1993 - JAMB Mathematics Past Questions and Answers - page 5

The two equations are y = x(2 - x) and y = (x - 1)(x - 3) The root of these equations are points where the graph of the equations cuts the x axis; but at these points = 0 put y = 0; 0 = x(2 - x) 0 = (x - 1)(x - 3) x = 0 or x = 2; x = 1 or x = 3 The values of x at P, Q, R are increasing towards the positive direction of x-axis at P, x = 1 at Q, x = 2 at R, x = 3 P, Q, R are respectively (1, 2, 3)

< PSO = \(\frac{1}{2}\) < SOQ = \(\frac{1}{2}\)(180) = 90^o

< RTS = < PQS = 32^o (Alternative angle)

< PSQ = 90 - < PSQ = 90^o - 32^o

= 58^o

OQ = OR = radii

< ROQ = 180 - 86 = 84^o

\(\bigtriangleup\)OQR = Isosceles

R = Q

R + Q + 84 = 180(angle in a \(\bigtriangleup\))

2R = 96 since R = Q

R = 48^o

ORQ = 48^o

Construction joins R to P such that SRP = straight line

R = 180^o - 107^o

< p = 180^o - (107^o - 34^o)

108 - 141^o = 39^o

Angle < S = 39^o (corr. Ang.) But in \(\bigtriangleup\)SRT

< S = < T = 39^o

SRT = 180 - (39^o + 39^o)

= 180^o - 78^o

= 102^o

Since QRU= X^o

RSU = X^o, But RSU = 180^o - (50^o + 35^o)

= 180^o - 85^o

= 95^o

x = 95^o

Using angle bisector theorem: line PS bisects angle QPR QS/QP = SR/PR 3/4 = SR/g 4SR = 27 SR = \(\frac{27}{4}\) = 6\(\frac{3}{4}\)cm

SQ² + OS² = OQ² + 5² = 14²

SQ² = 14² - 5²

196 - 25 = 171

ST² + TQ² = SQ²

ST² + 2² = 171

ST² = 171 - 4

= 167

ST = \(\sqrt{167}\)

= 12.92 = 12.9cm

Since area of square PQRS = 100cm²

each lenght = 10cm

Also TUYS : XQVU = 1 : 16

lengths are in ratio 1 : 4, hence TU : UV = 1: 4

Let TU = x

UV = 1: 4

hence TV = x + 4x = 5x = 10cm

x = 2cm

TU = 2cm

UV = 8cm

But TU = SY and UV = YR

YR = 8cm

TH = tan 45^o, TH = QH

\(\frac{TH}{5 + QH}\) = tan 30^o

TH = (b + QH) tan 30^o

QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)

QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)

QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)

= \(\frac{5}{\sqrt{3} - 1}\)