2009 - JAMB Mathematics Past Questions and Answers - page 4
31
What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?
A
A circle of radius 4 cm and the center 0
B
The perpendicular bisector of the chords
C
A straight line passing through 0
D
A circle of radius 6 cm and with center 0
correct option: a
x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16
x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0
Users' Answers & Commentsx2 + 9 = 25
x2 = 25 – 9
x2 = 16
x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0
32
What is the value of p if the gradient of the line joining (-1,p) and (p, 4) is
?
| 2 |
| 3 |
A
-2
B
-1
C
1
D
2
33
What is the value of r if the distance between the point (4,2) and (1,r) is 3 units?
A
1
B
2
C
3
D
4
correct option: b
A(4,2) and B(1,r), AB = 3 units
3 = √ (x2- x1)2 + (y2-y1)2
3 = √ (4-1)2 + (2-r)2
3 = √ 32 + (2-r)2
3 = √ 9 + 4 – 4r + r2
3 = √ r2 - 4r + 13
9 = r2 - 4r + 13 By squaring both sides
r2 - 4r + 4 = 0
(r-2)(r-2) = 0
r = 2
Users' Answers & Comments3 = √ (x2- x1)2 + (y2-y1)2
3 = √ (4-1)2 + (2-r)2
3 = √ 32 + (2-r)2
3 = √ 9 + 4 – 4r + r2
3 = √ r2 - 4r + 13
9 = r2 - 4r + 13 By squaring both sides
r2 - 4r + 4 = 0
(r-2)(r-2) = 0
r = 2
34
If y = 3 cos 4x, dy/dx equals
A
6 sin 8x
B
-24 sin 4x
C
12 sin 4x
D
-12 sin 4x
35
Find the value of sin 45o - cos 30o
A
\(\frac{2+\sqrt{3}}{4}\)
B
\(\frac{\sqrt{2}+\sqrt{3}}{4}\)
C
\(\frac{\sqrt{2}+\sqrt{3}}{2}\)
D
\(\frac{\sqrt{2}-\sqrt{3}}{2}\)
correct option: d
\(Sin 45 - cos 30\\
\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\\
=\frac{\sqrt{2}-\sqrt{3}}{2}\)
Users' Answers & Comments36
A cliff on the bank of a river is 300 meter high. if the angle of depression of a point on the opposite side of the river is 60o, find the width of the river.
A
100m
B
75√3 m
C
100√3m
D
200√3m
correct option: c
| Tan 60 = | 300 |
| x |
| x = | 300 |
| Tan 60 |
| x = | 300 |
| √3 |
x = (300/√3) x (√3/√3)
| x = | 300√3 |
| .....3 |
x = 100√3
37
If s = (2 + 3t)(5t - 4), find s/dt when t = 4/5 secs
A
0 units per sec
B
15 units per sec
C
22 unit per sec
D
26 units per sec
correct option: c
x = (2+3t)(5t-4)
Let u = 2+3t ∴du/dt = 3
and v = 5t-4 ∴dv/dt = 5
dx/dt = Vdu/dt + Udv/dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x4/5 - 2
= 24 - 2
= 22
Users' Answers & CommentsLet u = 2+3t ∴du/dt = 3
and v = 5t-4 ∴dv/dt = 5
dx/dt = Vdu/dt + Udv/dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x4/5 - 2
= 24 - 2
= 22
38
What value of s will make the function x(4-x) a maximum
A
4
B
3
C
2
D
1
correct option: c
x(4-x) = 4x-x2
dy/dx = 4-2x
As dy/dx = 0
4-2x = 0
2x = 4
x = 2
Users' Answers & Commentsdy/dx = 4-2x
As dy/dx = 0
4-2x = 0
2x = 4
x = 2
39
The distance travelled by a particle from a fixed point is given as s = (t3 - t2 - t + 5)cm. Find the minimum distance that the particle can cover from the fixed point.
A
2.3 cm
B
4.0 cm
C
5.2 cm
D
6.0 cm
correct option: c
S = t3 - t2 - t + 5
ds/dt = 3t2 - 2t - 1
As ds = 0
3t2 - 2t - 1 = 0
(3t+1)(t-1) = 0
∴ t = 1 or -1/3
At min pt t = -1/3
S = t3 - t2 - t + 5 put t = -1/3
= (-1/3)3 - (-1/3)2 - (-1/3) + 5
= -1/27 - 1/9 + 1/3 + 5
= 140 / 27
= 5.2
Users' Answers & Commentsds/dt = 3t2 - 2t - 1
As ds = 0
3t2 - 2t - 1 = 0
(3t+1)(t-1) = 0
∴ t = 1 or -1/3
At min pt t = -1/3
S = t3 - t2 - t + 5 put t = -1/3
= (-1/3)3 - (-1/3)2 - (-1/3) + 5
= -1/27 - 1/9 + 1/3 + 5
= 140 / 27
= 5.2
40
Evaluate ∫sec2θ dθ
A
sec θ tan θ + k
B
tan θ + k
C
2sec θ + k
D
sec θ + k
correct option: b
∫sec2θ dθ = ∫ 1/cos2 dθ
∫(cos)-2 dθ,
let u = cos θ
∴∫u-2 = 1/u + c
∫cos θ = sin θ + c
∫sec-2θ = 1/u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c
Users' Answers & Comments∫(cos)-2 dθ,
let u = cos θ
∴∫u-2 = 1/u + c
∫cos θ = sin θ + c
∫sec-2θ = 1/u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c