$y = x^2(1 + 4x)^{1/2}$ $\Rightarrow \frac{dy}{dx} = x^2 \cdot \frac{1}{2}(1+4x)^{-1/2} \cdot 4 + 2x(1+4x)^{1/2} = 2x(1+4x)^{-1/2}(x+1+4x) = \frac{2x(5x+1)}{\sqrt{1+4x}}$ | M1, B1, A1, M1, A1 [5] | product rule with $u = x^2, v = \sqrt{1+4x}$; ½ (...) -1/2 soi; correct expression; factorising or combining fractions; NB AG (need not factor out the (2x) must have evidence of $x + 1 + 4x$ oe or $2x(5x + 1)(1 + 4x)^{-1/2}$ or $2x(5x + 1)/(1 + 4x)^{1/2}$ | (need not factor out the (2x)) must have evidence of $x + 1 + 4x$ oe or $2x(5x + 1)(1 + 4x)^{-1/2}$ or $2x(5x + 1)/(1 + 4x)^{1/2}$

## Question 6(i)

$\sin(\pi/3) + \cos(\pi/6) = \sqrt{3}/2 + \sqrt{3}/2 = \sqrt{3}$ | B1 [1] | must be exact, must show working | Not just $\sin(\pi/3) + \cos(\pi/6) = \sqrt{3}$, if substituting for y and solving for x (or vv) must evaluate $\sin \pi/3$ e.g. not $\arccos(\sqrt{3} - \sin \pi/3)$

## Question 6(ii)

$2\cos 2x - \sin y \frac{dy}{dx} = 0$; $\Rightarrow 2\cos 2x = \sin y \frac{dy}{dx}$; $\Rightarrow \frac{dy}{dx} = \frac{2\cos 2x}{\sin y}$; When $x = \pi/6, y = \pi/6$; $\Rightarrow \frac{dy}{dx} = \frac{2\cos \pi/3}{{\sin \pi/6}} = 2$ | M1, A1, A1cao, M1dep, A1, [5] | Implicit differentiation; correct expression; substituting dep 1st M1; www | allow one error, but must have (±) sin y dy/dx. Ignore dy/dx = ... unless pursued. 2cos 2x dx – sin y dy = 0 is M1A1 (could differentiate wrt y, get dy/dx, etc.); $-2\cos 2x / \sin y$ is A0; or 30°

## Question 7(i)

$(3^2 + 1)(3^n - 1) = (3)^2 - 1$ or $3^{2n} - 1$ | B1 [1] | mark final answer | or $9^n - 1$; penalise $3^n$ if it looks like 3 to the power $n^2$

## Question 7(ii)

$3^n$ is odd $\Rightarrow 3^n + 1$ and $3^n - 1$ both even. As consecutive even nos. one must be divisibly by 4, so product is divisible by 8. | M1, M1, A1 [3] | $3^n$ is odd; $3^n + 1$ and $3^n - 1$ both even; completion | Induction: If true for $n, 3^{2(n+1)} - 1 = 9x(8k + 1) - 1 = 72k + 8 = 8(9k + 1)$ so div by 8. A1. When $n = 1, 3^2 - 1 = 8$ div by 8, true A1 (or similar with $9^n$)

## Question 8(i)

$f(-x) = \frac{1}{e^{-x} + e^{-(-x)} + 2} = f(x)$, [≒ f is even *]; Symmetrical about Oy | M1, A1, B1 [3] | substituting $-x$ for x in f(x); condone 'reflection in y-axis'; | Can imply that $e^{(-x)} = e^x$ from $f(-x) = \frac{1}{e^{-x} + e^x + 2}$. Must mention axis

## Question 8(ii)

$f(x) = (e^x + e^{-x} + 2)^{-1}(e^x - e^{-x})$ or $= \frac{(e^x + e^{-x} + 2.0 - (e^x - e^{-x})}{(e^x + e^{-x} + 2)^2} = \frac{(e^x - e^x)}{(e^x + e^{-x} + 2)^2}$ | B1, M1, A1 [3] | $d/dx (e^x) = e^x$ and $d/dx(e^{-x}) = -e^x$ soi; chain or quotient rule; condone missing bracket on top if correct thereafter; o.e. mark final answer | If differentiating $\frac{e^x}{(e^x+1)^2}$ withhold A1 (unless result in (iii) proved here). e.g. $\frac{e^x}{(e^x+1)^2} \times(e^x - e^x)$ M1, $= \frac{1}{e^x + e^{-x} + 2} \times(e^x - e^x)$ A1; condone no $e^{2x} = (e^x)^2$, for both M1 and A1

## Question 8(iii)

$f(x) = \frac{e^x}{e^{2x} + 1 + 2e^x} = \frac{e^x}{(e^x + 1)^2}$ | M1, A1 [2] | × top and bottom by $e^x$ (correctly); condone $e^2x$ for M1 but not A1; NB AG | or $\frac{e^x}{(e^x + 1)^2} = \frac{e^x}{e^{2x} + 2e^x + 1}$ M1, $= \frac{1}{e^x + e^{-x} + 2}$ A1; condone no $e^{2x} = (e^x)^2$, for both M1 and A1

## Question 8(iv)

$A = \int_0^1 \frac{1}{(e^x + 1)^2} dx$; let $u = e^x + 1, du = e^x dx$; when $x = 0, u = 2$; when $x = 1, u = e + 1$; $\Rightarrow A = \int_2^{1+e} \frac{1}{u^2} du = \left[-\frac{1}{u}\right]_2^{1+e} = -\frac{1}{1+e} + \frac{1}{2} = \frac{1}{2} - \frac{1}{1+e}$ | B1, M1, A1, M1, A1cao [5] | correct integral and limits; let $u = e^x + 1, du = e^x dx$; substituting correct limits (dep 1st M1 and integration); o.e. mark final answer. Must be exact. Don't allow $e^x$. | condone no dx, must use f(x) = $\frac{e^x}{(e^x+1)^2}$. Limits may be implied by subsequent work. If 0.231.. unsupported, allow 1st B1 only; or by inspection $\left[\frac{k}{e^x+1}\right]$ M1 $\left[-\frac{1}{e^x+1}\right]$ A1; upper–lower: 2 and 1+e (or 3.7..)for u, or 0 and 1 for x if substituted back (correctly); e.g. $\frac{e-1}{2(1+e)}$. Can isw 0.231, which may be used as evidence of M1. Can isw numerical ans (e.g. 0.231) but not algebraic errors

## Question 8(v)

Curves intersect when $f(x) = \frac{1}{4}e^x$; $\Rightarrow (e^x + 1)^2 = 4$; $\Rightarrow e^x = 1$ or $-3$; so as $e^x > 0$, only one solution $e^x = 1 \Rightarrow x = 0$ wwu (for this value); when $x = 0, y = 1/4$ wwu (for the y value) | M1, M1, A1, B1, B1 [5] | soi; or equivalent quadratic – must be correct; getting $e^x = 1$ and discounting other sol^n; x = 0 wwu (for this value); y = ¼ wwu (for the y value) | www e.g. $e^x = -1 [or e^x + 1 = -2]$ not possible; www unless verified. Do not allow unsupported. A sketch is not sufficient.

## Question 9(i)

When $x = 0, f(x) = a = 2^*$; When $x = \pi, f(\pi) = 2 + \sin b\pi = 3$; $\sin b\pi = 1$; $b\pi = \pi/2, b = 1/2^*$ or $1 = a + \sin(-\pi b) = a - \sin \pi b$; $\Rightarrow \sin(-b\pi) = -1$ condone using degrees; $\Rightarrow b = 1/2^*$ | M1, M1, A1, M1 for both points substituted; A1 solving for b or a; A1 substituting to get (or b) | or equiv transformation arguments: e.g. 'curve is shifted up 2 so a = 2'; e.g. period of sine curve is 4π, or stretched by sf. 2 in x-direction (not squeezed or squashed by ½); $\Rightarrow b = 1/2$ If verified allow M1A0. If y = 2 + sin ½ x verified at two points, SC2

## Question 9(ii)

$f'(x) = 1/2 \cos 1/2 x$; $\Rightarrow f'(0) = 1/2$; Maximum value of $\cos 1/2 x$ is 1; $\Rightarrow$ max value of gradient is $1/2$ | M1, A1, A1, M1, A1 [5] | ±k cos ½ x; cao; www; or $f''(x) = -1/4 \sin 1/2 x$; $f''(x) = 0 \Rightarrow x = 0$, so max val of $f'(x)$ is $1/2$

## Question 9(iii)

$y = 2 + \sin 1/2 x \leftrightarrow x = 2 + \sin 1/2 y$; $\Rightarrow x - 2 = \sin 1/2 y$; $\Rightarrow \arcsin(x - 2) = 1/2 y$; $\Rightarrow y = f^{-1}(x) = 2\arcsin(x - 2)$; Domain $1 \leq x \leq 3$; Range $-\pi \leq x \leq \pi$; Gradient at (2, 0) is 2 | M1, A1, A1, B1, B1, M1ft [6] | Attempt to invert formula; or $\arcsin(y - 2) = 1/2 x$; must be $y = \ldots$ or $f^{-1}(x) = \ldots$; or [1, 3]; or $[-\pi, \pi]$ or –π ≤ x ≤ π. Penalise '<' (or '1 to 3','–π to π') once only; or by differentiating $\arcsin(x - 2)$ condone no bracket for 1st A1 only; or sin⁻¹(x - 2), condone $f^{-1}(x)$, must have bracket in final ans; but not 1 ≤ y ≤ 3; but not –π ≤ x ≤ π. Penalise '<'s (or '1 to 3','–π to π') once only; or by differentiating arcsin(x - 2) or implicitly

## Question 9(iv)

$A = \int_0^{\pi} (2 + \sin \frac{1}{2}x) dx = \left[2x - k\cos \frac{1}{2}x\right]_0^{\pi}$ where k is positive; $= 2\pi - (-2) = 2\pi + 2 (= 8.2831\ldots)$ | M1, M1, A1, A1cao [4] | correct integral and limits; $\left[2x - k\cos \frac{1}{2}x\right]$ where k is positive; k = 2; answers rounding to 8.3 | soi from subsequent work, condone no dx but not 180. Unsupported correct answers score 1st M1 only.