2010 - JAMB Mathematics Past Questions and Answers - page 5

(\left[\begin{array}{cc}

x & 3\

2 & 7\end{array}\right])= 15
x x 7 - 2 x 3 = 15

7x - 6 = 15

7x = 15+6

7x = 21
x = ²¹/₇
x = 3

∠RTS = 90^o (∠ in semi circle)

∠UTS = 25^o (∠ in alternate segment)
X + 90 + 25 = 180(sum of ∠s on a str line)
X + 115 = 180
X = 180 - 115
X = 65^o

TUVW is a cyclic quad

3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)

3χ + 108 = 180

3χ = 180 - 108

3χ = 72
χ = ⁷²/₃
χ = 24^o

Area of a triangle = ¹/₂ base x height

= ¹/₂ x 4 x 5

= 10 cm²

= ⁵⁰/₃₆₀ x ²²/₇ x ³⁶/₁

= ¹¹⁰/₇

Locus of a point equidistant from two fixed points is the perpendicular bisector of the straight line joining the points Gradients of PQ = (5-3 / 3-1) = ²/₂ = 1

Gradient bisector = -1

Equation of perpendicular bisector
y - y₁ = m(χ - χ₁
y - 4 = -χ + 2
y = -χ + 6

Area of a triangle = ¹/₂ ab Sinθ

12√3 = ¹/₂ x 8 x q sin 60

12√3 = 4 x q x ^√3/₂

12√3 = 2q√3

q = 6

y = (2x + 1)³

dy/dx = 3(2x + 1)^3-1 x 2

= 3(2x + 1)² x 2

= 6(2x + 1)²

∫(sin χ + 2)dχ = (∫sin χ + ∫2)dx

= -cos χ + 2χ + K