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Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
If x + 1 is a factor of x3 + 3x2 + kx + 4, find the value of k
A
6
B
-6
C
8
D
-8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x + 1 is a factor of x3 + 3x2 + kx + 4
Let f(x) = x3 + 3x2 + kx + 4
∴ f(-1) = (-1)3 + 3(-1)2 + k(-1) + 4 = 0
-1 + 3 - k + 4 = 0
∴ k = 6
Source: Myschool Past Questions Archive (Record #31316)
Mathematics
UTME / JAMB 1992
Linear, Simultaneous & Quadratic Equations
Solution
Resolve \(\frac{3}{x^2 + x - 2}\) into partial fractions
A
\(\frac{1}{x - 1} - \frac{1}{x + 2}\)
B
\(\frac{1}{x + 1} + \frac{1}{x - 2}\)
C
\(\frac{1}{x + 1} - \frac{1}{x - 2}\)
D
\(\frac{1}{x - 2} + \frac{1}{x + 2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\)
\(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\)
A(x + 2) + B(x - 1) = 3
when x = 1, 3A = 3 \(\to\) a = 1
when x = -2, -3B = 3 \(\to\) B = -1
= \(\frac{1}{x - 1} - \frac{1}{x + 2}\)
Source: Myschool Past Questions Archive (Record #31319)
Mathematics
UTME / JAMB 1992
Inequalities & Linear Programming
Solution
Find all values of x satisfying the inequality -11 \(\leq\) 4 - 3x \(\leq\) 28
A
-5 \(\leq\) x \(\geq\) 8
B
5 \(\leq\) x \(\leq\) 8
C
-8 \(\leq\) x \(\leq\) 5
D
-5 < x \(\leq\) 8
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
To solve -11 \(\leq\) 4 - 3x \(\leq\) 28
-11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28
- 11 - 4\(\leq\) - 3x and -3x \(\leq\) 28 - 4
-15\(\leq\) -3x and -3x \(\leq\) 24
i.e. x \(\leq\) 5 and x \(\geq\) -8
∴ -8 \(\leq\) x \(\leq\) 5
Source: Myschool Past Questions Archive (Record #31320)
Mathematics
UTME / JAMB 1992
Sequences & Series (AP & GP)
Solution
Find the sum to infinity to the following series 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
A
1270
B
190
C
18
D
9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
a = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9
Source: Myschool Past Questions Archive (Record #31321)
Mathematics
UTME / JAMB 1992
Sequences & Series (AP & GP)
Solution
For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms
A
157
B
187
C
197
D
200
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
a = 2, d = 3 and n = 11
To find Sn/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)]
= \(\frac{11}{2}\) [2(2) + (11 - 1) 3]
= \(\frac{11}{2}\)n [4 + 10(3)]
= \(\frac{11}{2}\)(34)
= 11 x 17
= 187
Source: Myschool Past Questions Archive (Record #31328)
Mathematics
UTME / JAMB 1992
Binary Operations, Matrices & Determinants
Solution
p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)
Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\) Where pT is the transpose P calculate /pT/ when x = 0, y = 1 and z = 2
A
48
B
24
C
-24
D
-48
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
PT = \(\begin{vmatrix}0 & 2 & 4 \\ 2 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)
/pT/ = \(\begin{vmatrix}0 & 2 & 4 \\ 3 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)
= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]
= 0 - 12 + 36 = 24
Source: Myschool Past Questions Archive (Record #31332)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)
Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\)
PQ is equivalent to
A
PPT
B
pp-1
C
qp
D
pp
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = pT
pq = ppT
Source: Myschool Past Questions Archive (Record #31333)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
If the angles of quadrilateral are (p + 10)^\circ, (2p - 30)^\circ, (3p + 20)^\circ and 4p^\circ, find p.
A
63
B
40
C
36
D
28
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The sum of angles in a quadrilateral = 360^\circ
\(\therefore (p + 10) + (2p - 30) + (3p + 20) + 4p = 360\)
\(10p = 360^\circ \implies p = \frac{360}{10} = 36^\circ\)
Source: Myschool Past Questions Archive (Record #31335)
Mathematics
UTME / JAMB 1992
Trigonometry, Elevation & Bearings
Solution
Determine the distance on the earth's surface between two town P (lat 60^\circN, Long 20^\circE) and Q(Lat 60^\circN, Long 25^\circW) (Radius of the earth = 6400km)
A
\(\frac{800\pi}{9}\)km
B
\(\frac{800\sqrt{3\pi}}{9}\)km
C
800\(\pi\) km
D
800\(\sqrt{3\pi}\) km
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Angular difference (\(\theta\))= 25^\circ + 20^\circ = 45^\circ
\(\alpha\) = common latitude = 60^\circ
\(S = \frac{\theta}{360^\circ} \times 2\pi R \cos \alpha\)
\(S = \frac{45^\circ}{360^\circ} \times 2 \pi \times 6400 \times \cos 60^\circ\)
= \(\frac{6400\pi}{8} = 800\pi km\)
Source: Myschool Past Questions Archive (Record #31337)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Diagram
If in the diagram, FG is parallel to KM, find the value of x
Attached Figure / Diagram
A
75o
B
95o
C
105o
D
125o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #31338)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
The locus of a point which is equidistant from two given fixed points is the
A
perpendicular bisector of the straight line joining them
B
parallel line to the straight line joining them
C
transverse to the straight line joining them
D
angle bisector of 90o which the straight line joining them makes with the horizontal
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #31341)
Mathematics
UTME / JAMB 1992
Indices, Logarithms & Surds
Solution
What is the perpendicular distance of a point (2, 3) from the line 2x - 4y + 3 = 0?
A
\(\frac{\sqrt{5}}{2}\)
B
\(\frac{\sqrt{5}}{20}\)
C
\(\frac{5}{\sqrt{13}}\)
D
6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
2x - 4y + 3 = 0
Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)
= \(\frac{4 - 12 + 3}{\sqrt{20}}\)
= \(\frac{-5}{-2\sqrt{5}}\)
= \(\frac{\sqrt{5}}{2}\)
Source: Myschool Past Questions Archive (Record #31342)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
find then equation line through (5, 7) parallel to the line 7x + 5y = 12
A
5x + 7y = 120
B
7x + 5y = 70
C
x + y = 7
D
15x + 17y = 90
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Equation (5, 7) parallel to the line 7x + 5y = 12
5Y = -7x + 12
y = \(\frac{-7x}{5}\) + \(\frac{12}{5}\)
Gradient = \(\frac{-7}{5}\)
∴ Required equation = \(\frac{y - 7}{x - 5}\) = \(\frac{-7}{5}\) i.e. 5y - 35 = -7x + 35
5y + 7x = 70
Source: Myschool Past Questions Archive (Record #31347)
Mathematics
UTME / JAMB 1992
Trigonometry, Elevation & Bearings
Solution
Given that \(\theta\) is an acute angle and sin \(\theta\) = \(\frac{m}{n}\), find cos \(\theta\)
A
\(\frac{\sqrt{n^2 - m^2}}{m}\)
B
\(\frac{\sqrt{(n + m)(n - m)}}{n}\)
C
\(\frac{m}{\sqrt{n^2 - m^2}}\)
D
\(\sqrt{\frac{n}{n^2 - m^2}}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
sin \(\theta\) = \(\frac{m}{n}\)
Opp = m; Hyp = n
Adj = \(\sqrt{n^{2} - m^{2}}\)
\(\cos \theta = \frac{\sqrt{n^{2} - m^{2}}}{n}\)
= \(\frac{\sqrt{(n + m)(n - m)}}{n}\)
Source: Myschool Past Questions Archive (Record #31349)
Mathematics
UTME / JAMB 1992
Linear, Simultaneous & Quadratic Equations
Solution
Evaluate \(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
A
7
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}\)
\(\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}\)
= \(\frac{(x^{2} + 3x - 2)}{x + 2}\)
\(\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}\)
= \(\frac{2^{2} + 3(2) - 2}{2 + 2}\)
= \(\frac{4 + 6 - 2}{4}\) = 2
Source: Myschool Past Questions Archive (Record #31350)