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Mathematics
UTME / JAMB 1998
Linear, Simultaneous & Quadratic Equations
Divide 2x\(^{3}\) + 11x\(^2\) + 17x + 6 by 2x + 1.
A
x2 + 5x + 6
B
2x2 + 5x - 6
C
2x2 + 5x + 6
D
x2 - 5x + 6
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 #30687)
Mathematics
UTME / JAMB 1998
Linear, Simultaneous & Quadratic Equations
Solution
Express in partial fractions \(\frac{11x + 2}{6x^2 - x - 1}\)
A
\(\frac{1}{3x - 1}\) + \(\frac{3}{2x + 1}\)
B
\(\frac{3}{3x + 1}\) - \(\frac{1}{2x - 1}\)
C
\(\frac{3}{3x + 1}\) - \(\frac{1}{2x - 1}\)
D
\(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{11x + 2}{6x^2 - x - 1}\) = \(\frac{11x + 2}{(3x + 1)(2x - 1)}\)
= \(\frac{A}{3x + 1}\) + \(\frac{B}{2x - 1}\)
11x + 2 = A(2x - 1) + B(3x + 1)
put x = \(\frac{1}{2}\)
\(\frac{15}{2} = \frac{5}{2}B\)
B = 3.
put x = \(-\frac{1}{3}\)
\(-\frac{5}{3} = \frac{-5}{3}\)A \(\implies\) A = 1
∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)
Source: Myschool Past Questions Archive (Record #30696)
Mathematics
UTME / JAMB 1998
Linear, Simultaneous & Quadratic Equations
Solution
If x is a positive real number, find the range of values for which \(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)
A
x > -\(\frac{1}{6}\)
B
x > 0
C
0 < x < 4
D
0 < x < \(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)
= \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4x(2 + 3x) > 6x = 12x\(^2\) - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0
= x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution)
Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\)
Combining solutions in cases(1) and (2)
= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #30704)
Mathematics
UTME / JAMB 1998
Sequences & Series (AP & GP)
Solution
If p + 1, 2P - 10, 1 - 4p2are three consecutive terms of an arithmetic progression, find the possible values of p
A
-4, 2
B
-3, \(\frac{4}{11}\)
C
-\(\frac{4}{11}\), 2
D
5, -3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
2p - 10 = \(\frac{p + 1 + 1 - 4P^2}{2}\) (Arithmetic mean)
= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -\(\frac{4}{11}\)
Source: Myschool Past Questions Archive (Record #30708)
Mathematics
UTME / JAMB 1998
Sequences & Series (AP & GP)
Solution
The sum of the first three terms of a geometric progression is half its sum to infinity. Find the positive common ratio of the progression.
A
\(\frac{1}{4}\)
B
\(\frac{1}{\sqrt[3]{2}}\)
C
\(\frac{1}{\sqrt[3]{3}}\)
D
\(\frac{1}{\sqrt{2}}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let the first term of the geometric progression (GP) be \( a \) and the common ratio be \( r \). The sum of the first three terms is: a + ar + ar\(^2\) = a(1 + r + r\(^2\))
\(S_3 = a(1 + r + r^2)\)
The sum to infinity is:
\(S_{\infty} = \frac{a}{1 - r}\)
But: \(S_3 = \frac{1}{2} S_{\infty}\)
Substituting the expressions gives:
\(a(1 + r + r^2) = \frac{1}{2} \cdot \frac{a}{1 - r}\)
Dividing by \( a \):
\(1 + r + r^2 = \frac{1}{2(1 - r)}\)
Cross-multiplying:
\(2(1 + r + r^2)(1 - r) = 1\)
Expanding and simplifying leads to:
\(2(1 - r^3) = 1\)
Thus: \(r^3 = \frac{1}{2}\)
Taking the cube root:
\(r = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}\)
Therefore, the positive common ratio of the progression is:
\(\frac{1}{\sqrt[3]{2}}\).
Source: Myschool Past Questions Archive (Record #30712)
Mathematics
UTME / JAMB 1998
Binary Operations, Matrices & Determinants
Solution
The identity element with respect to the multiplication shown in the diagram below is \(\begin{array}{c|c} \otimes & p & p & r & s \\ \hline p & r & p & r & p
\\ q & p & q & r & s\\ r & r & r & r & r\\ s & q & s & r & q\end{array}\)
A
p
B
q
C
r
D
s
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The identity element (e) under an operation, say \(\otimes\), is the element such that for any given element under the operation, say a,
\(a \otimes e = e \otimes a = a\)
From the table, q is the identity element.
\(p \otimes q = q \otimes p = p\)
Same as all through.
Source: Myschool Past Questions Archive (Record #30713)
Mathematics
UTME / JAMB 1998
Binary Operations, Matrices & Determinants
Solution
The binary operation \(\ast\) is defined by x \(\ast\) y = xy - y - x for all real values x and y. If x \(\ast\) 3 = 2\(\ast\) x, find x
A
-1
B
4
C
1
D
5
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x \(\ast\) y = xy - y - x, x \(\ast\) 3 = 3x - 3 - x = 2x - 3
2 \(\ast\) x = 2x - x - 2 = x - 2
∴ 2x - 3 = x - 2
x = -2 + 3
= 1
Source: Myschool Past Questions Archive (Record #30715)
Mathematics
UTME / JAMB 1998
Binary Operations, Matrices & Determinants
Solution
The determinant of matrix \(\begin{pmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4\end{pmatrix}\) in terms of x is
A
-3x2 - 17
B
-3x2 + 9x - 1
C
3x2 + 17
D
3x2 - 9x + 5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\begin{vmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4\end{vmatrix}\) = x\(\begin{vmatrix}2 & 3 \\ 1+x & 4\end{vmatrix}\) - \(\begin{vmatrix}1-x & 3 \\ 1 & 4\end{vmatrix}\) = 0
= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]
= 5x - 3x2 -1 + 4x
= -3x2 + 9X - 1
Source: Myschool Past Questions Archive (Record #30716)
Mathematics
UTME / JAMB 1998
Binary Operations, Matrices & Determinants
Solution
Let I = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) p = \(\begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}\) Q = \(\begin{pmatrix} u & 4+u \\ -2v & v \end{pmatrix}\) be 2 x 2 matrices such that PQ = I. Find (u, v)
A
(-\(\frac{5}{2}\) - 1)
B
(-\(\frac{5}{2}\) - \(\frac{3}{2}\))
C
(-\(\frac{5}{6}\) - 1)
D
(\(\frac{5}{2}\) - \(\frac{3}{2}\))
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
PQ = \(\begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}\)\(\begin{pmatrix} u & 4+u \\ -2v & v \end{pmatrix}\)
= \(\begin{pmatrix} (2u-6v & 2(4+u) +3v)\\ 4u-10v & 4(4+u)+5v \end{pmatrix}\)
= \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
\(\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}\)
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -\(\frac{5}{2}\)
∴ (U, V) = (-\(\frac{5}{2}\) - 1)
Source: Myschool Past Questions Archive (Record #30718)
Mathematics
UTME / JAMB 1998
Plane Geometry, Polygons & Circle Theorems
The locus of all points at a distance 8cm from a point N passes through points T and S. If S is equidistant from T and N, find the area of triangle STN.
A
4\(\sqrt{3cm^2}\)
B
16\(\sqrt{3cm^2}\)
C
32cm2
D
64cm2
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 #30722)
Mathematics
UTME / JAMB 1998
Indices, Logarithms & Surds
Solution
If the distance between the points (x, 3) and (-x, 2) is 5. Find x
A
6.0
B
2.5
C
\(\sqrt{6}\)
D
\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
d2 = (y - y)2 + (x - x)2
5 = 4x2 + 1 = 25= 4x2 + 1
= 4x2 = 25 - 1= 24
x2 = \(\frac{24}{4}\)
x = \(\sqrt{6}\)
Source: Myschool Past Questions Archive (Record #30724)
Mathematics
UTME / JAMB 1998
Coordinate Geometry
Solution
The midpoint of the segment of the line y = 4x + 3, which lies between the x-axis and the y-axis
A
(\(\frac{3}{2}\), \(\frac{3}{2}\))
B
(\(\frac{2}{3}\), \(\frac{3}{2}\))
C
(\(\frac{3}{8}\), \(\frac{3}{2}\))
D
(-\(\frac{3}{8}\), \(\frac{3}{2}\))
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = 4x + 3
when x = 0, y = 3 \(\to\) (0, 3)
when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0)
mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{2}\)
-\(\frac{3}{8}\), \(\frac{3}{2}\)
Source: Myschool Past Questions Archive (Record #30725)
Mathematics
UTME / JAMB 1998
Inequalities & Linear Programming
Solution
solve the equation cos x + sin x = \(\frac{1}{cos x - sinx}\) for values of such that 0 \(\leq\) x < 2\(\pi\)
A
\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
B
\(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
C
0, \(\frac{\pi}{3}\)
D
0, \(\pi\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
cos x + sin x = \(\frac{1}{cos x - sinx}\)
= (cosx + sinx)(cosx - sinx) = 1
= cos\(^2\)x - sin\(^2\)x = 1
cos\(^2\)x - (1 - cos\(^2\)x) = 1
= 2cos\(^2\)x = 2
cos\(^2\)x = 1
= cosx = \(\pm\)1
x = cos\(^{-1}\) (\(\pm\)1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)
Source: Myschool Past Questions Archive (Record #30729)
Mathematics
UTME / JAMB 1998
Trigonometry, Elevation & Bearings
Solution
From the top of a vertical mast 150m high., two huts on the same ground level are observed. One due east and the other due west of the mast. Their angles of depression are 60o and 45o respectively. Find the distance between the huts
A
150(1 + \(\sqrt{3}\))m
B
50(3 + \(\sqrt{3}\))m
C
150 \(\sqrt{3}\)m
D
\(\frac{50}{\sqrt{3}}\)m
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Tan60º = \(\frac{150}{y}\)
\(\sqrt{3}\) = \(\frac{150}{y}\)
y = \(\frac{150\sqrt3}{3}\)
y = 50\(\sqrt3\)
Tan45º = \(\frac{150}{z}\)
1 = \(\frac{150}{z}\)
z = 150
But x = z + y
x = 150 + 50\(\sqrt3\)
x = 50(3 +\(\sqrt{3}\))m
Source: Myschool Past Questions Archive (Record #30734)
Mathematics
UTME / JAMB 1998
Differentiation & Applications
Solution
If y = 243(4x + 5)-2, find \(\frac{dy}{dx}\) when x = 1
A
\(\frac{-8}{3}\)
B
\(\frac{3}{8}\)
C
\(\frac{9}{8}\)
D
-\(\frac{8}{9}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
y = 243(4x + 5)-2, find \(\frac{dy}{dx}\)
= -1944(4x + 5)-3
= 1944(9)-3
\(\frac{dy}{dx}\) when x = 1
= -\(\frac{1944}{9^3}\)
= -\(\frac{1944}{729}\)
= \(\frac{-8}{3}\)
Source: Myschool Past Questions Archive (Record #30737)