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Mathematics
UTME / JAMB 2012
Sequences & Series (AP & GP)
Solution
The sum to infinity of a geometric progression is \(-\frac{1}{10}\) and the first term is \(-\frac{1}{8}\). Find the common ratio of the progression.
A
\(-\frac{1}{5}\)
B
\(-\frac{1}{4}\)
C
\(-\frac{1}{3}\)
D
\(-\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Sr = \(\frac{a}{1 - r}\)
\(-\frac{1}{10}\) = \(\frac{1}{8} \times \frac{1}{1 - r}\)
\(-\frac{1}{10}\) = \(\frac{1}{8(1 - r)}\)
\(-\frac{1}{10}\) = \(\frac{1}{8 - 8r}\)
cross multiply...
-1(8 - 8r) = -10
-8 + 8r = -10
8r = -2
r = -1/4
Source: Myschool Past Questions Archive (Record #27329)
Mathematics
UTME / JAMB 2012
Binary Operations, Matrices & Determinants
Solution
The binary operation * is defined on the set of integers such that p * q = pq + p - q. Find 2 * (3 * 4)
A
11
B
13
C
15
D
22
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
p * q = pq + p - q
First execute for 3 * 4
==> 3(4) + 3 - 4 = 12 + 3 - 4 = 11
Now we execute for 2 * 11
==> 2(11) + 2 - 11 = 22 + 2 - 11 = 13
Source: Myschool Past Questions Archive (Record #27330)
Mathematics
UTME / JAMB 2012
Binary Operations, Matrices & Determinants
Solution
The binary operation on the set of real numbers is defined by m*n = \(\frac{mn}{2}\) for all m, n \(\in\) R. If the identity element is 2, find the inverse of -5
A
\(-\frac{4}{5}\)
B
\(-\frac{2}{5}\)
C
4
D
5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
m * n = \(\frac{mn}{2}\)
Identify, e = 2
Let a \(\in\) R, then
a * a\(^{-1}\) = e
a * a\(^{-1}\) = 2
-5 * a\(^{-1}\) = 2
\(\frac{-5 \times a^{-1}}{2} = 2\)
\(a^{-1} = \frac{2 \times 2}{-5}\)
\(a^{-1} = -\frac{4}{5}\)
Source: Myschool Past Questions Archive (Record #27331)
Mathematics
UTME / JAMB 2012
Plane Geometry, Polygons & Circle Theorems
Solution
If \(\begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix}\), find the value of x
A
3
B
4
C
5
D
7
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix}\) = \(\begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix}\)
10 - 3x = 15 - 20
-3x = 15 - 20 - 10
-3x = -15
x = 5
Source: Myschool Past Questions Archive (Record #27332)
Mathematics
UTME / JAMB 2012
Binary Operations, Matrices & Determinants
Solution
Given that I3 is a unit matrix of order 3, find |I3|
A
-1
B
O
C
1
D
2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Recall that a unit matrice is a diagonal matrix in which the elements in the leading diagonal is unity. Therefore,
I3 = \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
I3 = \(+1\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} \)
I3 = +1(1 - 0) - 0(0 - 0) + 0(0 - 0)
= 1(1)
= 1
Source: Myschool Past Questions Archive (Record #27333)
Mathematics
UTME / JAMB 2012
Plane Geometry, Polygons & Circle Theorems
Solution
The angles of a polygon are given by x, 2x, 3x, 4x and 5x respectively. Find the value of x.
A
24o
B
30o
C
33o
D
36o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Since there are 5 angles given, the polygon is a pentagon.
Sum of interior angles of a pentagon = (2(5) - 4) x 90^\circ = 540^\circ
\(\therefore\) x + 2x + 3x + 4x + 5x = 15x
15x = 540^\circ
\(x = \frac{540}{15} = 36^\circ\)
Source: Myschool Past Questions Archive (Record #27334)
Mathematics
UTME / JAMB 2012
Mensuration (Areas, Perimeters & Volumes)
Solution
Calculate the volume of a cuboid of length 0.76cm, breadth 2.6cm and height 0.82cm.
A
3.92cm3
B
2.13cm3
C
1.97cm3
D
1.62cm3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Volume of cuboid = L x b x h
= 0.76cm x 2.6cm x 0.82cm
= 1.62cm3
Source: Myschool Past Questions Archive (Record #27335)
Mathematics
UTME / JAMB 2012
Coordinate Geometry
Solution
The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is
A
\(\frac{1}{2}\)
B
\(\frac{2}{5}\)
C
\(-\frac{4}{7}\)
D
\(-\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
PQ = \(\frac{y_1 - y_0}{x_1 - x_0}\) = \(\frac{-3 - (-7)}{-2 - 5}\) = \(\frac{-3 + 7}{-2 - 5}\) = \(\frac{4}{-7}\)
Source: Myschool Past Questions Archive (Record #27337)
Mathematics
UTME / JAMB 2012
Indices, Logarithms & Surds
Solution
The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is
A
\(\sqrt{13}\)
B
\(3\sqrt{2}\)
C
\(\sqrt{26}\)
D
\(10\sqrt{5}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
P1 (4, 3), P2 (x, y)
y = 2x + 4 .....(1)
y = 7 - x .....(2)
Substitute (2) in (1)
7 - x = 2x + 4
7 - 4 = 2x + x
3 = 3x
x = 1
Substitute in eqn (2)
y = 7 - x
y = 7 - 1
y = 6
P2 (1, 6)
Distance between 2 points is given as
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)
D = \(\sqrt{(-3)^2 + (3)^2}\)
D = \(\sqrt{9 + 9}\)
D = \(\sqrt{18}\)
D = \(\sqrt{9 \times 2}\)
D = \(3\sqrt{2}\)
Source: Myschool Past Questions Archive (Record #27338)
Mathematics
UTME / JAMB 2012
Coordinate Geometry
Solution
Find the equation of the line through the points (-2, 1) and (-\(\frac{1}{2}\), 4)
A
y = 2x - 3
B
y = 2x + 5
C
y = 3x - 2
D
y = 2x + 1
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\)
\(\frac{y - 1}{x - -2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)
= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)
y = 2x + 5
Source: Myschool Past Questions Archive (Record #27339)
Mathematics
UTME / JAMB 2012
Indices, Logarithms & Surds
Solution
If angle \(\theta\) is 135o, evaluate cos\(\theta\)
A
\(\frac{1}{2}\)
B
\(\frac{\sqrt{2}}{2}\)
C
\(-\frac{\sqrt{2}}{2}\)
D
\(-\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\theta\) = 135o
Cos 135o = Cos(90 + 45)o
= cos90ocos45o - sin90osin45o
= 0cos45o - (1 x \(\frac{\sqrt{2}}{2}\))
= \(-\frac{\sqrt{2}}{2}\)
Source: Myschool Past Questions Archive (Record #27340)
Mathematics
UTME / JAMB 2012
Plane Geometry, Polygons & Circle Theorems
Solution
If y = x2 - \(\frac{1}{x}\), find \(\frac{\delta y}{\delta x}\)
A
2x - \(\frac{1}{x^2}\)
B
2x + x2
C
2x - x2
D
2x + \(\frac{1}{x^2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = x2 - \(\frac{1}{x}\)
y = x2 - x-1
\(\frac{\delta y}{\delta x}\) = 2x + x-2
\(\frac{\delta y}{\delta x}\) = 2x + \(\frac{1}{x^2}\)
Source: Myschool Past Questions Archive (Record #27341)
Mathematics
UTME / JAMB 2012
Integration & Applications
Solution
Evaluate \(\int^2_1(x^2 - 4x)dx\)
A
\(\frac{11}{3}\)
B
\(\frac{3}{11}\)
C
\(-\frac{3}{11}\)
D
\(-\frac{11}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\int^2_1(x^2 - 4x)dx\)
\((\frac{x^3}{3} - \frac{4x^2}{2})\)
substituting integrate values
\([\frac{8}{3} - \frac{4 \times 4}{2}] - [\frac{1}{2} - 2]\)
= \(-\frac{11}{3}\)
Source: Myschool Past Questions Archive (Record #27342)
Mathematics
UTME / JAMB 2012
Integration & Applications
Solution
Evaluate \(\int^{\frac{\pi}{4}}_0sec^2 \theta d \theta\)
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\int^{\frac{\pi}{4}}_0sec^2 \theta d \theta\)
= \([\tan \theta]_{0} ^{\frac{\pi}{4}}\)
= \(\tan \frac{\pi}{4} - \tan 0\)
= \(1 - 0\)
= 1.
Source: Myschool Past Questions Archive (Record #27343)
Mathematics
UTME / JAMB 2012
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The mean of seven numbers is 96. If an eighth number is added, the mean becomes 112. Find the eighth number.
A
126
B
180
C
216
D
224
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{x}{7} = \frac{96}{1}\) ==> \(\frac{672 + x}{8} = 112\)
Therefore x = 224
Source: Myschool Past Questions Archive (Record #27344)