WAEC Further Mathematics Past Questions & Answers - Page 9

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

x - 1 = 2
x = 3
f(2) = (3)$^3$ + 3(3)$^2$ + 4(3) - 5
f(2) = 27 + 27 + 12 - 5

= 61

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

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

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