$(x + y)^2 = 4x$ | M1, A1 [4] | Implicit differentiation of LHS correct expression = 4

$\Rightarrow \quad 2(x + y)( \frac{dy}{dx}) = 4$ | |

$\Rightarrow \quad 1 + \frac{dy}{dx} = \frac{2}{2(x + y)} = \frac{2}{x + y}$ | A1 |

$\Rightarrow \quad \frac{dy}{dx} = \frac{2}{x + y} - 1$ * | A1 | www (AG) | A0 if missing brackets in earlier working

or $x^2 + 2xy + y^2 = 4x$ | M1dep, A1 [4] | Implicit differentiation of LHS dep correct expansion correct expression = 4 (oe after re-arrangement)

$\Rightarrow \quad 2x + 2x\frac{dy}{dx} + 2y + 2y\frac{dy}{dx} = 4$ | |

$\Rightarrow \quad \frac{dy}{dx}(2x + 2y) = 4 - 2x - 2y$ | A1 | www (AG) | allow 1 error provided 2 $xdy/dx$ and 2 $ydy/dx$ are correct, but must expand $(x + y)^2$ correctly for M1 (so $x^2 + y^2 = 4x$ is M0) ignore superfluous $dy/dx = \ldots$ for M1, and for both A1s if not pursued

$\Rightarrow \quad \frac{dy}{dx} = \frac{4 - 2x - 2y}{2x + 2y} = \frac{4}{2x + 2y} - 1 = \frac{2}{x + y} - 1$ * | B1 [4] | (AG) oe (e.g. from $x + y = 2$) | or e.g. $2/(x + y) - 1 = 0 \Rightarrow x + y = 2, \Rightarrow 4 = 4x, \Rightarrow x = 1, y = 1$ (oe)

When $x = 1, y = 1$, $\frac{dy}{dx} = \frac{2}{1 + 1} - 1 = 0$ * | |

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# Question 7(i)

bounds $-\pi + 1, \pi + 1$ | B1 B1, B1cao [3] | or $\ldots < y < \ldots$ or $(-\pi + 1, \pi + 1)$ | not $\ldots < x < \ldots$, not 'between ...'

$\Rightarrow \quad -\pi + 1 < f(x) < \pi + 1$ | |

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# Question 7(ii)

$y = 2 \arctan x + 1 \Leftrightarrow y$ | M1 | attempt to invert formula | one step is enough, i.e. $y - 1 = 2 \arctan x$ or $y - 1 = 2 \arctan y$

$x = 2 \arctan y + 1$ | |

$\Rightarrow \quad \frac{x-1}{2} = \arctan y$ | A1 | or $\frac{y-1}{2} = \arctan x$ | need not have interchanged $x$ and $y$ at this stage

$\Rightarrow \quad y = \tan(\frac{x-1}{2}) \Rightarrow f^{-1}(x) = \tan(\frac{x-1}{2})$ | A1 | allow $y = \ldots$

| B1 | reasonable reflection in $y = x$ | curves must cross on $y = x$ line if present (or close enough to imply intention) curves shouldn't touch or cross in the third quadrant

| B1 | $(1, 0)$ intercept indicated. | [5]

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# Question 8(i)

$\int_0^1 \frac{x^3}{1+x} dx$ let $u = 1 + x$, $du = dx$ | B1, B1 | $a = 1, b = 2$ $(u - 1)^3/u$

when $x = 0$, $u = 1$, when $x = 1$, $u = 2$ | |

$= \int_1^2 \frac{(u-1)^3}{u} du$ | M1 | expanding (correctly)

$= \int_1^2 \frac{u^3 - 3u^2 + 3u - 1}{u} du$ | A1 dep | dep $du = dx$ (o.e.) AG | e.g. $du/dx = 1$, condone missing $dx$'s and $du$'s, allow $du = 1$

$= \int_1^2 (u^2 - 3u + 3 - \frac{1}{u}) du$ * | B1 | $[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u]_1^2$ | [7]

$\int_1^1 \frac{x^3}{1+x} dx = [\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u]_1^2$ | M1 | substituting correct limits dep integrated | upper – lower; may be implied from 0.140...

$= (\frac{8}{3} - 6 + 6 - \ln 2) - (\frac{1}{3} - \frac{3}{2} + 3 - \ln 1)$ | |

$= \frac{5}{6} - \ln 2$ | A1cao [7] | must have evaluated ln $1 = 0$ | must be exact – must be 5/6

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# Question 8(ii)

$y = x^2 \ln(1 + x)$ | M1, B1, A1 | Product rule $d/dx (\ln(1 + x)) = 1/(1 + x)$

$\Rightarrow \quad \frac{dy}{dx} = x^2 \cdot \frac{1}{1+x} + 2x \ln(1 + x)$ | | cao (oe) mark final ans | or $d/dx (\ln u) = 1/u$ where $u = 1 + x$ ln1+x is A0 but condone missing bracket in $\ln(1+x)$

$= \frac{x^2}{1+x} + 2x \ln(1 + x)$ | M1 | substituting $x = 0$ into correct deriv www | when $x = 0$, $dy/dx = 0$ with no evidence of substituting M1A0 but condone missing bracket in $\ln(1+x)$

When $x = 0$, $dy/dx = 0 + 0 \ln 1 = 0$ | A1cao [5] |

($\Rightarrow$ Origin is a stationary point) | |

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# Question 8(iii)

$A = \int_0^1 x^2 \ln(1 + x) d x$ | B1 | Correct integral and limits | condone no $dx$, limits (and integral) can be implied by subsequent work

let $u = \ln(1 + x)$, $dv/dx = x^2$ | M1 | parts correct | $u$, $du/dx$, $dv/dx$ and $v$ all correct (oe)

$\frac{du}{dx} = \frac{1}{1 + x}$, $v = \frac{1}{3}x^3$ | |

$\Rightarrow \quad A = [\frac{1}{3}x^3 \ln(1 + x)] - \int_0^1 \frac{1}{3}x^3 \frac{1}{1+x} dx$ | A1 | | condone missing brackets

$= \frac{1}{3} \ln 2 - (\frac{5}{18} - \frac{1}{3} \ln 2)$ | B1 | $= \frac{1}{3} \ln 2 - \ldots$ | $\ldots - 1/3$ (result from part (i)) | condone missing bracket, can re-work from scratch

$= \frac{1}{3} \ln 2 - \frac{5}{18} - \frac{1}{3} \ln 2$ | B1 ft [6] | $\ldots - 1/3$ (result from part (i)) | condone missing bracket, can re-work from scratch

$= \frac{2}{3} \ln 2 - \frac{5}{18}$ | A1 [6] | cao | oe e.g., $\frac{12\ln 2 - 5}{18}$, $\frac{5}{18} - \frac{1}{3}\ln 4 - \frac{1}{3}$, etc but must have evaluated $\ln 1 = 0$ Must combine the two ln terms

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# Question 9(i)

$\frac{d}{dx} \frac{\sin x}{\cos x} = \frac{\cos x \cdot \cos x \cdot \cos x - \sin x \cdot (- \sin x)}{\cos^2 x}$ | M1, A1 [3] | Quotient (or product) rule | product rule: $\frac{1}{\cos x} \cos x + \sin x \cdot (- \sin x)(-\frac{1}{\cos^2 x}) (-\sin x)$ but must show evidence of using chain rule on $1/\cos x$ (or $d/dx$ (sec $x$) = sec $x$ tan $x$ used)

$= \frac{\cos^2 x + \sin^2 x}{cos^2 x} = \frac{1}{\cos^2 x}$ * | A1 [3] | (AG) |

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# Question 9(ii)

Area $= \int_0^{\pi/4} \frac{1}{\cos^2 x} dx$ | B1 | correct integral and limits (soi) | condone no $dx$; limits can be implied from subsequent work

$= [\tan x]_0^{\pi/4}$ | M1, A1 [3] | | unsupported scores M0

$= \tan(\pi/4) - \tan 0 = 1$ | |

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# Question 9(iii)

$f(0) = 1/\cos^2(0) = 1$ | B1 | must show evidence

$g(x) = 1/2\cos^2(x + \pi/4)$ | M1, A1 [3] | 

$g(0) = 1/2\cos^2(\pi/4) = 1$ | |

($\Rightarrow$ $f$ and $g$ meet at $(0, 1)$) | |

| | or $f(π/4) = 1/\cos^2(π/4) = 2$ so $g(0) = ½ f(π/4) = 1$

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# Question 9(iv)

Translation in $x$-direction through $-\pi/4$ | M1, A1, M1, A1 [8] | must be in $x$-direction, or $(-\pi/4)$ ($0$) | 'shift' or 'move' for 'translation' M1 A0; $(-π/4)$ alone SC1

Stretch in $y$-direction scale factor $\frac{1}{2}$ | | must be in $y$-direction | 'contract' or 'compress' or 'squeeze' for 'stretch' M1A0; 'enlarge' M0

| B1 ft [8] | asymptotes correct | stated or on graph; condone no $x = \ldots$, ft $\pi/4$ to right only (viz. $-\pi/4$, $3\pi/4$) stated or on graph; ft $\pi/4$ to right only (viz. ($\pi/4$, $\frac{1}{2}$) )

| B1 ft | min point $(-\pi/4, \frac{1}{2})$ | 'y-values halved', or 'x-values reduced by $\pi/4$, are M0 (not geometric transformations), but for M1 condone mention of $x$- and $y$- values provided transformation words are used.

| B1 dep | curves intersect on $y$-axis correct curve, dep B3, with asymptote lines indicated and correct, and TP in correct position | [8]

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# Question 9(v)

Same as area in (ii), but stretched by s.f. $\frac{1}{2}$. So area = $\frac{1}{2}$ | B1 ft [1] | $\frac{1}{2}$ area in (ii) | or $\int_{-\pi/4}^{\pi/4} \frac{1}{2\cos^2(x + \pi/4)} dx = \frac{1}{2}[\tan(x + \pi/4)]_{-\pi/4}^{\pi/4} = \frac{1}{2}$ | allow unsupported