WAEC Past Questions and Answers - Page 1013

y = \(3x^2 + 2\)

\(y^1 = \frac{dy}{dx} = 6x\)

Evaluating this derivative at x = 1 (since the point of interest is (1, 5)) gives us the slope of the tangent line at that point:
\(^mtangent = y^1(1) = 6 (1) = 6\)
Slope of the Normal Line \( ^mnorma l= - \frac{1}{^mtangent}\)

\(^mnormal = - \frac{1}{6}\)

\(y−y_1​= ^mnormal⋅(x−x_1)\)

=y-5=-\(\frac{1}{6}(x-1)\)

=y-5=-\(\frac{1}{6}x+\frac{1}{6}\)

=y=-\(\frac{1}{6}x+\frac{1}{6}+5\)

Multiply through by 6

=6y=-x+1+30

∴6y+ x - 31=0

Using the dot product:

a.b = |a||b|cos θ

a.b = 5(-2) + 12(3) = -10 + 36 = 26

|a| = √(52 + 122) = √(25 + 144)

|a| = √169 = 13

|b| = \(√((-2)^2 + 3^2)\) = √(4 + 9)

|b| = √13

= 26 = 13√13 x cos θ

= \(\frac{26}{13\sqrt13}\) = cos θ

=\(\frac{2}{\sqrt13}\) = cos θ

= θ = \(cos^{-1} (\frac{2}{\sqrt13})\)

∴ θ = 56.3º

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