## Question 3(a)

### Way 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{rhs} = 2\cosh^2 x - 1 = 2\!\left(\frac{e^x+e^{-x}}{2}\right)^2 - 1$ | M1 | Substitutes the correct exponential form into the rhs |
| $= 2\!\left(\frac{e^{2x}+2+e^{-2x}}{4}\right)-1$ | dM1 | Squares correctly to obtain expression in $e^{2x}$ and $e^{-2x}$. Dependent on previous mark |
| $= \frac{e^{2x}+e^{-2x}}{2}+1-1 = \frac{e^{2x}+e^{-2x}}{2} = \cosh 2x = \text{lhs}$ * | A1* | Complete proof with no errors |

### Way 2

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{lhs} = \cosh 2x = \frac{e^{2x}+e^{-2x}}{2}$ | M1 | Substitutes the correct exponential form |
| $= 2\!\left(\frac{(e^x+e^{-x})^2-2}{4}\right)$ | dM1 | Completes the square correctly to obtain expression in $e^x$ and $e^{-x}$. Dependent on previous mark |
| $2\!\left(\frac{e^x+e^{-x}}{2}\right)^2 - 1 = 2\cosh^2 x - 1 = \text{rhs}$ * | A1* | Complete proof with no errors |

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## Question 3(b)

### Way 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| $29\cosh x - 3(2\cosh^2 x - 1) = 38$ | M1 | Substitutes the result from part (a) |
| $6\cosh^2 x - 29\cosh x + 35 = 0 \Rightarrow \cosh x = \ldots$ | M1 | Forms a 3-term quadratic and attempts to solve for $\cosh x$. May apply General Principles for solving a 3TQ |
| $\cosh x = \frac{7}{3}$ or $\cosh x = \frac{5}{2}$ | A1 | Both correct (or equivalent values) |
| $\cosh x = \alpha \Rightarrow x = \ln\!\left(\alpha + \sqrt{\alpha^2-1}\right)$ or $\ln\!\left(\alpha - \sqrt{\alpha^2-1}\right)$, or $\frac{e^x+e^{-x}}{2}=\alpha \Rightarrow x = \ldots$ | M1 | Uses correct $\ln$ form for arcosh to find at least one value of $x$ for $\alpha>1$, or uses correct exponential form for cosh and solves resulting 3TQ in $e^x$ to find at least one value of $x$ for $\alpha>1$ |
| $x = \ln\!\left(\frac{7}{3} \pm \sqrt{\frac{40}{9}}\right)$ and $x = \ln\!\left(\frac{5}{2} \pm \sqrt{\frac{21}{4}}\right)$ | A1A1 | A1: Any 2 of the 4 solutions. Penalise lack of brackets once where necessary (first time). Penalise lack of simplification once (first time). A1: All 4 correct |

**Equivalent forms accepted:**
- $x = \ln\frac{7\pm 2\sqrt{10}}{3}$ and $x = \ln\frac{5\pm\sqrt{21}}{2}$
- $x = \pm\ln\!\left(\frac{7+2\sqrt{10}}{3}\right)$ and $x = \pm\ln\!\left(\frac{5+\sqrt{21}}{2}\right)$
- $x = \ln(7\pm 2\sqrt{10})-\ln 3$ and $x = \ln(5\pm\sqrt{21})-\ln 2$

**Note:** Decimal answers are $\pm 1.49\ldots,\ \pm 1.56\ldots$

# Question 3 (Way 2):

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $29\left(\frac{e^x+e^{-x}}{2}\right)-3\left(\frac{e^{2x}+e^{-2x}}{2}\right)=38$ or $6\left(\frac{e^x+e^{-x}}{2}\right)^2-29\left(\frac{e^x+e^{-x}}{2}\right)+35=0$ | M1 | Substitutes the correct exponential forms |
| $3e^{4x}-29e^{3x}+76e^{2x}-29e^x+3=0$ | M1A1 | M1: Multiplies by $e^{2x}$ or $e^{-2x}$ to obtain a quartic in $e^x$ or $e^{-x}$; A1: Correct quartic in any form (not necessarily all on one side) |
| $(3e^{2x}-14e^x+3)(e^{2x}-5e^x+1)=0 \Rightarrow x=\ldots$ | M1 | Solves their quartic to find at least one value for $x$ |
| $x=\ln\left(\frac{7}{3}\pm\sqrt{\frac{40}{9}}\right)$ and $x=\ln\left(\frac{5}{2}\pm\sqrt{\frac{21}{4}}\right)$ | A1A1 | A1: First pair correct; A1: All 4 correct. Equivalent exact forms accepted e.g. $x=\ln\frac{7\pm2\sqrt{10}}{3}$, $x=\pm\ln\left(\frac{7+2\sqrt{10}}{3}\right)$, $x=\ln(7\pm2\sqrt{10})-\ln 3$ |

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