Year : 
1999
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

11 - 20 of 45 Questions

# Question Ans
11.

The sum of two numbers is twice their difference. If the difference of the numbers is P, find the larger of the two numbers

A. p/2

B. 3p/2

C. 5p/2

D. 3p

Detailed Solution

Let the numbers be x and y
x+y = 2p.....(i)
x-y = p......(ii)
2x = 3p
x = 3p/2
12.

A binary operation * is defined by a*b = ab+a+b for any real number a and b. if the identity element is zero, find the inverse of 2 under this operation.

A. 2/3

B. 1/2

C. -1/2

D. -2/3

Detailed Solution

\(a*a^{-1} = aa^{-1} + a + a^{-1} = e\)

if e = 0

\(2.2^{-1} + 2 + 2^{-1} = 0\)
collecting like terms, we have:
\(3.2^{-1} + 2 = 0\)

= \(2^{-1}\) = -\(\frac{2}{3}\)
13.

Factorize completely \(x^{2} + 2xy + y^{2} + 3x + 3y - 18\).

A. (x+y+6)(x+y-3)

B. (x-y-6)(x-y+3)

C. (x-y+6)(x-y-3)

D. (x+y-6)(x+y+3)

A
14.

Three consecutive positive integers k, l and m are such that l\(^2\) = 3(k+m). Find the value of m.

A. 4

B. 5

C. 6

D. 7

Detailed Solution

l\(^2\) = 3 (k + m)
Since they are consecutive positive numbers, we have
l = k+1, m = k+2.
\(\to\) (k+1)\(^2\) = 3(k + k + 2)
k\(^2\) + 2k + 1 = 3(2k + 2)
k\(^2\) + 2k + 1 = 6k + 6
k\(^2\) + 2k - 6k + 1 - 6 = 0
k\(^2\) - 4k - 5 = 0
k\(^2\) - 5k + k - 5 = 0
k(k - 5) + 1(k - 5) = 0
k = -1 or 5
Since k, l and m are positive, then k = 5.
m = k + 2 = 5 + 2
= 7.
15.

Express \(\frac{1}{x^{3}-1}\) in partial fractions

A. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)

B. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{x - 2}{x^{2} + x + 1})\)

C. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 2)}{x^{2} + x + 1})\)

D. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 1)}{x^{2} - x - 1})\)

Detailed Solution

\(\frac{1}{x^{3} - 1}\)
\(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\)
\(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\)
\(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\)
Comparing the two sides of the equation,
\(A + B = 0 ... (1)\)
\(A - B + C = 0 ... (2)\)
\(A - C = 1 ... (3)\)
From (3), \(C = A - 1\), putting that in (2),
\(A - B = -C \implies A - B = 1 - A\)
\(2A - B = 1 ... (4)\)
(1) + (4): \(3A = 1 \implies A = \frac{1}{3}\)
\(A = -B \implies B = -\frac{1}{3}\)
\(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\)
\(\therefore \frac{1}{x^{3} - 1} = \frac{1}{3(x -
16.

The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the G.P if its sum to infinity is 8.

A. 8/5

B. 8/3

C. 72/25

D. 56/9

Detailed Solution

Le the common ratio be r so that the first term is 2r.

Sum, s = a/(1-r)
ie. 8 = 2r/(1-r)

8(1-r) = 2r,
8 - 8r = 2r
8 = 2r + 8r
8 = 10r
r = 4/5.

where common ratio (r) = \(frac{second term(n_2)}{first term(a)}\),
r = \(frac{n_2}{2r}\)
r = 4/5 and a = 2r or 8/5
4/5 \(times\) 8\5 = n\(_2\)
\(\frac{32}{25}\) = n\(_2\)
The sum of the first two terms = a + n\(_2\)
= \(\frac{8}{5}\) + \(\frac{32}{25}\)
= \(\frac{40 + 32}{25}\)
= \(\frac{72}{25}\)
17.

Divide 4x\(^3\) - 3x + 1 by 2x - 1

A. 2x2-x+1

B. 2x2-x-1

C. 2x2+x+1

D. 2x2+x-1

D
18.

Find a positive value of \(\alpha\) if the coordinate of the centre of a circle x\(^2\) + y\(^2\) - 2\(\alpha\)x + 4y - \(\alpha\) = 0 is (\(\alpha\), -2) and the radius is 4 units.

A. 1

B. 2

C. 3

D. 4

Detailed Solution

\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\)
r = 4 units; centre (\(\alpha\), -2).
\((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\)
\((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\)
\(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\)
\(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\)
\(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\)
\(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\)
\(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\)
\((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)
19.

A man 1.7m tall observes a bird on top of a tree at an angle of 30°. if the distance between the man's head and the bird is 25m, what is the height of the tree?

A. 26.7m

B. 14.2m

C. \(1.7+(25\frac{\sqrt{3}}{3}m\)

D. \(1.7+(25\frac{\sqrt{2}}{2}m\)

Detailed Solution

Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30.
x = 12.5m
The height of the tree = 12.5 + 1.7
= 14.2 m
20.

In ∆MNO, MN = 6 units, MO = 4 units and NO = 12 units. If the bisector of and M meets NO at P, calculate NP.

A. 4.8 units

B. 7.2 units

C. 8.0 units

D. 18.0 units

Detailed Solution

Hint: start with making a sketch of the figure described.
11.

The sum of two numbers is twice their difference. If the difference of the numbers is P, find the larger of the two numbers

A. p/2

B. 3p/2

C. 5p/2

D. 3p

Detailed Solution

Let the numbers be x and y
x+y = 2p.....(i)
x-y = p......(ii)
2x = 3p
x = 3p/2
12.

A binary operation * is defined by a*b = ab+a+b for any real number a and b. if the identity element is zero, find the inverse of 2 under this operation.

A. 2/3

B. 1/2

C. -1/2

D. -2/3

Detailed Solution

\(a*a^{-1} = aa^{-1} + a + a^{-1} = e\)

if e = 0

\(2.2^{-1} + 2 + 2^{-1} = 0\)
collecting like terms, we have:
\(3.2^{-1} + 2 = 0\)

= \(2^{-1}\) = -\(\frac{2}{3}\)
13.

Factorize completely \(x^{2} + 2xy + y^{2} + 3x + 3y - 18\).

A. (x+y+6)(x+y-3)

B. (x-y-6)(x-y+3)

C. (x-y+6)(x-y-3)

D. (x+y-6)(x+y+3)

A
14.

Three consecutive positive integers k, l and m are such that l\(^2\) = 3(k+m). Find the value of m.

A. 4

B. 5

C. 6

D. 7

Detailed Solution

l\(^2\) = 3 (k + m)
Since they are consecutive positive numbers, we have
l = k+1, m = k+2.
\(\to\) (k+1)\(^2\) = 3(k + k + 2)
k\(^2\) + 2k + 1 = 3(2k + 2)
k\(^2\) + 2k + 1 = 6k + 6
k\(^2\) + 2k - 6k + 1 - 6 = 0
k\(^2\) - 4k - 5 = 0
k\(^2\) - 5k + k - 5 = 0
k(k - 5) + 1(k - 5) = 0
k = -1 or 5
Since k, l and m are positive, then k = 5.
m = k + 2 = 5 + 2
= 7.
15.

Express \(\frac{1}{x^{3}-1}\) in partial fractions

A. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)

B. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{x - 2}{x^{2} + x + 1})\)

C. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 2)}{x^{2} + x + 1})\)

D. \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 1)}{x^{2} - x - 1})\)

Detailed Solution

\(\frac{1}{x^{3} - 1}\)
\(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\)
\(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\)
\(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\)
Comparing the two sides of the equation,
\(A + B = 0 ... (1)\)
\(A - B + C = 0 ... (2)\)
\(A - C = 1 ... (3)\)
From (3), \(C = A - 1\), putting that in (2),
\(A - B = -C \implies A - B = 1 - A\)
\(2A - B = 1 ... (4)\)
(1) + (4): \(3A = 1 \implies A = \frac{1}{3}\)
\(A = -B \implies B = -\frac{1}{3}\)
\(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\)
\(\therefore \frac{1}{x^{3} - 1} = \frac{1}{3(x -
16.

The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the G.P if its sum to infinity is 8.

A. 8/5

B. 8/3

C. 72/25

D. 56/9

Detailed Solution

Le the common ratio be r so that the first term is 2r.

Sum, s = a/(1-r)
ie. 8 = 2r/(1-r)

8(1-r) = 2r,
8 - 8r = 2r
8 = 2r + 8r
8 = 10r
r = 4/5.

where common ratio (r) = \(frac{second term(n_2)}{first term(a)}\),
r = \(frac{n_2}{2r}\)
r = 4/5 and a = 2r or 8/5
4/5 \(times\) 8\5 = n\(_2\)
\(\frac{32}{25}\) = n\(_2\)
The sum of the first two terms = a + n\(_2\)
= \(\frac{8}{5}\) + \(\frac{32}{25}\)
= \(\frac{40 + 32}{25}\)
= \(\frac{72}{25}\)
17.

Divide 4x\(^3\) - 3x + 1 by 2x - 1

A. 2x2-x+1

B. 2x2-x-1

C. 2x2+x+1

D. 2x2+x-1

D
18.

Find a positive value of \(\alpha\) if the coordinate of the centre of a circle x\(^2\) + y\(^2\) - 2\(\alpha\)x + 4y - \(\alpha\) = 0 is (\(\alpha\), -2) and the radius is 4 units.

A. 1

B. 2

C. 3

D. 4

Detailed Solution

\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\)
r = 4 units; centre (\(\alpha\), -2).
\((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\)
\((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\)
\(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\)
\(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\)
\(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\)
\(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\)
\(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\)
\((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)
19.

A man 1.7m tall observes a bird on top of a tree at an angle of 30°. if the distance between the man's head and the bird is 25m, what is the height of the tree?

A. 26.7m

B. 14.2m

C. \(1.7+(25\frac{\sqrt{3}}{3}m\)

D. \(1.7+(25\frac{\sqrt{2}}{2}m\)

Detailed Solution

Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30.
x = 12.5m
The height of the tree = 12.5 + 1.7
= 14.2 m
20.

In ∆MNO, MN = 6 units, MO = 4 units and NO = 12 units. If the bisector of and M meets NO at P, calculate NP.

A. 4.8 units

B. 7.2 units

C. 8.0 units

D. 18.0 units

Detailed Solution

Hint: start with making a sketch of the figure described.