| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Function properties proof |
| Difficulty | Moderate -0.3 This is a straightforward multi-part algebraic proof question requiring routine manipulation to combine fractions (part a), completing the square or using the discriminant to show a quadratic is always positive (part b), and combining these results (part c). All techniques are standard C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(x) = \frac{(x+2)^2 - 3(x+2) + 3}{(x+2)^2} = \frac{x^2 + 4x + 4 - 3x - 6 + 3}{(x+2)^2} = \frac{x^2 + x + 1}{(x+2)^2}\) | M1, A1, A1 | cso (4 marks) |
| (b) \(x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} > 0\) for all values of \(x\) | M1, A1, A1 | (3 marks) |
| (c) Numerator is positive from (b); \(x \neq -2 \Rightarrow (x+2)^2 > 0\) (Denominator is positive); Hence \(f(x) > 0\) | B1 | (1 mark) |
| Alternative to (b): \(\frac{d}{dx}(x^2 + x + 1) = 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \Rightarrow x^2 + x + 1 = \frac{3}{4}\); A parabola with positive coefficient of \(x^2\) has a minimum \(\Rightarrow x^2 + x + 1 > 0\); Accept equivalent arguments | M1, A1, A1 | (3 marks) |
(a) $f(x) = \frac{(x+2)^2 - 3(x+2) + 3}{(x+2)^2} = \frac{x^2 + 4x + 4 - 3x - 6 + 3}{(x+2)^2} = \frac{x^2 + x + 1}{(x+2)^2}$ | M1, A1, A1 | cso (4 marks)
(b) $x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} > 0$ for all values of $x$ | M1, A1, A1 | (3 marks)
(c) Numerator is positive from (b); $x \neq -2 \Rightarrow (x+2)^2 > 0$ (Denominator is positive); Hence $f(x) > 0$ | B1 | (1 mark)
**Alternative to (b):** $\frac{d}{dx}(x^2 + x + 1) = 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \Rightarrow x^2 + x + 1 = \frac{3}{4}$; A parabola with positive coefficient of $x^2$ has a minimum $\Rightarrow x^2 + x + 1 > 0$; Accept equivalent arguments | M1, A1, A1 | (3 marks)
---2.
$$f ( x ) = 1 - \frac { 3 } { x + 2 } + \frac { 3 } { ( x + 2 ) ^ { 2 } } , x \neq - 2$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = \frac { x ^ { 2 } + x + 1 } { ( x + 2 ) ^ { 2 } } , x \neq - 2$.
\item Show that $x ^ { 2 } + x + 1 > 0$ for all values of $x$.
\item Show that $\mathrm { f } ( x ) > 0$ for all values of $x , x \neq - 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q2 [8]}}