C3 June 2011 Q7
7. \[\mathrm{f}(x) = \frac{4x - 5}{(2x + 1)(x - 3)} - \frac{2x}{x^2 - 9}, \qquad x \neq \pm 3,\ x \neq -\frac{1}{2}\]
(a) Show that \[\mathrm{f}(x) = \frac{5}{(2x + 1)(x + 3)}\] (5)
The curve \(C\) has equation \(y = \mathrm{f}(x)\). The point \(P\left(-1, -\dfrac{5}{2}\right)\) lies on \(C\).
(b) Find an equation of the normal to \(C\) at \(P\). (8)
| Scheme | Marks |
|---|---|
| \(x^2 - 9 = (x + 3)(x - 3)\) | B1 |
| \(\dfrac{4x - 5}{(2x + 1)(x - 3)} - \dfrac{2x}{(x + 3)(x - 3)}\) | |
| \(= \dfrac{(4x - 5)(x + 3)}{(2x + 1)(x - 3)(x + 3)} - \dfrac{2x(2x + 1)}{(2x + 1)(x + 3)(x - 3)}\) | M1 |
| \(= \dfrac{5x - 15}{(2x + 1)(x - 3)(x + 3)}\) | M1A1 |
| \(= \dfrac{5\cancel{(x - 3)}}{(2x + 1)\cancel{(x - 3)}(x + 3)} = \dfrac{5}{(2x + 1)(x + 3)}\) | A1* |
| (5) |
| Scheme | Marks |
|---|---|
| \(f(x) = \dfrac{5}{2x^2 + 7x + 3}\) | |
| \(f^{\prime}(x) = \dfrac{-5(\boldsymbol{4x + 7})}{(2x^2 + 7x + 3)^2}\) | M1M1A1 |
| \(f^{\prime}(-1) = -\dfrac{15}{4}\) | M1A1 |
| Uses \(m_1m_2 = -1\) to give gradient of normal \(= \dfrac{4}{15}\) | M1 |
| \(\dfrac{y - \left(-\frac{5}{2}\right)}{(x - -1)} = \text{their } \dfrac{4}{15}\) | M1 |
| \(y + \dfrac{5}{2} = \dfrac{4}{15}(x + 1)\) or any equivalent form | A1 |
| (8) | |
| (13 marks) |