$g(f(x)) = e^{\ln x} = e^{\ln x^2} = x^2$ | M1, M1, A1 [3] | Forming g(f(x)) (soi)

## Question 3(i)

$\frac{dy}{dx} = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{x^4} = \frac{x - 2x \ln x}{x^4} = \frac{1 - 2\ln x}{x^3}$ | M1, B1, A1, A1 [4] | quotient rule with $u = \ln x$ and $v = x^2$; $\frac{d}{dx}(x) = 1/x$ soi; correct expression (o.e.); o.e. cao, mark final answer, but must have divided top and bottom by x | Consistent with their derivatives; udv ± vdu in the quotient rule is M0. Condone $\ln x \cdot 2x = \ln 2x^2$ for this A1 (provided $\ln x.2x$ is shown). e.g. $\frac{1}{x^3} - \frac{2\ln x}{x^3}, x^{-3} - 2x^{-3}\ln x$

## Question 3 (alternative)

$\frac{dy}{dx} = -2x^{-3} \ln x + x^{-2}(\frac{1}{x})$ | M1, B1, A1 [4] | product rule with $u = x^{-2}$ and $v = \ln x$; $\frac{d}{dx}(\ln x) = 1/x$ soi; correct expression; o.e. cao, mark final answer, must simplify the $x^{-2}(1/x)$ term. | or vice-versa

## Question 3(ii)

$\int \frac{\ln x}{x^2} dx$ let $u = \ln x, du/dx = 1/x$; $dv/dx = 1/x^2, v = -x^{-1}$ $= -\frac{1}{x}\ln x + \int \frac{1}{x^2} dx = -\frac{1}{x}\ln x + \int \frac{1}{x^2} dx = -\frac{1}{x}\ln x - \frac{1}{x} + c = -\frac{1}{x}(\ln x + 1) + c^*$ | M1, A1, A1, A1, A1 [4] | Integration by parts with $u = \ln x, du/dx = 1/x, dv/dx = 1/x^2, v = -x^{-1}$ must be correct, condone + c; must be correct, condone missing c; at this stage. Need to see 1/x²; condone missing c; NB AG must have c shown in final answer | Must be correct. at this stage . Need to see 1/x²

## Question 4(i)

$h = a - bc^{-kt} \Rightarrow a = 10.5$ (their)a - bce^0 = 0.5$ $\Rightarrow b = 10$ | B1, M1, A1cao [3] | a need not be substituted

## Question 4(ii)

$h = 10.5 - 10e^{-kt}$; When $t = 8, h = 10.5 - 10e^{-8k} = 6$; $10e^{-8k} = 4.5$; $\Rightarrow -8k = \ln 0.45$; $k = \ln 0.45/(-8) = 0.09981\ldots = 0.10$ | M1, M1, M1, A1 [3] | ft their a and b (even if made up); taking lns correctly on a correct re-arrangement - ft a, b if not ceased cao (www) but allow 0.1 | allow M1 for $a - be^{-8k} = 6$; allow a and b unsubstituted; allow their 0.45 (or 4.5) to be negative