| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Complete the square |
| Difficulty | Moderate -0.3 This is a standard multi-part question on completing the square and inverse functions. Part (i) is routine algebraic manipulation, parts (ii)-(iii) require understanding of stationary points and domain restrictions for inverses, and part (iv) involves finding an inverse function—all standard P1 techniques with no novel problem-solving required. Slightly easier than average due to straightforward structure. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(25 - 2(x+3)^2\) | B1 B1 | Mark expression if present: B1 for 25 and B1 for \(-2(x+3)^2\). If no expression award \(a = 25\) B1 and \(b = 3\) B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((-3, 25)\) | B1FT | FT from answers to (i) or by calculus |
| Answer | Marks | Guidance |
|---|---|---|
| \((k) = -3\) also allow \(x\) or \(k \geq -3\) | B1FT | FT from answer to (i) or (ii) NOT \(x = -3\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 25 - 2(x+3)^2 \rightarrow 2(x+3)^2 = 25 - y\) | *M1 | Makes their squared term containing \(x\) the subject or equivalent with \(x/y\) interchanged first. Condone errors with +/- signs |
| \(x + 3 = (\pm)\sqrt{\frac{1}{2}(25-y)}\) | DM1 | Divide by \(\pm 2\) and then square root, allow \(\pm\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 7 - 2x^2 - 12x \rightarrow 2x^2 + 12x + y - 7\ (=0)\) | *M1 | Rearranging equation of the curve |
| \(x = \frac{-12 \pm \sqrt{12^2 - 8(y-7)}}{4}\) | DM1 | Correct use of their \(a\), \(b\) and \(c\) in quadratic formula. Allow just \(+\) in place of \(\pm\) |
| \(g^{-1}(x) = \sqrt{\left(\frac{25-x}{2}\right)} - 3\) oe, isw if substituting \(x = -3\) | A1 | \(\pm\) gets A0. Must now be a function of \(x\). Allow \(y =\) |
## Question 7(i):
$25 - 2(x+3)^2$ | B1 B1 | Mark expression if present: B1 for 25 and B1 for $-2(x+3)^2$. If no expression award $a = 25$ B1 and $b = 3$ B1
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## Question 7(ii):
$(-3, 25)$ | B1FT | FT from answers to (i) or by calculus
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## Question 7(iii):
$(k) = -3$ also allow $x$ or $k \geq -3$ | B1FT | FT from answer to (i) or (ii) **NOT** $x = -3$
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## Question 7(iv):
**EITHER method:**
$y = 25 - 2(x+3)^2 \rightarrow 2(x+3)^2 = 25 - y$ | *M1 | Makes their squared term containing $x$ the subject or equivalent with $x/y$ interchanged first. Condone errors with +/- signs
$x + 3 = (\pm)\sqrt{\frac{1}{2}(25-y)}$ | DM1 | Divide by $\pm 2$ and then square root, allow $\pm$
**OR method:**
$y = 7 - 2x^2 - 12x \rightarrow 2x^2 + 12x + y - 7\ (=0)$ | *M1 | Rearranging equation of the curve
$x = \frac{-12 \pm \sqrt{12^2 - 8(y-7)}}{4}$ | DM1 | Correct use of their $a$, $b$ and $c$ in quadratic formula. Allow just $+$ in place of $\pm$
$g^{-1}(x) = \sqrt{\left(\frac{25-x}{2}\right)} - 3$ oe, isw if substituting $x = -3$ | A1 | $\pm$ gets A0. Must now be a function of $x$. Allow $y =$7 The function f is defined by $\mathrm { f } : x \mapsto 7 - 2 x ^ { 2 } - 12 x$ for $x \in \mathbb { R }$.\\
(i) Express $7 - 2 x ^ { 2 } - 12 x$ in the form $a - 2 ( x + b ) ^ { 2 }$, where $a$ and $b$ are constants.\\
(ii) State the coordinates of the stationary point on the curve $y = \mathrm { f } ( x )$.\\
The function g is defined by $\mathrm { g } : x \mapsto 7 - 2 x ^ { 2 } - 12 x$ for $x \geqslant k$.\\
(iii) State the smallest value of $k$ for which g has an inverse.\\
(iv) For this value of $k$, find $\mathrm { g } ^ { - 1 } ( x )$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q7 [7]}}