Question Library

x\(^2\) - 3x + 2 = 0 x\(^2\) - 2x - x + 2 = 0 x(x - 2) - 1(x - 2) = 0 (x - 1)(x - 2) = 0 Therefore, x = 1 or 2 Hence x = 1 is the smallest value.

x(a - c) + y(c - a) ==> -x(c - a) + y(c - a) = (y - x)(c - a) = -(a - c)(y - x) = (a - c)(x - y)

y ∝ \(\frac{1}{x^2}\) y = \(\frac{\text{k}}{x^2}\) When x = 3, y = 4. y = \(\frac{\text{k}}{x^2}\) 4 = \(\frac{\text{k}}{9}\) k = 36. When x = 2, y = \(\frac{36}{4}\) = 9

3x\(^2\) + 25x - 18 = 0; 3x\(^2\) - 2x + 27x - 18 = 0 x(3x - 2) + 9(2x + 2) = 0 (x + 9)(3x - 2) = 0 Either 3x - 2 = 0 or x + 9 = 0 3x = 2 or x = -9 ∴ x = \(\frac{2}{3}\) or -9

(x + 2)(x - 7) = 0; either x + 2 = 0 or x - 7 = 0 x = -2 or 7

x + y = \(\frac{3}{2}\), x - y = \(\frac{5}{2}\) x + y = \(\frac{3}{2}\), Multiply through by 2 2x + 2y = 3 ---------------- 1 x - y = \(\frac{5}{2}\) Multiply through by 2 2x - 2y = 5 --------------- 2 Add eqn (1 and 2) 4x = 8 x = \(\frac{8}{4}\) = 2 Put x = 2 into either Eqn 1 or Eqn 2 2x - 2y = 5 --------------- 2 2(2) - 2y = 5 4 - 2y = 5 - 2y = 5 - 4 = 1 y = -\(\frac{1}{2}\) 2y + x. = 2 x -\(\frac{1}{2}\) + 2 = -1 + 2 = 1.

The gradient of the line passing through points (0,3) and (1,0) m = \(\frac{y_2 - y_1}{x_2 - x_1}\) m = \(\frac{0 - 3}{1 - 0}\) = \(\frac{-3}{1}\) = - 3 Equation of this line ⇒ y = mx + c y = -3x + 3; since the line is solid; The equation represented by the graph is y \le -3x + 3

The minimum value can be read directly from the y-axis of the graph to be the lowest attainable value of the function at the point where it turns. And this occurs at y = -7.

The roots of the equation are the values of x at the point where the graph crosses the x-axis. And this can be read from the graph to be 1.4 and -1.7.

Given that AC = 16 cm and BD = 12 cm. If O is the center. The diagonals of a rhombus meet at 90º. AO = \(\frac{\text{AC}}{2}\) = 8cm; BO = \(\frac{\text{BD}}{2}\) = 6 cm AB\(^2\) = AO\(^2\) + OB\(^2\) (pythagoras theorem) AB\(^2\) = 64 + 36 = 100; AB = 10 cm Area of △ABO = \(\frac{1}{2}\) \times 8 \times 6 = 24 cm\(^2\) Area of rhombus = 4 \times area of △ABO ⇒ 4 \times 24 = 96 cm\(^2\).

Total surface area = 2\pir\(^2\) + 2\pirh = 2\pir(r + h) ⇒ 2 \times \(\frac{22}{7}\) \times 7 \times (7 + 5) ⇒ 44 \times 12 = 528 cm\(^2\)

Total surface area of cone = \pirl + \pir\(^2\) = \pir(l + r) r = 3 and l = 4 Area = \(\frac{22}{7}\) \times 3 (4 + 3) = \(\frac{22}{7}\) \times 3 \times 7 = 66 cm\(^2\)

Capacity of water tank = (\(\frac{1}{2}\) \times \(\frac{3}{2}\) \times \(\frac{3}{2}\)) m\(^3\) = \(\frac{9}{8}\) m\(^3\) = \(\frac{9}{8}\) \times 1000 litres = 1125 litres. Water tankers capacity = 1500 litres Amount of water remaining = (1500 - 1125) litres = 375 litres

Volume of sphere = \(\frac{4}{3}\)\pir\(^3\) Surface area of sphere = 4\pir\(^2\) Since volume and area = k cm\(^3\) and k cm\(^2\) \(\frac{4}{3}\)\pir\(^3\) = 4\pir\(^2\) \(\frac{4}{3}\)r\(^3\) = 4r\(^2\) 4\pir\(^3\) = 12\pir\(^2\) r\(^3\) - 3r\(^2\) = 0 r\(^2\)(r - 3) = 0 ⇒ r = 3 cm d = 2 x 3 = 6cm.

Triangles OAD is similar to OBC \(\frac{\text{OA}}{\text{AD}}\) = OB But \(\frac{\text{OB}}{\text{BC}}\) = OA + OB Height of the cone is OB Let OA = x; OB = 12 + x \(\frac{\text{x}}{6}\) = \(\frac{(12 + x)}{12}\) Hence \(\frac{\text{x}}{6}\) = \(\frac{(12 + x)}{12}\) ⇒ 12x = 6x + 72 6x = 72 ⇒ x = 12 cm Height of cone is DB = 12 + x = (12 + 12) cm = 24 cm