2006 - JAMB Mathematics Past Questions and Answers - page 6
51
The binary operation \(\oplus\) defined on the set of real numbers is such that x \(\oplus\) y = \(\frac{xy}{6}\) for all x, y \(\epsilon\) R. Find the inverse of 20 under this operation when the identity element is 6
A
\(\frac{9}{5}\)
B
\(\frac{1}{20}\)
C
\(\frac{10}{3}\)
D
\(\frac{1}{12}\)
correct option: b
\(\frac{\frac{1}{20} \times 6}{6}\) = \(\frac{6}{20}\) x \(\frac{1}{6}\) = \(\frac{1}{20}\)
Users' Answers & Comments52
If p varies inversely as the cube of q and q varies directly as the square of r, what is the relationship between P and r?
A
p varies directly as r3
B
p varies directly as r6
C
p varies as 6\(\sqrt{r}\)
D
p varies inversely as r6
correct option: d
p \(\alpha\) \(\frac{1}{q^3}\), q \(\alpha\) r2, p \(\alpha\) (r2)3, p \(\alpha\) \(\frac{1}{r^6}\)
Users' Answers & Comments53
a binary operation \(\oplus\) o the set of rational numbers is defined as x \(\oplus\) y = \(\frac{x^2 - y^2}{2xy}\). Find -5 \(\oplus\) 3
A
\(\frac{8}{15}\)
B
-\(\frac{17}{15}\)
C
\(\frac{17}{15}\)
D
\(\frac{8}{-15}\)
correct option: d
\(\frac{(-5)^2 - (3^2)}{2(-5)(+3)}\)
= \(\frac{25 - 9}{-30}\)
= \(\frac{16}{-30}\)
= \(\frac{8}{-15}\)
Users' Answers & Comments= \(\frac{25 - 9}{-30}\)
= \(\frac{16}{-30}\)
= \(\frac{8}{-15}\)
54
If T = 2\(\pi\) \(\sqrt{\frac{L}{g}}\), make g the subject of the formula
A
\(\frac{4\pi L^2}{T}\)
B
\(\frac{4\pi^2 L^2}{T}\)
C
\(\frac{4\pi ^2 L}{T}\)
D
\(\frac{2\pi \sqrt{T}}{T}\)
correct option: c
T = 2\(\pi\) \(\sqrt{\frac{L}{g}}\)
T2 = 4\(\pi^2 \frac{L}{g}\)
g = \(\frac{4\pi ^2 L}{T}\)
Users' Answers & CommentsT2 = 4\(\pi^2 \frac{L}{g}\)
g = \(\frac{4\pi ^2 L}{T}\)
55
The sum of the first n positive integers is
A
\(\frac{1}{2}\)(n + 1)
B
n \(\frac{1}{2}\)(n + 1)
C
n(n - 1)
D
n(n + 1)
56
If x = \(\begin{vmatrix} 1 & 0 & 1 \ 2 & -1 & 0 \ -1 & 0 & 1\end{vmatrix}\) and y = \(\begin{vmatrix} -1 & 1 & 2 \ 0 & -1 & -1 \ 2 & 0 & 1\end{vmatrix}\)
find 2x - y
find 2x - y
A
\(\begin{vmatrix} 3 & -1 & 0 \ 4 & -3 & -1 \ -4 & 1 & 1\end{vmatrix}\)
B
\(\begin{vmatrix} 3 & -1 & 0 \ 4 & -1 & 1 \ -4 & 1 & 1\end{vmatrix}\)
C
\(\begin{vmatrix} 3 & -1 & 0 \ 4 & -3 & 1 \ -4 & 1 & 1\end{vmatrix}\)
D
\(\begin{vmatrix} 3 & -1 & 0 \ 4 & 1 & 1 \ -4 & -1 & 1\end{vmatrix}\)
correct option: b
2(10 - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4
2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1
2(10 - (20) = 0 2(0) - (-1) = 1 2(1) - (1) = 1
\(\begin{vmatrix} 3 & -1 & 0 \ 4 & -1 & 1 \ -4 & 1 & 1\end{vmatrix}\)
Users' Answers & Comments2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1
2(10 - (20) = 0 2(0) - (-1) = 1 2(1) - (1) = 1
\(\begin{vmatrix} 3 & -1 & 0 \ 4 & -1 & 1 \ -4 & 1 & 1\end{vmatrix}\)
57
Find the value of k if the expression kx3 + x2 - 5x - 2 leaves a remainder 2 when it is is divided by 2X = 1
A
10
B
8
C
-8
D
-10
correct option: d
Users' Answers & Comments58
How many terms of the series 3, -6, 12, 24,... are needed to make a total of 1 - 28?
A
9
B
8
C
12
D
10
correct option: b
Sn = \(\frac{a(1 - (r)^n)}{1 - r}\)
255 = \(\frac{3(1 - (-2))^n}{1 - (2)}\)
255 = \(\frac{3(1-(-2)^n}{3}\)\)
2n = 256
2n = 28
n = 8
Users' Answers & Comments255 = \(\frac{3(1 - (-2))^n}{1 - (2)}\)
255 = \(\frac{3(1-(-2)^n}{3}\)\)
2n = 256
2n = 28
n = 8
59
If y = x2 - x - 12, find the range of values of x for which y \(\geq\) 0,
A
-3 < x \(\leq\) 4
B
x , -3 or x > 4
C
-3 \(\leq\) x \(\leq\) 4
D
x \(\leq\) -3 or x \(\geq\) 4
correct option: c
y = x2 - x - 12, let y = 0
x2 - x - 12 = 0, (x + 3)(x - 4) = 0
x = -3 or 4
ranges are -3 and 4
Users' Answers & Commentsx2 - x - 12 = 0, (x + 3)(x - 4) = 0
x = -3 or 4
ranges are -3 and 4
60
Find the roots of x3 - 2x2 - 5x + 6 = 0
A
1, -2, 3
B
1, 2, -3,
C
-1, -2, 3
D
-1, 2, -3
correct option: a
Users' Answers & Comments