JAMB Mathematics Past Questions & Answers - Page 266

x + 1 is a factor of x³ + 3x² + kx + 4

Let f(x) = x³ + 3x² + kx + 4

∴ f(-1) = (-1)³ + 3(-1)² + k(-1) + 4 = 0

-1 + 3 - k + 4 = 0

∴ k = 6

Let the amounts invested at 4% and 5% respectively be x and y.

\(\therefore x + y = 280 ... (i)\)

Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\)

Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\)

\(\therefore 0.04x + 0.05y = 12.80\)

\(\implies 4x + 5y = 1280 ... (ii)\)

From (i), \(x = 280 - y\).

Put into (ii), \(4(280 - y) + 5y = 1280\)

\(1120 - 4y + 5y = 1280\)

\(1120 + y = 1280 \implies y = 1280 - 1120 = N160\)

\(\therefore\) N160 was invested at the rate of 5% per annum.

Given S = ut + \(\frac{1}{2} at^2\)

S = ut + \(\frac{1}{2} at^2\)

∴ 2S = 2ut + at²

= at² + 2ut - 2s = 0

t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\)

= -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\)


= \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))

\(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\)

Squaring both sides,

\(y^{2} + 48y + 576 = 121y\)

\(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\)

\(y^{2} - 64y - 9y + 576 = 0\)

\(y(y - 64) - 9(y - 64) = 0\)

\((y - 9)(y - 64) = 0\)

\(\therefore \text{y = 64 or y = 9}\)

\(9p^{2} - q^{2} + 6qr - 9r^{2}\)

= \(9p^{2} - (q^{2} - 6qr + 9r^{2})\)

= \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\)

= \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\)

= \(9p^{2} - (q - 3r)^{2}\)

= \((3p + (q - 3r))(3p - (q - 3r))\)

= \((3p + q - 3r)(3p - q + 3r)\)