**Answer:** $x^{2n} - 1 = (x^n - 1)(x^n + 1)$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** one of $2^n−1$, $2^{n+1}$ is divisible by three; $2^n − 1$, $2^n$ and $2^n + 1$ are consecutive integers; one must therefore be divisible by 3; but $2^n$ is not, so one of the other two is | **Marks:** M1 | **Guidance:** award notwithstanding false reasoning; condone 'factor' for 'multiple'

**Answer:** (correct reasoning must be given if justified) | **Marks:** A1 | **Guidance:** (none)

**Answer:** **or** $2^n$ is not div by 3, and so has remainder 1 or 2 when divided by 3; if remainder is 1, $2^n − 1$ is div by 3; if remainder is 2, then $2^n +1$ is div by 3 [so $2^n − 1$ is divisible by 3] | **Marks:** A2 | **Guidance:** (none)

**Total Marks:** [4]

---

## Question 8(i)

**Answer:** $\frac{dy}{dx} = \frac{(x+4)^{-1/2} \cdot 1 - x \cdot (x+4)^{-3/2}}{[(x+4)^{-1/2}]^2}$ | **Marks:** M1 | **Guidance:** quotient rule: $v \times$ their $u' - u \times$ their $v'$, and correct denominator

**Answer:** $\frac{1}{2}u^{-1/2}$ soi | **Marks:** B1 | **Guidance:** or $-\frac{1}{2}u$ (PR); PR: $x(-\frac{1}{2})(x+4)^{-3/2} +(x+4)^{-1/2}$

**Answer:** correct expression | **Marks:** A1 | **Guidance:** (none)

**Answer:** $= \frac{x+4-\frac{1}{2}x}{(x+4)^{3/2}} = \frac{\frac{1}{2}x+4}{(x+4)^{3/2}} = \frac{x+8}{2(x+4)^{3/2}}$ * | **Marks:** M1 | **Guidance:** factoring out $(x+4)^{-1/2}$ o.e. NB AG

**Answer:** (continued) | **Marks:** A1 | **Guidance:** (none)

**Total Marks:** [5]

---

## Question 8(ii)

**Answer:** [asymptote is] $x = -4$ | **Marks:** B1 | **Guidance:** soi; but from correct working

**Answer:** gradient of tangent at O= $8/(2 \cdot 4^{3/2}) = \frac{1}{2}$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** eqn of tangent is $y = \frac{1}{2}x$ | **Marks:** B1 | **Guidance:** o.e. e.g. using gradient

**Answer:** When $x = -4$, $y = -2$, so $(-4, -2)$ | **Marks:** B1 | **Guidance:** (none)

**Total Marks:** [4]

---

## Question 8(iii)

**Answer:** let $u = x + 4$, $du = dx$ | **Marks:** B1 | **Guidance:** or $dv/du = 1$ or $2v dy/dx = 1$ or $2v dy = dx$ oe e.g. $dy/dx = \frac{1}{2}(x+4)^{-1/2}$

**Answer:** $\int_0^5 \frac{x}{(x+4)^{1/2}} dx = \int_4^9 \frac{u-4}{u^{1/2}} du$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** $= \int_4^9 (u^{1/2} - 4u^{-1/2}) du$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** $= [\frac{2}{3}u^{3/2} - 8u^{1/2}]_4^9$ | **Marks:** B1 | **Guidance:** $u^{1/2} - 4u^{-1/2}$ or $u^{1/2} - 4/u^{1/2}$, or $\sqrt{u} - 4/\sqrt{u}$

**Answer:** $= (18 - 24) - (16/3 - 16)$ | **Marks:** M1 | **Guidance:** substituting correct limits (upper − lower); 0, 5 for x; 4,9 for u; 2,3 for v

**Answer:** $= 14/3$ | **Marks:** A1cao | **Guidance:** (none)

**Answer:** **or** (following first 2 marks) let $v = u - 4$, $v' = u^{-1/2}$, $v' = 1$, $w = 2u^{1/2}$ | **Marks:** M1 | **Guidance:** by parts with no substitution: $u = x$, $u' = 1$, $v' = (x+4)^{-1/2}$, $v = 2(x+4)^{1/2}$ M1 $= [2x(x+4)^{1/2}] - \int_0^5 2(x+4)^{1/2} du$

**Answer:** (continued) | **Marks:** A1 | **Guidance:** (none)

**Answer:** $= [2x(x+4)^{1/2} - \frac{4}{3}(x+4)^{3/2}]_0^5$ | **Marks:** A1 | **Guidance:** (none)

**Answer:** (continued) | **Marks:** A1cao | **Guidance:** =14/3 A1 (so max of 4/6)

**Answer:** $y$-coordinate of $Q$ is $2\frac{1}{2}$ | **Marks:** B1 | **Guidance:** (soi)

**Answer:** Area of triangle = $\frac{1}{2} \times 5 \times 5/2 = 25/4$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** Enclosed area = $25/4 - 14/3 = 1\frac{7}{12}$ | **Marks:** B1 | **Guidance:** or 19/12, or 1.583; isw from correct exact answer

**Answer:** **or** $\int_0^5 \frac{1}{2}x \, dx$ | **Marks:** M1 | **Guidance:** (none)

**Answer:** (continued) $= [\frac{1}{4}x^2]_0^5 = 25/4$ | **Marks:** A1 | **Guidance:** (none)

**Total Marks:** [9]

---

## Question 9(i)

**Answer:** $1 + k$ | **Marks:** B1 | **Guidance:** (none)

**Total Marks:** [1]

---

## Question 9(ii)

**Answer:** $f'(x) = 2e^{2x} - 2ke^{-2x}$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** $f'(x) = 0 \Rightarrow 2e^{2x} - 2ke^{-2x} = 0$ | **Marks:** M1 | **Guidance:** their derivative = 0

**Answer:** $\Rightarrow e^{2x} = ke^{-2x}$ | **Marks:** A1 | **Guidance:** NB AG

**Answer:** $\Rightarrow e^{4x} = k$, $4x = \ln k$, $x = \frac{1}{4}\ln k$ * | **Marks:** A1 | **Guidance:** (none)

**Answer:** $y = e^{(1/2 \ln k)} + ke^{(-1/2 \ln k)}$ | **Marks:** M1 | **Guidance:** substituting $x = \frac{1}{4}\ln k$ into $f(x)$

**Answer:** $= \sqrt{k} + k/\sqrt{k} = 2\sqrt{k}$ | **Marks:** A1cao | **Guidance:** or $2k^{1/2}$

**Total Marks:** [5]

---

## Question 9(iii)

**Answer:** Area = $\int_0^{\frac{1}{4}\ln k} (e^{2x} + ke^{-2x})[d x]$ | **Marks:** B1 | **Guidance:** correct integral and limits (soi)

**Answer:** $= \int_0^{\frac{1}{4}\ln k} (e^{2x} + ke^{-2x}) d x = [\frac{1}{2}e^{2x} - \frac{1}{2}ke^{-2x}]_0^{\frac{1}{4}\ln k}$ | **Marks:** B1 | **Guidance:** (none)

**Answer:** $= \frac{1}{2}k - \frac{1}{2} - \frac{1}{2} + \frac{1}{2}k$ | **Marks:** M1 | **Guidance:** (none)

**Answer:** $= k - 1$ | **Marks:** A1 | **Guidance:** (none)

**Total Marks:** [4]

---

## Question 9(iv)(A)

**Answer:** $g(x) = e^{(2x + 1/2 \ln k)} + ke^{(2x - 1/2 \ln k)}$ | **Marks:** M1 | **Guidance:** Substitute $x + \frac{1}{4}\ln k$ for $x$ in $f(x)$

**Answer:** $= e^{2x}e^{(1/2 \ln k)} + ke^{2x}e^{(-1/2 \ln k)}$ | **Marks:** M1 | **Guidance:** $e^{xy} = e^x e^y$ used

**Answer:** $= (e^{\ln k})^{1/2}e^{2x} + k(e^{\ln k})^{-1/2}e^{-2x}$ | **Marks:** (none) | **Guidance:** (none)

**Answer:** $= k^{1/2}e^{2x} + kk^{-1/2}e^{-2x}$ | **Marks:** (none) | **Guidance:** (none)

**Answer:** $= \sqrt{k}(e^{2x} + e^{-2x})$ * | **Marks:** A1 | **Guidance:** NB AG – must show enough working; e.g. $k^{1/2}e^{2x} + kk^{-1/2}e^{-2x}$

**Total Marks:** [3]

---

## Question 9(iv)(B)

**Answer:** $g(-x) = \sqrt{k}(e^{-2x} + e^{2x})$ | **Marks:** M1 | **Guidance:** substituting $-x$ for $x$

**Answer:** $= g(x)$ so $g$ is even | **Marks:** A1 | **Guidance:** must include $g(−x) = g(x)$, either define an even function or conclude that $g$ is even

**Total Marks:** [2]

---

## Question 9(iv)(C)

**Answer:** $g(x)$ is symmetrical about the $y$-axis, and $f(x)$ is $g(x)$ translated $\frac{1}{4}k$ in $x$-direction | **Marks:** B1 | **Guidance:** allow 'shift' or 'move'

**Answer:** so $f(x)$ is symmetrical about $x = \frac{1}{4}\ln k$ | **Marks:** B1 | **Guidance:** allow final B1 even if unsupported

**Total Marks:** [3]