# Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(1+\frac{x}{p}\right)^q = 1 + q\frac{x}{p} + \frac{q(q-1)}{2!}\left(\frac{x}{p}\right)^2 + \cdots$ | M1* | One of $\frac{q}{p}x$ or $\frac{q(q-1)}{2!}\left(\frac{x}{p}\right)^2$ (soi), e.g. $\frac{q}{p}=-1$ scores M1 A1 |
| $\frac{q}{p} = -1 \qquad \frac{q(q-1)}{2p^2} = \frac{3}{4}$ | A1 A1 | Allow $x$'s on both sides of equations (if correct) |
| $q = -p \Rightarrow \frac{-p(-p-1)}{2p^2} = \frac{3}{4}$ or $\frac{q(q-1)}{2q^2} = \frac{3}{4}$ | M1dep* | Eliminating $p$ (or $q$) from simultaneous equations (not involving $x$) involving both variables |
| $\Rightarrow p = 2$ | A1 | $p=2$ www (or $q=-2$) |
| $\Rightarrow q = -2$ | A1ft | $q=-2$ (or $p=2$) for second value, ft their $p$ or $q$ |
| Valid for $\left|\frac{x}{2}\right| < 1 \Rightarrow |x| < 2$ | A1 | or $-2 < x < 2$ www, allow $-2 < |x| < 2$ but not say $x < 2$. **SC** If M0 M0 and no wrong working: B1 for $p=2$ **and** $q=-2$, B1 for $-2 < x < 2$ (max 2 marks) |

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