CAIE P1 2012 November — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyModerate -0.8 This is a straightforward inverse function question requiring algebraic manipulation to rearrange y = √((x+3)/2) + 1 for x in terms of y. The steps are routine: isolate the square root, square both sides, and rearrange to quadratic form. Finding the domain of f^(-1) from the range of f is standard. Easier than average as it follows a well-practiced procedure with no conceptual challenges.
Spec1.02v Inverse and composite functions: graphs and conditions for existence

2 A function f is such that \(\mathrm { f } ( x ) = \sqrt { } \left( \frac { x + 3 } { 2 } \right) + 1\), for \(x \geqslant - 3\). Find
  1. \(\mathrm { f } ^ { - 1 } ( x )\) in the form \(a x ^ { 2 } + b x + c\), where \(a , b\) and \(c\) are constants,
  2. the domain of \(\mathrm { f } ^ { - 1 }\).

Question 2:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(x) = \sqrt{\frac{x+3}{2}} + 1\), for \(x \geq -3\)
Make \(x\) the subject or interchange \(x, y\)M1 Attempt at \(x\) as subject and removes \(+1\)
\(\rightarrow 2(x-1)^2 - 3\)M1 Squares both sides and deals with "\(+3\)" and "\(\div 2\)"
\(\rightarrow 2x^2 - 4x - 1\)A1 co
[3]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Domain of \(f^{-1}\) is \(\geq 1\)B1 co, condone \(>1\)
[1]
## Question 2:

### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = \sqrt{\frac{x+3}{2}} + 1$, for $x \geq -3$ | | |
| Make $x$ the subject or interchange $x, y$ | M1 | Attempt at $x$ as subject and removes $+1$ |
| $\rightarrow 2(x-1)^2 - 3$ | M1 | Squares both sides and deals with "$+3$" and "$\div 2$" |
| $\rightarrow 2x^2 - 4x - 1$ | A1 | co |
| **[3]** | | |

### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Domain of $f^{-1}$ is $\geq 1$ | B1 | co, condone $>1$ |
| **[1]** | | |

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2 A function f is such that $\mathrm { f } ( x ) = \sqrt { } \left( \frac { x + 3 } { 2 } \right) + 1$, for $x \geqslant - 3$. Find\\
(i) $\mathrm { f } ^ { - 1 } ( x )$ in the form $a x ^ { 2 } + b x + c$, where $a , b$ and $c$ are constants,\\
(ii) the domain of $\mathrm { f } ^ { - 1 }$.

\hfill \mbox{\textit{CAIE P1 2012 Q2 [4]}}