| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Moderate -0.3 This is a straightforward C3 composite functions question requiring completing the square for range, direct substitution for composition evaluation, and solving a simple quadratic equation. All techniques are standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02u Functions: definition and vocabulary (domain, range, mapping) |
| Answer | Marks |
|---|---|
| \(f(x) = (x - \frac{3}{2})^2 - \frac{9}{4} + 7 = (x - \frac{3}{2})^2 + \frac{19}{4}\) | M1 A1 |
| \(\therefore f(x) \geq \frac{19}{4}\) | A1 |
| Answer | Marks |
|---|---|
| \(= g(11) = 21\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(fg(x) = f(2x - 1) = (2x - 1)^2 - 3(2x - 1) + 7\) | M1 | |
| \(\therefore 4x^2 - 4x + 1 - 6x + 3 + 7 = 17\) | ||
| \(2x^2 - 5x - 3 = 0\) | A1 | |
| \((2x + 1)(x - 3) = 0\) | M1 | |
| \(x = -\frac{1}{2}, 3\) | A1 | (9 marks) |
**(a)**
$f(x) = (x - \frac{3}{2})^2 - \frac{9}{4} + 7 = (x - \frac{3}{2})^2 + \frac{19}{4}$ | M1 A1 |
$\therefore f(x) \geq \frac{19}{4}$ | A1 |
**(b)**
$= g(11) = 21$ | M1 A1 |
**(c)**
$fg(x) = f(2x - 1) = (2x - 1)^2 - 3(2x - 1) + 7$ | M1 |
$\therefore 4x^2 - 4x + 1 - 6x + 3 + 7 = 17$ | |
$2x^2 - 5x - 3 = 0$ | A1 |
$(2x + 1)(x - 3) = 0$ | M1 |
$x = -\frac{1}{2}, 3$ | A1 | (9 marks)2. The functions $f$ and $g$ are defined by
$$\begin{aligned}
& \mathrm { f } ( x ) \equiv x ^ { 2 } - 3 x + 7 , \quad x \in \mathbb { R } \\
& \mathrm {~g} ( x ) \equiv 2 x - 1 , \quad x \in \mathbb { R }
\end{aligned}$$
(a) Find the range of f .\\
(b) Evaluate $\operatorname { gf } ( - 1 )$.\\
(c) Solve the equation
$$\mathrm { fg } ( x ) = 17$$
\begin{enumerate}
\setcounter{enumi}{2}
\item $f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 13 x ^ { 2 } + 26 x - 17 } { x ^ { 2 } - 3 x + 3 } , x \in \mathbb { R }$.\\
(a) Find the values of the constants $A$, $B$, $C$ and $D$ such that
\end{enumerate}
$$f ( x ) = x ^ { 2 } + A x + B + \frac { C x + D } { x ^ { 2 } - 3 x + 3 }$$
The point $P$ on the curve $y = \mathrm { f } ( x )$ has $x$-coordinate 1.\\
(b) Show that the normal to the curve $y = \mathrm { f } ( x )$ at $P$ has the equation
$$x + 5 y + 9 = 0$$
\begin{enumerate}
\setcounter{enumi}{3}
\item (a) Given that
\end{enumerate}
$$x = \sec \frac { y } { 2 } , \quad 0 \leq y < \pi ,$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x \sqrt { x ^ { 2 } - 1 } } .$$
(b) Find an equation for the tangent to the curve $y = \sqrt { 3 + 2 \cos x }$ at the point where $x = \frac { \pi } { 3 }$.\\
\hfill \mbox{\textit{Edexcel C3 Q2 [9]}}