FP2 June 2014 Q3
3. \[y = \sqrt{8 + \mathrm{e}^x}, \quad x \in \mathbb{R}\]
Find the series expansion for \(y\) in ascending powers of \(x\), up to and including the term in \(x^2\), giving each coefficient in its simplest form. (8)
| Scheme | Marks |
|---|---|
| \(y = \sqrt{8 + \mathrm{e}^x}\) | |
| \(y = \left(8 + \mathrm{e}^x\right)^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac{1}{2}\left(8 + \mathrm{e}^x\right)^{-\frac{1}{2}} \times \mathrm{e}^x\) M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = k\left(8 + \mathrm{e}^x\right)^{-\frac{1}{2}} \times \mathrm{e}^x\) A1: Correct differentiation | M1A1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{1}{2}\left(8 + \mathrm{e}^x\right)^{-\frac{1}{2}} \times \mathrm{e}^x - \dfrac{1}{4}\left(8 + \mathrm{e}^x\right)^{-\frac{3}{2}} \times \mathrm{e}^{2x}\) M1: Correct use of the product rule \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = k\left(8 + \mathrm{e}^x\right)^{-\frac{1}{2}} \times \mathrm{e}^x \pm K\left(8 + \mathrm{e}^x\right)^{-\frac{3}{2}} \times \mathrm{e}^{(2)x}\) A1: Correct second derivative with \(\mathrm{e}^x \times \mathrm{e}^x\) or \(\mathrm{e}^{2x}\) | M1A1 |
| \(\mathrm{f}(0) = 3\) May only appear in the expansion | B1 |
| \(\mathrm{f}^{\prime}(0) = \dfrac{1}{6},\ \mathrm{f}^{\prime\prime}(0) = \dfrac{17}{108}\) Attempt both \(\mathrm{f}^{\prime}(0)\) and \(\mathrm{f}^{\prime\prime}(0)\) with their derivatives found above | M1 |
| \((y =)\ 3 + \dfrac{1}{6}x + \dfrac{17}{216}x^2\) M1: Uses the correct Maclaurin series with their values. Accept 2 or 2! in \(x^2\) term A1: Correct expression | M1 A1cso |
| (8) | |
| (8 marks) |
Alternative Methods
| Scheme | Marks |
|---|---|
| \(\mathrm{e}^x = 1 + x + \dfrac{x^2}{2!} + \ldots\) 2 or 2! | |
| \(y = \left(9 + x + \dfrac{x^2}{2}\ldots\right)^{\frac{1}{2}}\) M1: Subst corrrect expansion | M1 |
| \(= 3\left(1 + \dfrac{x}{9} + \dfrac{x^2}{9 \times 2} + \ldots\right)^{\frac{1}{2}}\) B1: for 3 A1: for bracket | B1 A1 |
| \(= 3\left(1 + \dfrac{1}{2}\left(\dfrac{x}{9} + \dfrac{x^2}{2 \times 9}\right) + \dfrac{\frac{1}{2} \times \left(-\frac{1}{2}\right)}{2!}\left(\dfrac{x}{9} + \dfrac{x^2}{2 \times 9}\right)^2\right)\) M1: Binomial expansion up to at least the squared term, 2 or 2! With squared term A1: Correct expansion ie contents of bracket correct | M1A1 |
| \(= 3 + \dfrac{x}{6} + \dfrac{x^2}{12} - \dfrac{3}{8} \times \dfrac{x^2}{81}\) M1 Remove all brackets | M1 |
| \((y =)\ 3 + \dfrac{1}{6}x + \dfrac{17}{216}x^2\) M1: Combine \(x^2\) terms and obtain a 3 term quadratic A1: Correct expression with or without \(y = \ldots\) | M1A1 |
By implicit differentiation: For the first 4 marks (rest as first method)
\(y^2 = 8 + \mathrm{e}^x\)
M1A1 \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\) M1A1 \(2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \mathrm{e}^x\)