JAMB Mathematics Past Questions & Answers - Page 257

$\frac{dm}{dt}$ = 4 - 0.6t

$\int$dm = $\int$(4 - 0.6t)dt

m = $4t - 0.3t^2 + c$, when t = 0, m = 2g

∴ c = 2

m = $4t - 0.3t^2 + 2$, when t = 5 minutes

m = $4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2$

= 14.5

$y = x \sin x$

$\frac{\mathrm d y}{\mathrm d x} = x \cos x + \sin x$

$\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = x (- \sin x) + \cos x + \cos x$

= $2 \cos x - x \sin x$

$\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}$

$\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}$

= $\frac{(x^{2} + 3x - 2)}{x + 2}$

$\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}$

= $\frac{2^{2} + 3(2) - 2}{2 + 2}$

= $\frac{4 + 6 - 2}{4} = 2$

sin $\theta$ = $\frac{m}{n}$ 

Opp = m; Hyp = n

Adj = $\sqrt{n^{2} - m^{2}}$

$\cos \theta = \frac{\sqrt{n^{2} - m^{2}}}{n}$

= $\frac{\sqrt{(n + m)(n - m)}}{n}$

Equation (5, 7) parallel to the line 7x + 5y = 12

5Y = -7x + 12

y = $\frac{-7x}{5}$ + $\frac{12}{5}$

Gradient = $\frac{-7}{5}$

∴ Required equation = $\frac{y - 7}{x - 5}$ = $\frac{-7}{5}$ i.e. 5y - 35 = -7x + 35

5y + 7x = 70