## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use identity $\sec^2 A = 1 + \tan^2 A$ | B1 | |
| Attempt solution of three-term quadratic equation to obtain two values of $\tan A$ | M1 | Implied if correct values obtained; allow M1 for incorrect factorisation provided expansion would give correct first and third terms; allow M1 for incorrect use of formula if only one error present |
| Obtain $\tan A = -3$ and $\tan A = 4$ | A1 | And no others; implied by $A = \tan^{-1}-3$ and $\tan^{-1}4$ | $A=-3, 4$ is A0 here unless subsequent work shows values used correctly |
| Use correct identities to produce equation in $\tan B$ only | M1 | Equation might be $t^3 = 27\ldots$ | $\ldots$ or $t^5+t^3-27t^2-27=0$ |
| State $\tan B = 3$ | A1 | And no others |
| Substitute at least one pair of non-zero numerical values into $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ | M1 | Must be the correct identity |
| Obtain one of $\frac{1}{13}$ and $\frac{3}{4}$ or exact equiv | A1 | |
| Obtain the other exact value or equiv | A1 [8] | And no others |

## Question 5i:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Either: State $e^{2x}=8e^{-x}$ and so $e^{3x}=8$ | B1 | | Verifying by substitution of $\ln 2$ in each equation earns B0B0 |
| Obtain $e^x=2$ and hence $x=\ln 2$ | B1 | AG; necessary detail needed |
| Or 1: State $e^{2x}=8e^{-x}$ and so $e^{3x}=8$; State $3x=\ln 8$, $x=\ln 8^{\frac{1}{3}}$ and hence $x=\ln 2$ | B1, B1 | AG; necessary detail needed | Going immediately from $x=\frac{1}{3}\ln 8$ to $x=\ln 2$ does not earn the second B1 |
| Or 2: State $e^{2x}=8e^{-x}$ and $2x=\ln 8 - x$; State $3x=\ln 8$, $x=\ln 8^{\frac{1}{3}}$ and hence $x=\ln 2$ | B1, B1 [2] | AG; necessary detail needed | Going immediately from $x=\frac{1}{3}\ln 8$ to $x=\ln 2$ does not earn the second B1 |

## Question 5ii:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain $k_1e^{-x}$ and $k_2e^{2x}$ | M1 | Any non-zero constants $k_1$ and $k_2$ |
| Obtain correct $-8e^{-x} - \frac{1}{2}e^{2x}$ or, if done separately, $-8e^{-x}$ and $\frac{1}{2}e^{2x}$ | A1 | |
| Apply limits 0 and $\ln 2$ correctly to their integral(s) | M1 | Condone one sign slip; earned by sight of $-8e^{-\ln 2} - \frac{1}{2}e^{2\ln 2} + 8 + \frac{1}{2}$ (or equivs if integrals treated separately) | M1 also implied by sight only of $-4-2+8+\frac{1}{2}$ (or equivs...) |
| Obtain at least $-4-2+8+\frac{1}{2}$ (or equivs) | *A1 | |
| Obtain $\frac{5}{2}$ or equiv | A1 [5] | Final A1 dependent on *A1 |