ) = \frac { 1 } { 7 }\).
Question 9(i):
AnswerMarks
\(gg(x) = g(2x-3) = 2(2x-3) - 3 = 4x - 9\)M1A1
Question 9(ii):
AnswerMarks Guidance
\(y = \dfrac{1}{x^2 - 9} \rightarrow x^2 = \dfrac{1}{y} + 9\) OEM1 Invert; add 9 to both sides or with \(x/y\) interchanged
\(f^{-1}(x) = \sqrt{\dfrac{1}{x} + 9}\)A1
Attempt soln of \(\sqrt{\dfrac{1}{x} + 9} > 3\) or attempt to find range of f. \((y > 0)\)M1
Domain is \(x > 0\) CAOA1 May simply be stated for B2
Question 9(iii):
EITHER method:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{(2x-3)^2 - 9} = \frac{1}{7}\)(M1)
\((2x-3)^2 = 16\) or \(4x^2 - 12x - 7 = 0\)A1
\(x = \frac{7}{2}\) or \(-\frac{1}{2}\)A1
\(x = \frac{7}{2}\) onlyA1
OR method:
AnswerMarks Guidance
AnswerMarks Guidance
\(g(x) = f^{-1}\!\left(\frac{1}{7}\right)\)(M1)
\(g(x) = 4\)A1
\(2x - 3 = 4\)A1
\(x = \frac{7}{2}\)A1
Total4 
## Question 9(i):

$gg(x) = g(2x-3) = 2(2x-3) - 3 = 4x - 9$ | M1A1 |

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## Question 9(ii):

$y = \dfrac{1}{x^2 - 9} \rightarrow x^2 = \dfrac{1}{y} + 9$ OE | M1 | Invert; add 9 to both sides or with $x/y$ interchanged

$f^{-1}(x) = \sqrt{\dfrac{1}{x} + 9}$ | A1 |

Attempt soln of $\sqrt{\dfrac{1}{x} + 9} > 3$ or attempt to find range of f. $(y > 0)$ | M1 |

Domain is $x > 0$ CAO | A1 | May simply be stated for B2

## Question 9(iii):

**EITHER method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(2x-3)^2 - 9} = \frac{1}{7}$ | (M1) | |
| $(2x-3)^2 = 16$ or $4x^2 - 12x - 7 = 0$ | A1 | |
| $x = \frac{7}{2}$ or $-\frac{1}{2}$ | A1 | |
| $x = \frac{7}{2}$ only | A1 | |

**OR method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $g(x) = f^{-1}\!\left(\frac{1}{7}\right)$ | (M1) | |
| $g(x) = 4$ | A1 | |
| $2x - 3 = 4$ | A1 | |
| $x = \frac{7}{2}$ | A1 | |
| **Total** | **4** | |

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9 Functions f and g are defined for $x > 3$ by

$$\begin{aligned}
& \mathrm { f } : x \mapsto \frac { 1 } { x ^ { 2 } - 9 } \\
& \mathrm {~g} : x \mapsto 2 x - 3
\end{aligned}$$

(i) Find and simplify an expression for $\operatorname { gg } ( x )$.\\

(ii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state the domain of $\mathrm { f } ^ { - 1 }$.\\

(iii) Solve the equation $\operatorname { fg } ( x ) = \frac { 1 } { 7 }$.\\

\hfill \mbox{\textit{CAIE P1 2017 Q9 [10]}}
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