JAMB Mathematics Past Questions & Answers - Page 239

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

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

\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1

r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\)

4x² + 5x + Q

To make a complete square, the coefficient of x² must be 1

= x² + \(\frac{5x}{4}\) + \(\frac{Q}{4}\)

Then (half the coefficient of x²) should be added

i.e. x² + \(\frac{5x}{4}\) + \(\frac{25}{64}\)

∴ \(\frac{Q}{4}\) = \(\frac{25}{64}\)

Q = \(\frac{4 \times 25}{64}\)

= \(\frac{25}{16}\)

\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)

Let \(x^{-1}\) = a and \(y^{-1}\)= b

2a - 3b = 4 .......(i)

4a + b = 1 .........(ii)

(i) x 3 = 12a + 3b = 3........(iii)

2a - 3b = 4 ...........(i)

(i) + (iii) = 14a = 7

∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\)

From (i), 3b = 2a - 4

3b = 1 - 4

3b = -3

∴ b = -1

From substituting, \(2^{-1} = x^{-1}\)

∴ x = 2

\(y^{-1} = -1, y = -1\)

\(\frac{1}{2}\)(3y - 4x)² = (8x² + kxy + Ly²)

\(\frac{1}{2}\)(9y² - 24xy + 16x²) = 8x² + kxy + Ly²

\(\frac{9}{2}\)y² - 12xy) = kxy + Ly²

k = -12 ∴ L = \(\frac{9}{2}\)