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