2006 - JAMB Mathematics Past Questions and Answers - page 4

Locus of point equidistant from a given straight is the perpendicular bisector of the straight line

∴Gradient of the line ax + by + c = 0

implies by = -ax - c

y = -^a/_bx - c

the gradient (m) = ^a/_b

∴Gradient of the perpendicular (m) = ^b/_a

If P(x,y) is the point intersection of the two lines equation of the perpendicular becomes

y - y₁ = m(x - x₁)

y - y₁ = ^b/_a(x - x₁)

ay - ay₁ = bx - bx₁

ay - bx + bx₁ - ay₁ = 0

ay - bx + b(x₁ - y₁) = 0

implies ay - bx + q

Each ∠of a regular polygon

(=\frac{(n-2)180}{n}\

=\frac{(6-2)180}{6}\

=\frac{(4)180}{6})

= 120^o

ΔQSR is isosceles

∴ Q = S = 30^o

Also ΔTSR is isosceles

∴ T = R = 30^o

∠TSV + ∠VSR = 120

∠TSV + 30 = 120

∠TSV = 120 - 30

∠TSV = 90^o

∴∴VTS + ∴VST + ∴TVS = 180 (sum of ∠s of a Δ)

30 + 90 + ∠TVS = 180

120 + ∠TVS = 180

∠TVS = 180 - 120

∠TVS = 60^o

Locus of point equidistant from points P and q is a perpendicular bisector of the straight joining P and Q

∴PN = NQ

In ΔPST; cos 60 = ^ST/₈

ST = 8cos 60

ST = 8 x ¹/₂ = 4

TR = 10 cm (opp. sides of a rectangle PQRT)

SR = ST + TR

SR = 4 + 10

SR = 14 cm

Y = x implies y ≤ x

Y + x = 4 implies y = 4 – x

∴y = 4 – x

∴y ≤ x, y + x ≤ 4 and y ≥ 0

X ⊕ y = ^xy/₆ = ^xe/₆ = x

Xe = 6x

e = 6

X ⊕ y = ^xx1/₆ where x¹ = 20

Xx20/6 = 6

X = ³⁶/₂₀

X = ⁹/₅

P ∝ ¹/_q³

P = ^K/_q³

q³ = ^K/_P

q = ^K/_p^1/3

But q ∝ r²

q = Kr^2
K/_p^1/3 = Kr²

r² = ^K/_p^1/3 x ¹/_K

r² = ¹/_p^1/3

r = (¹/_p^1/3)²

r = ¹/_p^1/6

r ∝ ¹/_p^1/6

∴ r varies inversely as ⁶√P

(x \times y = \frac{x^2 - y^2}{2xy}find -5 \times 3\

=\frac{(x+y)(x-y)}{2xy}\

=\frac{(-5+3)(-5-3)}{2(-5\times3)}\

=\frac{-2 \times -8}{2(-5\times3)}\

=\frac{-8}{15})

F(x) = Q x D + R

Kx³ + x² - 5x – 2 = Q(2x+1)+R

If 2x+1 = 0 implies x = -¹/₂

∴k(-¹/₂)³ + (-¹/₂)² -5(-¹/₂) -2

= Q(2(-¹/₂) + 1) +2

K(-¹/₈) + ¹/₄ + ⁵/₂ - 2 = Q(-1+1)+2

-^k/₈ + ¹/₄ + ⁵/₂ - 2 = 0+2

(-k+2+20-16) / 8 = 2

(-k+6) / 8 = 2

-k+6 = 2*8

-k+6 = 16

-k = 16-6

-k = 10

∴k = -10

(\frac{x+4}{3}-\frac{x-3}{2} < 4\

\frac{(2(x+4))-(3(x-3))}{6}<4)

2(x+4) – 3(x-3) < 4 x 6

2x + 8 – 3x+9 < 24

-x + 17 < 24

-x < 24 – 17

-x < 7

X > -7