| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Complete the square |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard P1 techniques: tangency condition via discriminant, solving quadratic inequalities, composite functions, and completing the square. All parts are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.02v Inverse and composite functions: graphs and conditions for existence1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = g(x) \rightarrow 2x^2 + 6x + 1 + k\ (= 0)\) | \*M1 | Forms a quadratic with all terms on same side |
| Use of \(b^2 = 4ac\) | DM1 | Uses the discriminant \(= 0\) |
| \((k =)\ 3\frac{1}{2}\) | A1 | OE, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4x + 8 = 2\ (\rightarrow x = -1\frac{1}{2})\) | \*M1 | Differentiating, equating gradients and solving to give \(x =\) |
| Substitutes their \(x\) into \(2x^2 + 6x + 1 + k = 0\) OR curve to find \(y\left(= \frac{-13}{2}\right)\) then both values into line | DM1 | Substituting appropriately for their \(x\) and proceeding to find a value of \(k\) |
| \((k =)\ 3\frac{1}{2}\) | A1 | OE, WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x^2 + 6x - 8\ (< 0)\) | M1 | Forms a quadratic with all terms on same side |
| \(-4\) and \(1\) | A1 | |
| \(-4 < x < 1\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((g^{-1}(x)) = \dfrac{x-1}{2}\) | B1 | Needs to be in terms of \(x\) |
| \((g^{-1}f(x)) = \dfrac{2x^2 + 8x + 1 - 1}{2} = 0 \rightarrow (2x^2 + 8x = 0) \rightarrow x =\) | M1 | Substitutes \(f\) into \(g^{-1}\) and attempts to solve it \(= 0\) as far as \(x =\) |
| \(0,\ -4\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(x+2)^2 - 7\) | B1 B1 | or \(a = +2,\ b = -7\) |
| Least value of \(f(x)\) or \(y =)\ -7\) or \(\geqslant -7\) | B1FT | FT for their \(b\) from a correct form of the expression |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = g(x) \rightarrow 2x^2 + 6x + 1 + k\ (= 0)$ | \*M1 | Forms a quadratic with all terms on same side |
| Use of $b^2 = 4ac$ | DM1 | Uses the discriminant $= 0$ |
| $(k =)\ 3\frac{1}{2}$ | A1 | OE, WWW |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4x + 8 = 2\ (\rightarrow x = -1\frac{1}{2})$ | \*M1 | Differentiating, equating gradients and solving to give $x =$ |
| Substitutes their $x$ into $2x^2 + 6x + 1 + k = 0$ OR curve to find $y\left(= \frac{-13}{2}\right)$ then both values into line | DM1 | Substituting appropriately for their $x$ and proceeding to find a value of $k$ |
| $(k =)\ 3\frac{1}{2}$ | A1 | OE, WWW |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x^2 + 6x - 8\ (< 0)$ | M1 | Forms a quadratic with all terms on same side |
| $-4$ and $1$ | A1 | |
| $-4 < x < 1$ | A1 | CAO |
---
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(g^{-1}(x)) = \dfrac{x-1}{2}$ | B1 | Needs to be in terms of $x$ |
| $(g^{-1}f(x)) = \dfrac{2x^2 + 8x + 1 - 1}{2} = 0 \rightarrow (2x^2 + 8x = 0) \rightarrow x =$ | M1 | Substitutes $f$ into $g^{-1}$ and attempts to solve it $= 0$ as far as $x =$ |
| $0,\ -4$ | A1 | CAO |
---
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(x+2)^2 - 7$ | B1 B1 | or $a = +2,\ b = -7$ |
| Least value of $f(x)$ or $y =)\ -7$ or $\geqslant -7$ | B1FT | FT for their $b$ from a correct form of the expression |
---9 Functions f and g are defined by
$$\begin{aligned}
& \mathrm { f } ( x ) = 2 x ^ { 2 } + 8 x + 1 \quad \text { for } x \in \mathbb { R } \\
& \mathrm {~g} ( x ) = 2 x - k \quad \text { for } x \in \mathbb { R }
\end{aligned}$$
where $k$ is a constant.\\
(i) Find the value of $k$ for which the line $y = \mathrm { g } ( x )$ is a tangent to the curve $y = \mathrm { f } ( x )$.\\
(ii) In the case where $k = - 9$, find the set of values of $x$ for which $\mathrm { f } ( x ) < \mathrm { g } ( x )$.\\
(iii) In the case where $k = - 1$, find $\mathrm { g } ^ { - 1 } \mathrm { f } ( x )$ and solve the equation $\mathrm { g } ^ { - 1 } \mathrm { f } ( x ) = 0$.\\
(iv) Express $\mathrm { f } ( x )$ in the form $2 ( x + a ) ^ { 2 } + b$, where $a$ and $b$ are constants, and hence state the least value of $\mathrm { f } ( x )$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q9 [12]}}