JAMB Mathematics Past Questions & Answers - Page 123

611.

In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS

A.

120o

B.

70o

C.

60o

D.

40o

Correct answer is B

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180

RQT = 180o - (50 + 60)

= 180o - 110o

= 70o

Since RQT = RTS = 70o

612.

In the diagram, O is the centre of the circle. If SOQ is a diameter and < PRS is 38o, what is the value of < PSQ

A.

148o

B.

104o

C.

80o

D.

52o

Correct answer is D

< SRP = < SQP = 38o (angle in the same segment of a circle are equal)

But < SPQ = 90o (angle in a semicircle)

also < PSQ + < SQP + < SPQ = 180o (angles in a triangle = 180o)

< PSQ + 38o + 90o = 190o

< PSQ = 128o = 180o

PSQ = 180o - 128o

PSQ = 52o

613.

In the diagram, PQRS is a parallelogram. Find the value of < SQR

A.

30o

B.

50o

C.

80o

D.

100o

Correct answer is D

SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR

< STP = < RQV = 30o

But SQR + 30o + 50o = 180o

SQR = 180 - 80

= 100o

614.

Find the inequality which represents the shaded portion in the diagram

A.

2x - y - 2 \(\geq\) 0

B.

2x - y - 2 \(\leq\) 0

C.

2x - y - 2 < 0

D.

2x - y - 2 > 0

Correct answer is A

2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2

when x = 0, y = -2, when y = 0, x = 1

615.

The equation of the graph is

A.

y = (x - 3)3

B.

y = (x + 3)3

C.

y = x3 - 27

D.

y = -x3 + 27

Correct answer is C

y = x3 - 27, y = -27 \(\to\) (0, -27)

when y = 0, x = 3 (3, 0)