Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.
4√3
2√6
3√2
2√3
Correct answer is D
d = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
x1 = x, x2 = -x, y1 = 4, y2 = 3, d = 7
7 = \(\sqrt{(-x -x)^2 + (3 - 4)^2}\)
7 = \(\sqrt{(-2x)^2 + (-1)^2}\)
7 = ( \(\sqrt{4x^2 + 1}\)
square both sides
7\(^2\) = 4x\(^2\) + 1
collect like terms
4x\(^2\) = 49 - 1
4x\(^2\) = 48
x\(^2\) = \(\frac{48}{4}\)
x\(^2\) = 12
x = √12
x = 2√3
If f(x-1) = x\(^3\) + 3x\(^2\) + 4x - 5, find f(2)
61
25
20
13
Correct answer is A
x - 1 = 2
x = 3
f(2) = (3)\(^3\) + 3(3)\(^2\) + 4(3) - 5
f(2) = 27 + 27 + 12 - 5
= 61
17.55i + 13.78j
17.55j - 13.78i
-17.55i + 13.78j
-17.55i - 13.78j
Correct answer is B
Converting the forces to their rectangular forms
F = (10N, 060º)
Fx = 10cos60 = 5i
fy = 10sin60 = 8.66j
F = 5i + 8.66j
P = (15N, 120º)
Px = 15cos120 = -7.5i
py = 15sin120 = 12.99j
P = -7.5i + 12.99j
Q = (12N, 200º)
Qx = 12cos200 = -11.28i
Qy = 12sin200 = -4.1j
Q = -11.28i -4.1j
The resultant force = F + P + Q
R = 5i + 8.66j + (-7.5i + 12.99j) + (-11.28i -4.1j)
R = -13.78i + 17.55j
If α and β are roots of x\(^2\) + mx - n = 0, where m and n are constants, form the
equation | whose | roots | are | 1
α |
and | 1
β |
. |
mnx\(^2\) - n\(^2\) x - m = 0
mx\(^2\) - nx + 1 = 0
nx\(^2\) - mx + 1 = 0
nx\(^2\) - mx - 1 = 0
Correct answer is D
x\(^2\) + mx - n = 0
a = 1, b = m, c = -n
α + β = \(\frac{-b}{a}\) = \(\frac{-m}{1}\) = -m
αβ = \(\frac{c}{a}\) = \(\frac{-n}{1}\) = -n
the roots are = \(\frac{1}{α}\) and \(\frac{1}{β}\)
sum of the roots = \(\frac{1}{α}\) + \(\frac{1}{β}\)
\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{α+β}{αβ}\)
α + β = -m
αβ = -n
\(\frac{α+β}{αβ}\) = \(\frac{-m}{-n}\) → \(\frac{m}{n}\)
product of the roots = \(\frac{1}{α}\) * \(\frac{1}{β}\)
\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{1}{αβ}\) → \(\frac{1}{-n}\)
x\(^2\) - (sum of roots)x + (product of roots)
x\(^2\) - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx\(^2\) - mx - 1 = 0
0N
2N
3(2t + 3)N
6N
Correct answer is D
F = m * a
d = t\(^2\) + 3t.
a = \(\frac{d^2d}{dt^2}\)
\(\frac{d[d]}{dt}\) = 2t + 3
\(\frac{d^2d}{dt^2}\) = 2m/s\(^2\)
a = 2m/s\(^2\)
F = m * a
F = 3 × 2 = 6N