| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find minimum domain for inverse |
| Difficulty | Moderate -0.3 This is a standard multi-part question covering routine A-level techniques: finding tangent conditions via discriminant, completing the square, identifying range from vertex form, and finding the minimum domain for invertibility (vertex x-coordinate). All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02e 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 existence1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y + kx = 12\). Sim Eqns. | M1 | Complete elimination of \(y\) (or \(x\)) |
| \(\to 2x^2 - 8x + kx + 2 = 0\) | A1 | |
| Use of \(b^2 - 4ac\) | M1 | Uses \(b^2 - 4ac\) on eqn \(= 0\), no "\(x\)" in \(a, b, c\). |
| \(\to (k-8)^2 = 16 \to k = 12\) or \(4\). | A1 | co.co |
| [4] | ||
| (ii) \(2x^2 - 8x + 14 = 2(x-2)^2 + 6\) | B1 ×3 | |
| [3] | ||
| (iii) Range of \(f \geq 6\). | B1√ | √ for \(c\) or from calculus. |
| [1] | ||
| (iv) Smallest \(A = 2\) | B1√ | √ to answer to (ii). |
| [1] | ||
| (v) Makes \(x\) the subject | M1 | Could interchange \(x, y\) first. |
| Order of operations correct. | M1 | Order must be correct. |
| \(g^{-1}(x) = \sqrt{\frac{x-6}{2}} + 2\) | A1 | co |
| [3] |
$f: x \mapsto 2x^2 - 8x + 14$
(i) $y + kx = 12$. Sim Eqns. | M1 | Complete elimination of $y$ (or $x$)
$\to 2x^2 - 8x + kx + 2 = 0$ | A1 |
Use of $b^2 - 4ac$ | M1 | Uses $b^2 - 4ac$ on eqn $= 0$, no "$x$" in $a, b, c$.
$\to (k-8)^2 = 16 \to k = 12$ or $4$. | A1 | co.co
| [4] |
(ii) $2x^2 - 8x + 14 = 2(x-2)^2 + 6$ | B1 ×3 |
| [3] |
(iii) Range of $f \geq 6$. | B1√ | √ for $c$ or from calculus.
| [1] |
(iv) Smallest $A = 2$ | B1√ | √ to answer to (ii).
| [1] |
(v) Makes $x$ the subject | M1 | Could interchange $x, y$ first.
Order of operations correct. | M1 | Order must be correct.
$g^{-1}(x) = \sqrt{\frac{x-6}{2}} + 2$ | A1 | co
| [3] |10 The function $\mathrm { f } : x \mapsto 2 x ^ { 2 } - 8 x + 14$ is defined for $x \in \mathbb { R }$.\\
(i) Find the values of the constant $k$ for which the line $y + k x = 12$ is a tangent to the curve $y = \mathrm { f } ( x )$.\\
(ii) Express $\mathrm { f } ( x )$ in the form $a ( x + b ) ^ { 2 } + c$, where $a , b$ and $c$ are constants.\\
(iii) Find the range of f .
The function $\mathrm { g } : x \mapsto 2 x ^ { 2 } - 8 x + 14$ is defined for $x \geqslant A$.\\
(iv) Find the smallest value of $A$ for which g has an inverse.\\
(v) For this value of $A$, find an expression for $\mathrm { g } ^ { - 1 } ( x )$ in terms of $x$.
\hfill \mbox{\textit{CAIE P1 2010 Q10 [12]}}