## (a)

Substitute $t = 3$ in $|2t - 1|$ and obtain value 5 | B1 | not awarded for final $|5|$ nor for $\pm5$

Substitute $t = -3$ in $|2t - 1|$ and apply modulus correctly to any negative value to obtain a positive value | M1 | with no modulus signs remaining

Obtain value 7 as final answer | A1 | not awarded for final $|7|$ nor for $\pm7$

NB: substitutions in $|2t + 1|$ will give 5 and 7 – this is 0/3, not MR; a further step to $5 < t < 7$ – B1 M1 A0; answers $\pm5, \pm 7$ – this is B0 M0 A0

**Total: [3]**

## (b)

**Either** Attempt solution of linear equation or inequality with signs of $x$ different | M1 |

Obtain critical value $-\sqrt{2}$ | A1 | or equiv (exact or decimal approximation)

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**Or 1** Attempt to square both sides | M1 | obtaining at least 3 terms on each side

Obtain $x^2 - 2\sqrt{2}x + 2 > x^2 + 6\sqrt{2}x + 18$ | A1 | or equiv; or equation; condone $>$ here

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**Or 2** Attempt sketches of $y = |x - \sqrt{2}|$, $y = |x + 3\sqrt{2}|$ | M1 |

Obtain $x = -\sqrt{2}$ at point of intersection | A1 | or equiv

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Conclude with inequality of one of the following types: | M1 | any integer $k$

$$x < k\sqrt{2}, \quad x > k\sqrt{2}, \quad x < \frac{k}{\sqrt{2}}, \quad x > \frac{k}{\sqrt{2}}$$

Obtain $x < -\sqrt{2}$ or $-\sqrt{2} > x$ as final answer | A1 | final answer $x < -\frac{2}{\sqrt{2}}$ (or similar unsimplified version) is A0

**Total: [4]**

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