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p = (y : 2y \(\geq\) 6) 2y \(\geq\) 6 y \(\geq \frac{6}{2}\) y = \(\geq\) 3 and Q = (y : y -3 \(\leq\) 4) y - 3 \(\leq\) 4 y \(\leq\) 4 + 3 y \(\leq\) 7 therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....} P\(\cap\)Q = {3, 4, 5, 6, 7}

6k2 = 5k + 6 6k2 - 5k - 6 = 0 6k2 - 0k + 4k - 6 = 0 3k(2k - 3) + 2(2k - 3) = 0 (3k + 2)(2k - 3) = 0 3k + 2 = 0 or 2k - 3 = 0 3k = -2 or 2k = 3 k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\) k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))

y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\) 6 = k\(\sqrt{3 + 1}\) 6 = k\(\sqrt{4}\) 6 = 2k k = \(\frac{6}{2}\) = 3 y = \(\sqrt{(x + 1)}\) 9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3) \(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\) 3 = \(\sqrt{x + 1}\)(square both sides) 9 = x + 1 x = 9 - 1 x = 8

y = x2 + 2x + k at point(2,0) x = 2, y = 0 0 = (2)2 + 2(20 + k) 0 = 4 + 4 + k 0 = 8 + k k = -8

No expanded working has been attached for this question yet. The verified answer is Option B.

\(\frac{2}{x + 3} - \frac{1}{x - 2}\) = \(\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)}\) = \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\) = \(\frac{x -7}{x^2 + x - 6}\) = \(\frac{x - 7}{x^2 + x - 6}\)

Let the interior angle = xo interior angle = 5xo (sum of int. angle ann exterior) (angles = angle or straight line) 6x = 180 x = \(\frac{180}{6}\) x = 30o no. of sides = \(\frac{\text{sum of exterior angles}}{\text{exterior angle}}\) = \(\frac{360}{30}\) = 12

P = x2 + 4x - 2, Q = 2x - 1 Q - p = 2, (2x - 1) - (x2 + 4x - 2) = 2 2x - 1 - x2 - 4x + 2 = 2 -2x - x2 + 1 -x2 - 2x - 1 = 0 x2 + 2x + 1 = 0 x2 + x + x + 1 = 0 x(x + 1) + 1(x + 1) = 0 (x + 1)(x + 1) = 0 x + 1 = 0 or x + 1 = 0 x = -1 or x = -1 x = -1

Volume of pyramid = \(\frac{1}{3}\) x base area x height = \(\frac{1}{3} \times 124 \times 8 \times 14\) = 4 x 8 x 14 = 448m\(^3\)

Volume of a cone = \(\frac{1}{3} \pi r^2h\) h2 = 52 = 32 = 25 - 9 = 16 h = \(\sqrt{16}\) h = 4cm v = \(\frac{1}{3} \times \frac{22}{7} \times 3^2 \times 4\) \(\frac{1}{3} \times \frac{22}{7} \times 9 \times 4\) = \(\frac{22 \times 3 \times 4}{7}\) = 37.7cm3 = 38cm3

1km = 100,000cm on the map 1 cm represent every 10 km which is equal to (10 x 100,000cm) = 1,000,000cm the scale is 1:1,000,000

(x + 2)(x\(^2\) - 3x + 2) + 2(x + 2)(x - 1) = (x + 2)m (x + 2)[(x\(^2\) - 3x + 2) + 2(x - 1)] = (x + 2)M divide both side by (x + 2) (x\(^2\) - 3x + 2) + 2(x - 1) = M x\(^2\) - 3x + 2 + 2x - 2 = M x\(^2\) - 3x + 2 + 2x - 2 = M x\(^2\) - 3x + 2x = M x\(^2\) - x = M M = x\(^2\) - x

L2 = 962 + 282 = 9216 + 784 = 10000 L = \(\sqrt{10000}\) = 100cm curved surface area = \(\pi r l\) = \(\frac{22}{7} \times 28 \times 100\) = 8800cm2 area of cone = area of sector area of sector = 8800cm2

< PQR is a right-angled triangle PR\(^2\) = 24\(^2\) + 10\(^2\) = 676 PR = \(\sqrt{676}\) = 26cm PQRTS is a square and PR = 26cm. meaning the remaining sides of the square are 26cm each Therefore, the perimeter of PQRST = 26 + 26 + 26 + 24 + 10 = 112cm

by Pythagoras x2 = 62 + 82 36 + 64 = 100 x = \(\sqrt{100}\) x = 10