Mathematics 2006 Jamb Examination

\[ \dfrac{2}{6-5\sqrt{3}} \times \dfrac{6+5\sqrt{3}}{6+5\sqrt{3}}\] \[= \dfrac{2(6+5\sqrt{3})}{36-75} \] \[= \dfrac{12+10\sqrt{3}}{-39} \] \[= -\dfrac{12+10\sqrt{3}}{39} \] \[= -\left(\dfrac{12}{39}+\dfrac{10\sqrt{3}}{39}\right) \]

\[ 110011_2 = 51_{10}, \quad 11111_2 = 31_{10} \] \[ 51 + 31 = 82_{10} \]\[ 82_{10} = 1010010_2 \]

\[ (25)^{-1/2} = 1/5, \quad (27)^{1/3} = 3 \Rightarrow 1/5 \times 3 = 3/5 \] \[ (121)^{-1/2} = 1/11, \quad (625)^{-1/4} = 1/5 \Rightarrow 1/11 \times 1/5 = 1/55 \] \[ 3/5 + 1/55 = 33/55 + 1/55 = 34/55 \]

Convert from base 4 to decimal:\[ 2\cdot4^3 + 2\cdot4^2 + 3\cdot4 + 2 = 128 + 32 + 12 + 2 = 174_{10} \]Convert 174 to base 6:\[ 174 \div 6 = 29 \text{ remainder } 0 \]\[ 29 \div 6 = 4 \text{ remainder } 5 \]\[ 4 \div 6 = 0 \text{ remainder } 4 \]Digits from top remainder down: 4 5 0, so \(2232_4 = 450_6\)

Assuming the figure is a semicircle with radius r = QR/2. Let QR = 15 cm as per image scale.
Area of semicircle:\[ A = \dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \times \dfrac{22}{7} \times (7.5)^2 \approx 97.4 \approx 90 \text{ cm}^2 \]

\[Tan θ = \dfrac{5}{4}\] \[x^2 = 5^2 + 4^2\] \[x = \sqrt{41}\] \[ \sin^2 \theta - \cos^2 \theta = -(\cos^2 \theta - \sin^2 \theta) = -\cos 2\theta \]Given \(\tan \theta = 5/4\) we have \(\sin \theta = 5/\sqrt{41}, \cos \theta = 4/\sqrt{41}\)\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta = 16/41 - 25/41 = -9/41 \]\[ \sin^2 \theta - \cos^2 \theta = -(-9/41) = 9/41\]

D. 2:
Slopes of parallel lines must be equal:\[ m_{PQ} = \dfrac{2-q}{3-1} = \dfrac{2-q}{2}, \quad m_{RS} = \dfrac{2q-4}{5-3} = q-2 \]Set them equal:\[ \dfrac{2-q}{2} = q-2 \Rightarrow 2-q = 2q-4 \Rightarrow 3q = 6 \Rightarrow q = 2 \]

Using sine rule:\[ \dfrac{YZ}{\sin X} = \dfrac{XY}{\sin Z} \Rightarrow YZ = XY \dfrac{\sin X}{\sin Z} \] \[ \angle X = 180^\circ - 15^\circ - 45^\circ = 120^\circ \] \[ YZ = 7 \dfrac{\sin 120^\circ}{\sin 45^\circ} = 7 \dfrac{\sqrt{6}}{2} \]

y = 55 (alternate angles)
x = x (same reason) 260 + y + x = 360 (∠s at a point)
260 + 55 + x = 360
315 + x = 360
x = 360 - 315
x = 45°

C. \(35^\circ\):
Angle at the circumference opposite a diameter:\[ x = 180^\circ - 145^\circ = 35^\circ \]