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.

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

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

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

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

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