JAMB Mathematics Past Questions & Answers - Page 424

Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30.

x = 12.5m

The height of the tree = 12.5 + 1.7

= 14.2 m

\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\)

r = 4 units; centre (\(\alpha\), -2).

\((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\)

\((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\)

\(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\)

\(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\)

\((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)

\(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\)

\(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\)

\(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\)

No explanation has been provided for this answer.

Let the common ratio be r so that the first term is 2r.

Sum, s = \(\frac{a}{1-r}\)

ie. 8 = \(\frac{2r}{1-r}\)

8(1-r) = 2r,

8 - 8r = 2r

8 = 2r + 8r

8 = 10r

r = \(\frac{4}{5}\).

where common ratio (r) = \(\frac{second term(n_2)}{first term(a)}\),

r = \(\frac{n_2}{2r}\)

r = \(\frac{4}{5}\) and a = 2r or \(\frac{8}{5}\)

\(\frac{4}{5}\) * \(\frac{8}{5}\) = n\(_2\)

\(\frac{32}{25}\) = n\(_2\)

The sum of the first two terms = a + n\(_2\)

= \(\frac{8}{5}\) + \(\frac{32}{25}\)

= \(\frac{40 + 32}{25}\)

= \(\frac{72}{25}\)

\(\frac{1}{x^{3} - 1}\)

\(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\)

\(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\)

\(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\)

Comparing the two sides of the equation,

\(A + B = 0 ... (1)\)

\(A - B + C = 0 ... (2)\)

\(A - C = 1 ... (3)\)

From (3), \(C = A - 1\), putting that in (2),

\(A - B = -C \implies A - B = 1 - A\)

\(2A - B = 1 ... (4)\)

(1) + (4): \(3A = 1 \implies A = \frac{1}{3}\)

\(A = -B \implies B = -\frac{1}{3}\)\(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\)

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