CAIE P1 2010 November — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2010
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind function from composite
DifficultyStandard +0.3 This is a straightforward multi-part question on standard P1 topics: completing the square (routine), finding an inverse function (standard procedure), and finding a function from a composite (algebraic manipulation). All parts follow textbook methods with no novel insight required, making it slightly easier than average.
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

7 The function f is defined by $$\mathrm { f } ( x ) = x ^ { 2 } - 4 x + 7 \text { for } x > 2$$
  1. Express \(\mathrm { f } ( x )\) in the form \(( x - a ) ^ { 2 } + b\) and hence state the range of f .
  2. Obtain an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\). The function g is defined by $$\mathrm { g } ( x ) = x - 2 \text { for } x > 2$$ The function h is such that \(\mathrm { f } = \mathrm { hg }\) and the domain of h is \(x > 0\).
  3. Obtain an expression for \(\mathrm { h } ( x )\).

Question 7:
Part (i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((x-2)^2\)M1 Must be "\(-2\)" \(\pm k\)
\((x-2)^2 + 3\)A1 co
\(f(x) > 3\)B1\(\sqrt{}\) ft on their '3'
Part (ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x - 2 = (\pm)\sqrt{y-3}\)M1 \(\pm\) not required for M mark
\(f^{-1}(x) = 2 + \sqrt{x-3}\)A1 \(f(x)\) + removal of minus sign needed
domain is \(x > 3\)B1\(\sqrt{}\) ft domain of \(f^{-1}\) = range of \(f\) or for \(f^{-1}\)
Part (iii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(h(x) = x^2 + 3\)B1 co
## Question 7:

**Part (i)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x-2)^2$ | M1 | Must be "$-2$" $\pm k$ |
| $(x-2)^2 + 3$ | A1 | co |
| $f(x) > 3$ | B1$\sqrt{}$ | ft on their '3' |

**Part (ii)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x - 2 = (\pm)\sqrt{y-3}$ | M1 | $\pm$ not required for M mark |
| $f^{-1}(x) = 2 + \sqrt{x-3}$ | A1 | $f(x)$ + removal of minus sign needed |
| domain is $x > 3$ | B1$\sqrt{}$ | ft domain of $f^{-1}$ = range of $f$ or for $f^{-1}$ |

**Part (iii)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $h(x) = x^2 + 3$ | B1 | co |

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7 The function f is defined by

$$\mathrm { f } ( x ) = x ^ { 2 } - 4 x + 7 \text { for } x > 2$$

(i) Express $\mathrm { f } ( x )$ in the form $( x - a ) ^ { 2 } + b$ and hence state the range of f .\\
(ii) Obtain an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state the domain of $\mathrm { f } ^ { - 1 }$.

The function g is defined by

$$\mathrm { g } ( x ) = x - 2 \text { for } x > 2$$

The function h is such that $\mathrm { f } = \mathrm { hg }$ and the domain of h is $x > 0$.\\
(iii) Obtain an expression for $\mathrm { h } ( x )$.

\hfill \mbox{\textit{CAIE P1 2010 Q7 [7]}}