2008 - JAMB Mathematics Past Questions and Answers - page 8

3x + 2y + 1 = 0

y = mx + c

2y = -3x - 1

y = -(\frac{3}{2}) x -(\frac{1}{2})

m = -(\frac{3}{2})

L(\begin{pmatrix} x_1 & y_1 \ -1 & -6 \end{pmatrix}) m L(\begin{pmatrix} x_2 & y_2 \ -3 & -5 \end{pmatrix})

D = (\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2})

D = (\sqrt{(-3 - (-1)^2 + (-5 -(-6)^2})

D = (\sqrt{(-3 + 1)^2 + (-5 + 6)^2})

D = (\sqrt{(-2)^2 + 1^2})

D = (\sqrt{4 + 1})

D = (\sqrt{5})

sin (\theta) = (\frac{3}{5}), find tan (\theta)

sin (\theta) = (\frac{opp}{hyp}) = (\frac{3}{5})

5² = x² + 3²

25 = x² + 9

x² = 16

x = (\sqrt{16})

x = 4

tan = (\frac{3}{4})

Tan 20^o = (\frac{68m}{x})

x tan 20^o = 68

x = (\frac{68}{tan 20}) = (\frac{68}{0.364})

x = 186.8

= 187m

y = (\frac{x^7 - x^5}{x^4}) = (\frac{x^7 - x^5}{x^4})

Y = X³ - X

(\frac{dy}{dx}) = 3x^{3 - 1} - x^{1 - 1}

= 3x² - x^o

(\frac{dy}{dx}) = 3x² - 1

y = sin x - x cos x

let; U = -x

(\frac{dy}{dx}) = -1

V = cos x

(\frac{dx}{dx}) = -5x

-x (-sin x) + cos x (-1)

x sin x - cos x

(\frac{dy}{dx}) = x sin x + cos x - cos x

(\frac{dy}{dx}) = x sin x

y = x(1 + x)

y = x + x²

(\frac{dy}{dx}) = 1 + 2x

at minimum (\frac{dy}{dx}) = 0

therefore, 1 + 2x = 0 (\to) 2x = -1

x = -(\frac{1}{2})

(\int ^{2}_{1})(6x² - 2x)dx

= [(\frac{6x^3}{3} - \frac{2x^2}{2})]²

= [2x³ - x²]²

= (2(2)³ - 4 - 2 + 1

= 16 - 4 - 2 + 1

= 17 - 6

= 11

(\int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}})cosx

([sin x]^{\frac{\pi}{2}}_{\frac{\pi}{2}})

= sin(\frac{\pi}{2})- (-sin (\frac{\pi}{2}))

= sin (\frac{\pi}{2}) + sin (\frac{\pi}{2})

= 2 sin (\frac{\pi}{2})

= 2 sin 1.5714

= 2(0.2704)

= 0.5

= 1