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Mathematics
UTME / JAMB 1991
Indices, Logarithms & Surds
Solution
Rationalize \(\frac{2\sqrt{3} + 3 \sqrt{2}}{3\sqrt{2} - 2 \sqrt{3}}\)
A
5 - 2\(\sqrt{6}\)
B
5 + 2\(\sqrt{6}\)
C
5\(\sqrt{6}\)
D
5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{2\sqrt{3} + 3 \sqrt{2}}{3\sqrt{2} - 2 \sqrt{3}}\)
= \(\frac{2\sqrt{3} + 3 \sqrt{2}}{3\sqrt{2} - 2 \sqrt{3}}\) x \(\frac{3\sqrt{2} + 2 \sqrt{3}}{3\sqrt{2} - 2 \sqrt{3}}\)
\(\frac{4(3) + 9(2) + 2(6) \sqrt{6}}{9(2) - 4(3)}\)
\(\frac{12 + 18 + 12\sqrt{6}}{1`8 - 12}\)
= \(\frac{30 + 12\sqrt{6}}{6}\)
= 5 + 2\(\sqrt{6}\)
Source: Myschool Past Questions Archive (Record #31382)
Mathematics
UTME / JAMB 1991
Plane Geometry, Polygons & Circle Theorems
Solution
Multiply (x2 - 3x + 1) by (x - a)
A
x3 - (3 + a) x2 + (1 + 3a)x - a
B
x3 - (3 - a)x2 + 3ax - a
C
x3 - (3 - a)x2 - (1 = 3a) - a
D
x3 + (3 - a)x2 + (1 + 3a) - a
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(x2 - 3x + 1)(x - a) = x3 - 3x2 + x - ax2 + 3ax - a
= x3 - (3 + a) x2 + (1 + 3a)x - a
Source: Myschool Past Questions Archive (Record #31384)
Mathematics
UTME / JAMB 1991
Linear, Simultaneous & Quadratic Equations
Solution
Evaluate \(\frac{xy^2 - x^2y}{x^2 - xy}\) When x = -2 and y = 3
A
3
B
-\(\frac{3}{5}\)
C
\(\frac{3}{5}\)
D
-3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{xy^2 - x^2y}{x^2 - xy^1}\)
= \(\frac{(-2)(3)^2 - (-2)^2(3)}{(-2)^2 - (-2)(3)}\)
= \(\frac{-30}{10}\)
= -3
Source: Myschool Past Questions Archive (Record #31387)
Mathematics
UTME / JAMB 1991
Plane Geometry, Polygons & Circle Theorems
Solution
If g(x) = x2 + 3x + 4, find g(x + 1) - g(x)
A
(x + 2)
B
2(x + 2)
C
(2x + 1)
D
(x2 + 4)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
g(x) = x2 + 3x + 4
= g(x + 1) = (x + 1)^2 + 3(x + 1) + 4
= x2 + 1 + 2x + 3x + 3 + 4
= x2 + 5x + 8
g(x + 1) - g(x) = x2 + 5x + 8 - (x2 + 3x + 4)
= x2 + 5x + 8 - x2 + 3x + 4
= 2x + 4
= 2(x + 2)
Source: Myschool Past Questions Archive (Record #31403)
Mathematics
UTME / JAMB 1991
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize \(m^3 - 2m^2\) - m + 2
A
(m2 + 1)(m - 2)
B
(m - 1)(m + 1)(m + 2)
C
(m - 2)(m + 1)(m - 1)
D
(m2 + 2)(m - 1)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(m^3 - 2m^2\) - m + 2
Let f(m) = \(m^3 - 2m^2\) - m + 2
= f(1)
= 1 - 2 - 1 + 2 = 0
∴ m - 1 is factor \(\frac{m^3 - 2m^2 - m + 2}{m - 1}\)
= \(m^2\) - m - 2
= (\(m^2\) - 2m + m - 2)
= m(m - 2) + 1(m - 2)
= ( m + 1)( m - 2)
= (m - 1)(m + 1)(m - 2)
Source: Myschool Past Questions Archive (Record #31404)
Mathematics
UTME / JAMB 1991
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize 1 - (a - b)2
A
(1 - a - b)(1 - a + b)
B
(1 + a - b)(1 - a + b)
C
(1 - a + b)(1 - a + b)
D
(1 + a + b)(1 + a + b)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
1 - (a - b)2 = [1 + (a - b)][1 - a + b]
= (1 + a - b)(1 - a + b)
Source: Myschool Past Questions Archive (Record #31405)
Mathematics
UTME / JAMB 1991
Plane Geometry, Polygons & Circle Theorems
Solution
Which of the following is a factor of rs + tr - pt - ps?
A
(p - s)
B
(s - p)
C
r - p)
D
r + p)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
rs + tr - pt - ps = rs - ps - tr - pt
= (r - p)s + (r - p)t
= (r - p)(s + t),
hence r - p is a factor
Source: Myschool Past Questions Archive (Record #31406)
Mathematics
UTME / JAMB 1991
Linear, Simultaneous & Quadratic Equations
Solution
Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x\(^2\) + xy = 6.
A
-1 and 5
B
-5 and 1
C
1 and 5
D
1 and 1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(3x + y = 8 ... (i)\)
\(x^{2} + xy = 6 ... (ii)\)
From (i), \(y = 8 - 3x\)
From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\)
Equating the two values of y, we have
\(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\)
\(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\)
\(2x^{2} - 8x + 6 = 0\)
\(x^{2} - 4x + 3 = 0\)
\(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\)
\((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\)
\(y = 8 - 3x \)
When x = 1, \(y = 8 - 3(1) = 5\)
When x = 3, \(y = 8 - 3(3) = -1\)
\(\therefore \text{y = -1 or 5}\)
Source: Myschool Past Questions Archive (Record #31407)
Mathematics
UTME / JAMB 1991
Inequalities & Linear Programming
Solution
Find the range of values of x which satisfy the inequality \(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\) < 1
A
x < \(\frac{12}{13}\)
B
x < 13
C
x < 9
D
\(\frac{13}{12}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1
= \(\frac{6x + 4x + 3x < 12}{12}\)
i.e. 13 x < 12 = x < \(\frac{12}{13}\)
Source: Myschool Past Questions Archive (Record #31408)
Mathematics
UTME / JAMB 1991
Plane Geometry, Polygons & Circle Theorems
Solution
Find the positive number n, such that thrice its square is equal to twelve times the number
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
3n2 = 12n
= 3n2 - 12n = 0
= 3n(n - 4) = 0
∴ n = 4
Source: Myschool Past Questions Archive (Record #31409)
Mathematics
UTME / JAMB 1991
Sequences & Series (AP & GP)
Solution
What is the nth term of the progression 27, 9, 3,......?
A
27\(\frac{1}{3}\) n - 1
B
3n + 2
C
27 + 18(n - 1)
D
27 + 6(n - 1)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given 27, 9, 3,......this is a G.P
r = \(\frac{9}{27}\)
= \(\frac{1}{3}\)
T = arn - 1
= 27\(\frac{1}{3}\) n - 1
Source: Myschool Past Questions Archive (Record #31410)
Mathematics
UTME / JAMB 1991
Linear, Simultaneous & Quadratic Equations
Solution
Solve the equation (x - 2) (x - 3) = 12
A
2, 3
B
3, 6
C
-1, 6
D
1, -6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(x - 2) (x - 3) = 12
x\(^2\) - 3x - 2x + 6 = 12
x\(^2\) - 5x - 6 = 0
(x +1)(x - 6) = 0
x = -1 or 6
Source: Myschool Past Questions Archive (Record #31411)
Mathematics
UTME / JAMB 1991
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)
A
-2x - 2\(\sqrt{x (1 + x)}\)
B
1 + 2x + 2\(\sqrt{x (1 + x)}\)
C
\(\sqrt{x (1 + x)}\)
D
1 + 2x - 2\(\sqrt{x (1 + x)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)
= \((\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}) (\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}})\)
= \(\frac{(1 + x) + \sqrt{x(1 + x)} + \sqrt{x(1 + x)} + x}{(1 + x) - x}\)
= \(\frac{1 + 2x + 2\sqrt{x(1 + x)}}{1}\)
= \(1 + 2x + 2\sqrt{x(1 + x)}\)
Source: Myschool Past Questions Archive (Record #31412)
Mathematics
UTME / JAMB 1991
Linear, Simultaneous & Quadratic Equations
Solution
Evaluate x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\)
A
(x2 - 1)-\(\frac{1}{2}\)
B
(x2 - 1)1
C
(x2 - 1)
D
(x2 - 1)-1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\)
= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\)
= \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\)
= (x2 - 1)-\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #31413)
Mathematics
UTME / JAMB 1991
Algebraic Expressions, Polynomials & Factorisation
Solution
Find the gradient of the line passing through the points (-2, 0) and (0, -4)
A
2
B
-4
C
-2
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Given (-2, 0) and (0, -4)
Gradient = \(\frac{-4 - 0}{0 - (-2)}\)
= \(\frac{-4}{2}\)
= -2
Source: Myschool Past Questions Archive (Record #31414)