2006 - JAMB Mathematics Past Questions and Answers - page 3

\(\frac{2}{6-5\sqrt{3}} = \frac{2}{6-5\sqrt{3}} \times \frac{6+5\sqrt{3}}{6+5\sqrt{3}}\\ =\frac{2(6+5\sqrt{3})}{(6-5\sqrt{3})(6+5\sqrt{3})}\\ =\frac{12+10-\sqrt{3}}{36-25(3)}\\ =\frac{12+10-\sqrt{3}}{36-75}\\ =\frac{12+10-\sqrt{3}}{-39}\\ =-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\)

110011₂ + 11111₂ = 1010010₂

25)^-1/₂ x (27)^1/₃ + (121)^-1/₂ x (625)^-1/₄
5^2x-1/₂ x 3^3x1/₃ + 11^2x-1/₂ x 5^4x-1/₄
5^-1 x 3¹ x 11^-1 x 5^-1
1/₅ x ³/₁ + ¹/₁₁ x ¹/₅
³/₅ + ¹/₅ = ³³⁺¹/₅₅
=³⁴/₅₅

1^st convert to base 10
2232⁴ = 2x4³ + 2x4² + 3x4¹ + 2x4⁰
= 2x64 + 2x16 + 3x4 + 2x1
= 128 + 32 + 12 + 2
= 174 convert to base 6
6/174
6/29 R 0
6/4 R 5
6/0 R 5
= 450₆

Area of rectangle PQRS
= 10 x 7 = 70 cm²
Area of semi circle
QR = Dπ = 7 x ²²/₇ = 22cm²
∴Area of the figure = 70 + 22
= 92cm²
= 90cm²

Tan θ = ⁵/₄
x² = 5² + 4²
= 25 + 16
= 41
x = √41
sin²θ - cos²θ = (⁵/_√41)² - (⁴/_√41)²
= ²⁵/₄₁ - ¹⁶/₄₁
= ⁹/₄₁

Gradient PQ, P(1,q) and Q(3,2) \(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\) Gradient of RS : R(3,4) and S(5,2q) \(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\) = q-2 Since PQ and RS are parallel, their generation are equal \(∴ \frac{(2-q)}{2} = q-2\) 2-q = 2(q-2) 2-q = 2q-4 2+4 = 2q+q 6 = 3q q = 2

∠yxz = 180 - (45 + 15)
= 180 - 60
= 120^o
Using sine rule
\(\frac{x}{sinx}=\frac{7}{sinz}\\ \frac{x}{sin 120}=\frac{7}{sin 45}\\ x=\frac{7 sin 120}{sin45}\\ x=\frac{7sin(180-120)}{sin 45}\\ x=\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x =\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x = 7\left(\frac{\sqrt{6}}{2}\right)\)

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^o

∠PSQ = 90^o (∠ in semi circle)
∠PSQ + ∠QSR = 145^o
90 + ∠QSR = 145
∠QSR = 145 - 90
∠QSR = 55^o
BUt x^o = ∠QSR (∠s in the same segment)
∴x^o = 55^o