C3 January 2007 Q2
2. \[\mathrm{f}(x) = 1 - \frac{3}{x + 2} + \frac{3}{(x + 2)^2}, \quad x \neq -2.\]
(a) Show that \(\mathrm{f}(x) = \dfrac{x^2 + x + 1}{(x + 2)^2}\), \(x \neq -2\). (4)
(b) Show that \(x^2 + x + 1 \gt 0\) for all values of \(x\). (3)
(c) Show that \(\mathrm{f}(x) \gt 0\) for all values of \(x\), \(x \neq -2\). (1)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \dfrac{(x + 2)^2, -3(x + 2) + 3}{(x + 2)^2}\) | M1 A1, A1 |
| \(= \dfrac{x^2 + 4x + 4 - 3x - 6 + 3}{(x + 2)^2} = \dfrac{x^2 + x + 1}{(x + 2)^2}\) * cso | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(x^2 + x + 1 = \left(x + \dfrac{1}{2}\right)^2 + \dfrac{3}{4}\), \(\gt 0\) for all values of \(x\). | M1 A1, A1 |
| (3) |
Alternative to (b)
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x^2 + x + 1) = 2x + 1 = 0 \Rightarrow x = -\dfrac{1}{2} \Rightarrow x^2 + x + 1 = \dfrac{3}{4}\) | M1 A1 |
| A parabola with positive coefficient of \(x^2\) has a minimum \(\Rightarrow x^2 + x + 1 \gt 0\) Accept equivalent arguments | A1 (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \dfrac{\left(x + \dfrac{1}{2}\right)^2 + \dfrac{3}{4}}{(x + 2)^2}\) Numerator is positive from (b) \(x \neq -2 \Rightarrow (x + 2)^2 \gt 0\) (Denominator is positive) Hence \(\mathrm{f}(x) \gt 0\) | B1 |
| (1) | |
| (8 marks) |