2005 - JAMB Mathematics Past Questions and Answers - page 4
31
Find the derivative of y = sin(2x3 + 3x - 4)
A
cos (2x3 + 3x - 4)
B
-cos (2x3 + 3x - 4)
C
(6x2 + 3) cos (2x3 + 3x - 4)
D
-(6x2 + 3) cos (2x3 + 3x - 4)
correct option: c
y = sin (2x3 + 3x - 4)
let u = 2x3 + 3x - 4
∴du/dx = 6x2
y = sin U
dy/du = cos U
dy/dx = du/dx * dy/dy
∴dy/dx = (6x2 + 3) cos U
= (6x2 + 3)cos(2x3 + 3x - 4)
Users' Answers & Commentslet u = 2x3 + 3x - 4
∴du/dx = 6x2
y = sin U
dy/du = cos U
dy/dx = du/dx * dy/dy
∴dy/dx = (6x2 + 3) cos U
= (6x2 + 3)cos(2x3 + 3x - 4)
32
The radius r of a circular disc is increasing at the rate of 0.5cm/sec. At what rate is the area of the disc increasing when its radius is 6cm?
A
36 π cm2/sec
B
18 π cm2/sec
C
6 π cm2/sec
D
3 π cm2/sec
correct option: c
Increase in radius dr/dt = 0.5
Area of the circular disc = πr2
Increase in area: dA/dr = 2πr
Rate of increase in area: dA/dt = dA/dr * dr/dt
when r = 6cm
= 2πr * 0.5
= 2 * π * 6 * 0.5
2 * π * 6 * 1/2 = 6 π cm2/sec
Users' Answers & CommentsArea of the circular disc = πr2
Increase in area: dA/dr = 2πr
Rate of increase in area: dA/dt = dA/dr * dr/dt
when r = 6cm
= 2πr * 0.5
= 2 * π * 6 * 0.5
2 * π * 6 * 1/2 = 6 π cm2/sec
33
The maximum value of the function
f(x) = 2 + x - x2 is
f(x) = 2 + x - x2 is
A
9/4
B
7/4
C
3/2
D
1/2
correct option: a
f(x) = 2 + x - x2
dy/dx = 1-2x
As dy/dx = 0
1-2x = 0
2x = 1
x = 1/2
At x = 1/2
f(x) = 2 + x - x2
= 2 + 1/2 -(1/2)2
= 2 + 1/2 - 1/4
= (8+2-1) / 4 = 9/4
Users' Answers & Commentsdy/dx = 1-2x
As dy/dx = 0
1-2x = 0
2x = 1
x = 1/2
At x = 1/2
f(x) = 2 + x - x2
= 2 + 1/2 -(1/2)2
= 2 + 1/2 - 1/4
= (8+2-1) / 4 = 9/4
34
Find the area of the figure bounded by the given pair of curves y = x2 - x + 3 and y = 3
A
17/6 units (sq)
B
7/6 units (sq)
C
5/6 units (sq)
D
1/6 units (sq)
correct option: a
Area bounded by
y = x2 - x + 3 and y = 3
x2 - x + 3 = 3
x2 - x = 0
x(x -1) = 0
x= 0 and x = 1
\(\int^{1}_{0}\)(x2 - x + 3)dx = [1/3x3 - 1/2x2 + 3x]\(^1 _0\)
(1/3(1)3 - 1/2(1)2 + 3(1) - (1/3(0)3 - 1/2(0)2 + 3(0))
= (1/3 - 1/2 + 3)- 0
= \(\frac{2-3+18}{6}\)
= 17/6
Users' Answers & Commentsy = x2 - x + 3 and y = 3
x2 - x + 3 = 3
x2 - x = 0
x(x -1) = 0
x= 0 and x = 1
\(\int^{1}_{0}\)(x2 - x + 3)dx = [1/3x3 - 1/2x2 + 3x]\(^1 _0\)
(1/3(1)3 - 1/2(1)2 + 3(1) - (1/3(0)3 - 1/2(0)2 + 3(0))
= (1/3 - 1/2 + 3)- 0
= \(\frac{2-3+18}{6}\)
= 17/6
35
Evaluate \(\int_0^{\frac{\pi}{0}}sin2xdx\)
A
1
B
zero
C
-1/2
D
-1
correct option: a
\(\int_0^{\frac{\pi}{2}}\)sin 2x dx = [-1/2cos 2x + C]\(_0^{\frac{\pi}{2}}\)
=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1
Users' Answers & Comments=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1
36
A
16
B
12
C
8
D
6
correct option: b
Users' Answers & Comments37
A
7
B
8
C
9
D
12
correct option: a
Angle of Excellent
= 360 - (120+80+90)
= 360 - 290
= 70o
If 360o represents 36 students
1o will represent 36/360
50o will represent 36/360 * 70/1
= 7
Users' Answers & Comments= 360 - (120+80+90)
= 360 - 290
= 70o
If 360o represents 36 students
1o will represent 36/360
50o will represent 36/360 * 70/1
= 7
38
A
1
B
2
C
3
D
4
correct option: b
\(mean=\frac{60+5x}{18+x}\3.5=\frac{60+5x}{18+x}\\frac{7}{2}=\frac{60+5x}{18+x}\)
7(18+x) = 2(60+5x)
126 + 7x = 120 + 10x
10x - 7x = 126 - 120
3x = 6
x = 2
Users' Answers & Comments7(18+x) = 2(60+5x)
126 + 7x = 120 + 10x
10x - 7x = 126 - 120
3x = 6
x = 2
39
The model height and range of heights 1.35, 1.25, 1.35, 1.40, 1.35, 1.50, 1.35, 1.50, and 1.20 are m and r respectively. Find m+2r.
A
1.35
B
1.65
C
1.95
D
3.00
correct option: c
1.35, 1.25, 1.35, 1.40, 1.35, 1.50, 1.35, 1.50, and 1.20
Modal height = 1.35
Range (r) = 1.50 - 1.20
= 0.30
∴m + 2r = 1.35 + 2(0.30)
= 1.35 + 0.60
= 1.95
Users' Answers & CommentsModal height = 1.35
Range (r) = 1.50 - 1.20
= 0.30
∴m + 2r = 1.35 + 2(0.30)
= 1.35 + 0.60
= 1.95
40
Find the value of t if the standard deviation of 2t, 3t, 4t, 5t, and 6t is √2
A
1
B
2
C
3
D
4
correct option: a
\(mean(x) = \frac{20t}{5} = 4t\standard\hspace{1mm}deviation=\sqrt{\frac{10t^2}{5}}\\sqrt{2}=\sqrt{\frac{10t^2}{5}}\\sqrt{2}=\sqrt{2t^2}\2 = 2t^2\t^2=1\t=1\)
Users' Answers & Comments