JAMB Mathematics Past Questions & Answers - Page 276

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

Multiply both sides by LCM i.e. (1 - x(x + 2))

∴ x - 3 = p(x + 2) + Q(1 - x)

When x = +1

(+1) - 3 = p(+1 + 2) + Q(1 - 1)

-2 = 3p + 0(Q)

3p = -2

∴ p = $\frac{-2}{3}$

$\frac{1}{r - 1}$ + $\frac{2}{r + 1}$ = $\frac{3}{r}$

Multiply through by r(r -1) which is the LCM

= (r)(r + 1) + 2(r)(r - 1)

= 3(r - 1)(r + 1)

= r² + r + 2r² - 2r

3r² - 3 = 3r²

r = 3r² - 3

-r = -3

∴ r = 3

$\frac{a}{a - x}$ = $\frac{b}{x - b}$

$\frac{1}{1 - x}$ = $\frac{3}{x - 3}$

∴ 3(1 - x) = x - 3

3 - 3x = x - 3

Rearrange 6 = 4x; x = $\frac {6}{4}$

= $\frac{3}{2}$

Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero

$\therefore (x - 1) = 0$

x = 1

Substitute in the polynomial the value x = 1

= $p(1)^3 + q(1)^2 + 11(1) - 6 = 0$

p + q + 5 = 0 .....(i)

Also since x - 3 is a factor, $\therefore$ x - 3 = 0

x = 3

Substitute $p(3)^3 + q(3)^2 + 11(3) - 6 = 0$

27p + 9q = -27 ......(2)

Combine eqns. (i) and (ii)

Multiply equation (i) by 9 to eliminate q

9p + 9q = -45

Subtract (ii) from (i), $18p = 18$

$\therefore$ p = 1

Put p = 1 in (i), 

$1 + q = -5 \implies q = -6$

$(p, q) = (1, -6)$

a2x - b2y - b2x + a2y = a2x - b2x - b2y + a2y Rearrange

= x(a2 - b2) + y(a2 - b2)

= (x + y)(a2 - b2)

= (x + y)(a + b)(a - b)