WAEC Further Mathematics Past Questions & Answers - Page 8

\(sin x = \frac{4}{5}\) and \(cos y = \frac{12}{13}\)

x is obtuse i.e sin x = + ve while cos x = + ve

\(cos x=\frac{3}{5}==>cos x=-\frac {3}{5}(obtuse)\)

\(sin y= \frac{5}{13}\)

\(sin (x-y) = sin x\) \(cos y - cos x\) \(sin y\)

\(sin(x-y) = \frac{4}{5}\times\frac{12}{13}-(-\frac{3}{5})\times\frac{5}{13}\)

\(sin(x-y) = \frac{48}{65}-(-\frac{3}{13})\)

\(\therefore sin (x-y) = \frac{48}{65} + \frac{3}{13} = \frac{63}{65}\)

\(\int^1_0 x(x^2-2)^2 dx\)

\((x^2-2)^2=x^4-2x^2-2x^2+4\)

=\(x^4-4x^2+4\)

\(x(x^2-2)^2=x(x^4-4x^2+4)\)

=\(x^5-4x^3+4x\)

\(\int^1_0 x(x^2-2)^2 dx = \int^1_0 x^5 - 4x^3 + 4x dx\)

=\((\frac{x^6}{6} - x^4+2x^2)^1_0\)

= \((\frac{(1)^6}{6} - (1)^4 +2(1)^2)-(\frac{(0)^6}{6} - (0)^4+2(0)^2)\)

=\(\frac{7}{6} - 0 =\frac{7}{6}\)

\(\therefore 1\frac{1}{6}\)

\(S = 5t^3 - \frac{19}{2} t^2 + 6t - 4\)

\(v(t)=\frac{dS}{dt}=15t^2-19t+6\)

\(a(t)=\frac{dv}{dt}=30t-19\)

∴a(2)=30(2)-19=60-19=41 \(ms ^{-2}\)

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