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Mathematics
WAEC / WASSCE 1998
Algebra
Solution
solve the inequality \((Y-2)<\frac{y}{3}\)
A
y > -9
B
y < 3
C
y > 4
D
y < 9
E
y > 0
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((Y-2)<\frac{y}{3}\)
3y - 6 < y
2y< 6
y < \(\frac{6}{2}\)
y < 3.
Source: Myschool Past Questions Archive (Record #26336)
Mathematics
WAEC / WASSCE 1998
Algebra
Diagram
Solution
The graph is that of the quadratic expression
Attached Figure / Diagram
A
\(y=x^2 + x - 2\)
B
\(y=x^2 - x + 2\)
C
\(y=x^2 - x - 2\)
D
\(y=x + x + 2\)
E
\(y=x^2 - 2 x - 2\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The curve cuts the x-axis at x = -1 and x = 2
(x + 1)(x - 2) = 0
x\(^2\) - 2x + x - 2 = 0
x\(^2\) - x - 2 = 0
Source: Myschool Past Questions Archive (Record #26345)
Mathematics
UTME / JAMB 1998
Algebra
Solution
Given that log4(Y - 1) + log4(\(\frac{1}{2}\)x) = 1 and log2(y + 1) + log2x = 2, solve for x and y respectively
A
2, 3
B
3, 2
C
-2, -3
D
-3, -2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
log4(y - 1) + log4(\(\frac{1}{2}\)x) = 1
log4(y - 1)(\(\frac{1}{2}\)x) \(\to\) (y - 1)(\(\frac{1}{2}\)x) = 4 ........(1)
log2(y + 1) + log2x = 2
log2(y + 1)x = 2 \(\to\) (y + 1)x = 22 = 4.....(ii)
From equation (ii) x = \(\frac{4}{y + 1}\)........(iii)
put equation (iii) in (i) = y (y - 1)[\(\frac{1}{2}(\frac{4}{y - 1}\))] = 4
= 2y - 2
= 4y + 4
2y = -6
y = -3
x = \(\frac{4}{-3 + 1}\)
= \(\frac{4}{-2}\)
X = 2
therefore x = -2, y = -3
Source: Myschool Past Questions Archive (Record #30619)
Mathematics
UTME / JAMB 1998
Algebra
Solution
Solve for the equation \(\sqrt{x}\) - \(\sqrt{(x - 2)}\) - 1 = 0
A
\(\frac{3}{2}\)
B
\(\frac{2}{3}\)
C
\(\frac{4}{9}\)
D
\(\frac{9}{4}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\sqrt{x}\) - \(\sqrt{(x - 2)}\) - 1 = 0
= \(\sqrt{x}\) - \(\sqrt{(x - 2)}\) = 1
= (\(\sqrt{x}\) - \(\sqrt{(x - 2)}\))2 = 1
= x - 2 \(\sqrt{x(x - 2)}\) + x -2 = 1
= (2x - 3)2 = [2 \(\sqrt{x(x - 4)}\)]2
= 4x2 - 12x + 9
= 4(x2 - 2x)
= 4x2 - 12x + 9
= 4x2 - 8x
4x = 9
x = \(\frac{9}{4}\)
Source: Myschool Past Questions Archive (Record #30674)
Mathematics
UTME / JAMB 1998
Algebra
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
Algebra
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
Algebra
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
Algebra
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
Algebra
Solution
find the equation of the curve which passes through by 6x - 5
A
6x2 - 5x + 5
B
6x2 + 5x + 5
C
3x2 - 5x - 5
D
3x2 - 5x + 3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
m = \(\frac{dy}{dv}\) = 6x - 5
∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3
Source: Myschool Past Questions Archive (Record #30740)
Mathematics
UTME / JAMB 1998
Algebra
Solution
If x, y can take values from the set (1, 2, 3, 4), find the probability that the product of x and y is not greater than 6
A
\(\frac{5}{8}\)
B
\(\frac{5}{16}\)
C
\(\frac{1}{2}\)
D
\(\frac{3}{8}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\begin{vmatrix}& \hline 1 & 2 & 3 & 4\\\hline 1 & 1 & 2& 3 & 4\\2 & 2& 4 & 6 & 8\\ 3& 3& 6& 9 & 12\\4 & 4 & 8 & 12& 16\end{vmatrix}\)
p (product of x, and y \(\leq\) 6) = \(\frac{10}{16}\)
= \(\frac{5}{8}\)
Source: Myschool Past Questions Archive (Record #30758)
Mathematics
WAEC / WASSCE 1998
Algebra
Solution
(a) The value of the expression \(2Ax - Kx^{2}\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.
(b) Solve the equation \(x^{2} - 3x - 1 = 0\), giving your answers correct to 1 decimal place.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(2Ax - Kx^{2}\)
When x = 1, \(2A(1) - K(1^{2}) = 7\)
\(2A - K = 7 ... (1)\)
When x = 2, \(2A(2) - K(2^{2}) = 4\)
\(4A - 4K = 4 \implies A - K = 1 ... (2)\)
From (2), \(A = 1 + K\)
(1) becomes : \(2(1 + K) - K = 7\)
\(2 + 2K - K = 7 \implies 2 + K = 7\)
\(\therefore K = 5\)
\(A = 1 + K = 1 + 5 = 6\)
\(A, K = 6, 5\)
(b) \(x^{2} - 3x - 1 = 0\)
\(a = 1, b = -3, c = -1\)
Using the quadratic formula,
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
\(x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\)
\(x = \frac{3 \pm \sqrt{9 + 12}}{2}\)
\(x = \frac{3 \pm \sqrt{21}}{2} = \frac{3 \pm 4.583}{2}\)
\(x = \frac{3 + 4.583}{2} ; x = \frac{3 - 4.583}{2}\)
\(x = \frac{7.583}{2} = 3.7915 \approxeq 3.8\) (to 1 d.p)
\(x = \frac{-1.583}{2} = -0.7915 \approxeq -0.8\) (to 1 d.p)
Source: Myschool Past Questions Archive (Record #54431)
Mathematics
WAEC / WASSCE 1998
Algebra
Solution
(a) Copy and complete the table of values for the relation \(y = 5 - 7x - 6x^{2}\) for \(-3 \leq x \leq 2\).
x
-3
-2
-1
-0.5
0
1
2
y
-28
6
5
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the :
(i) graph of \(y = 5 - 7x - 6x^{2}\) ; (ii) line \(y = 3\) on the same axis.
(c) Use your graph to find the : (i) roots of the equation \(2 - 7x - 6x^{2} = 0\) ; (ii) maximum value of \(y = 5 - 7x - 6x^{2}\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
x
-3
-2
-1
-0.5
0
1
2
y
-28
-5
6
7
5
-8
-33
(b)(i)
(c)(i) \(2 - 7x - 6x^{2} = 0\)
\(2 + 3 - 7x - 6x^{2} = 3\)
\(5 - 7x - 6x^{2} = 3\)
\(\therefore \text{the roots at y = 3 are } x = -1.4; 0.2\)
(ii) Maximum value of \(y = 5 - 7x - 6x^{2}\) is y = 7.
Source: Myschool Past Questions Archive (Record #54448)
Mathematics
UTME / JAMB 1997
Algebra
Solution
Solve the simultaneous equations \(\frac{2}{x} - {\frac{3}{y}}\) = 2, \(\frac{4}{x} + {\frac{3}{y}}\) = 10
A
x = \(\frac{3}{2}\), y = \(\frac{3}{2}\)
B
x = \(\frac{1}{2}\), y = \(\frac{3}{2}\)
C
x = \(\frac{-1}{2}\), y = \(\frac{-3}{2}\)
D
x = \(\frac{1}{3}\), y = \(\frac{3}{2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{2}{x} - {\frac{3}{y}}\) = 2.....(1)
\(\frac{4}{x} + {\frac{3}{y}}\) = 10 ... (2)
(1) + (2):
\(\frac{6}{x}\) = 12 \(\to\) x = \(\frac{6}{12}\)
x = \(\frac{1}{2}\)
put x = \(\frac{1}{2}\) in equation (i)
= 4 - \(\frac{3}{y}\) = 2
= 4 - 2
= \(\frac{3}{y}\)
therefore y = \(\frac{3}{2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #30801)
Mathematics
WAEC / WASSCE 1997
Algebra
Solution
Solve me equation: 2/3 (x + 5) = 1/4(5x - 3)
A
11/7
B
18/23
C
3
D
43/7
E
7
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
2/3 (x + 5) = 1/4(5x - 3)
8x + 40 = 15x - 9
-7x = -49
x = 7
Source: Myschool Past Questions Archive (Record #23859)
Mathematics
WAEC / WASSCE 1997
Algebra
Solution
Which of the following equations has its roots as 4 and -5?
A
x2 + 4x - 20 = 0
B
x2 + x + 20 = 0
C
x2 - x + 20 = 0
D
x2 + x - 20 = 0
E
x2 - x - 20 = 0
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
(x - 4)(x + 5) = 0
x\(^2\) + 5x - 4x - 20 = 0
x\(^2\) + x - 20 = 0
Source: Myschool Past Questions Archive (Record #23860)