CAIE P1 2019 November — Question 2 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyModerate -0.3 This is a straightforward inverse function question requiring completing the square (a standard technique) and solving for x. The domain restriction x > 4 makes it routine. Slightly easier than average as it's a well-practiced procedure with clear steps and no conceptual surprises.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02v Inverse and composite functions: graphs and conditions for existence

2 The function g is defined by \(\mathrm { g } ( x ) = x ^ { 2 } - 6 x + 7\) for \(x > 4\). By first completing the square, find an expression for \(\mathrm { g } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { g } ^ { - 1 }\).

Question 2:
AnswerMarks Guidance
\((y =)\left[(x-3)^2\right][-2]\)\*B1, DB1 DB1 dependent on 3 in 1st bracket
\(x - 3 = (\pm)\sqrt{y+2}\) or \(y - 3 = (\pm)\sqrt{x+2}\)M1 Correct order of operations
\(\left(g^{-1}(x)\right) = 3 + \sqrt{x+2}\)A1 Must be in terms of \(x\)
Domain of \(g^{-1}\) is \((x) > -1\)B1 Allow \((-1, \infty)\). Do not allow \(y > -1\) or \(g(x) > -1\) or \(g^{-1}(x) > -1\)
## Question 2:

$(y =)\left[(x-3)^2\right][-2]$ | **\*B1, DB1** | DB1 dependent on 3 in 1st bracket

$x - 3 = (\pm)\sqrt{y+2}$ **or** $y - 3 = (\pm)\sqrt{x+2}$ | **M1** | Correct order of operations

$\left(g^{-1}(x)\right) = 3 + \sqrt{x+2}$ | **A1** | Must be in terms of $x$

Domain of $g^{-1}$ is $(x) > -1$ | **B1** | Allow $(-1, \infty)$. Do not allow $y > -1$ or $g(x) > -1$ or $g^{-1}(x) > -1$

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2 The function g is defined by $\mathrm { g } ( x ) = x ^ { 2 } - 6 x + 7$ for $x > 4$. By first completing the square, find an expression for $\mathrm { g } ^ { - 1 } ( x )$ and state the domain of $\mathrm { g } ^ { - 1 }$.\\

\hfill \mbox{\textit{CAIE P1 2019 Q2 [5]}}