Further Mathematics questions and answers

Further Mathematics Questions and Answers

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.

41.

The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.

A.

4√3

B.

2√6

C.

3√2

D.

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

42.

If f(x-1) = x\(^3\) + 3x\(^2\) + 4x - 5, find f(2)

A.

61

B.

25

C.

20

D.

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

43.

A particle is acted upon by forces F = (10N, 060º), P = (15N, 120º) and Q = (12N, 200º). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.

A.

17.55i + 13.78j

B.

17.55j - 13.78i

C.

-17.55i + 13.78j

D.

-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

44.

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
.

A.

mnx\(^2\) - n\(^2\) x - m = 0

B.

mx\(^2\) - nx + 1 = 0

C.

nx\(^2\) - mx + 1 = 0

D.

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

45.

A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t\(^2\) + 3t. Find the magnitude of F at time t.

A.

0N

B.

2N

C.

3(2t + 3)N

D.

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