## Question 1:

**EITHER Solution 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln(x+2) = \ln\!\left(2\!\left(1+\tfrac{1}{2}x\right)\right) = \ln 2 + \ln\!\left(1+\tfrac{1}{2}x\right)$ | M1 | Changes $\ln(x+2)$ or $\ln(x^2+5)$ so that the formula given in the list of formulae (MF19) can be applied |
| $\ln(x+2) = \ln 2 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \ldots$ | A1 | |
| $\ln(x^2+5) = \ln\!\left(5\!\left(1+\tfrac{1}{5}x^2\right)\right) = \ln 5 + \tfrac{1}{5}x^2 + \ldots$ | A1 | |
| $\left(\ln 2 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \ldots\right) + \left(\ln 5 + \tfrac{1}{5}x^2 + \ldots\right)$ | M1 | Sums power series |

**OR Solution 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = (x+2)^{-1} + 2x(x^2+5)^{-1}$ | (M1 A1) | Finds first derivative |
| $f''(x) = -(x+2)^{-2} - (2x)^2(x^2+5)^{-2} + 2(x^2+5)^{-1}$ | (A1) | Finds second derivative |
| $f(0) = \ln 10 \quad f'(0) = \tfrac{1}{2} \quad f''(0) = \tfrac{3}{20}$ | (M1) | Evaluates derivatives at zero |
| $\ln(x+2) + \ln(x^2+5) = \ln 10 + \tfrac{1}{2}x + \tfrac{3}{40}x^2$ | A1 | Accept $\ln 10$ written as $\ln 2 + \ln 5$ but do not accept decimals. WWW |

**Total: 5 marks**