JAMB Mathematics Past Questions & Answers - Page 169

(x² + x + 1)( x² - x + 1)

= x²(x² + x + 1) - x(x² + x + 1) + (x³ + x + 1)

= x⁴ - x³ + x² + x³ - x² - x + 1

= x⁴ + x² + 1

b = a + cp....(i)

r = ab + $\frac{1}{2}$cp².....(ii)

expressing b² in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i)

b - a = cp = $\frac{b - a}{c}$

sub. for p in eqn.(ii)

r = ab + $\frac{1}{2}$c$\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}$

2cr = 2ab + b² - 2ab + a²

b² = 2cr - a²

T = $\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}$ = $\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}$

= $\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}$

= $\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}$

= $\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}$

T = $\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}$

1st term a = 3, 5th term = 9, sum of n = 81

nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9

3 + 4d = 9

4d = 9 - 3

d = $\frac{6}{4}$

= $\frac{3}{2}$

= 6

S_n = $\frac{n}{2}$(6 + $\frac{3}{4}$n - $\frac{3}{2}$)

81 = $\frac{12n + 3n^2}{4}$ - 3n

= $\frac{3n^2 + 9n}{4}$

3n² + 9n = 324

3n² + 9n - 324 = 0

By almighty formula positive no. n = 9
= 3

$\frac{\sin^{2} x}{1 + \cos x} + \frac{\sin^{2} x}{1 - \cos x}$

$\frac{\sin^{2} x (1 - \cos x) + \sin^{2} x (1 + \cos x)}{1 - \cos^{2} x}$

= $\frac{\sin^{2} x - \cos x \sin^{2} x + \sin^{2} x + \sin^{2} x \cos x}{\sin^{2} x}$

(Note: $\sin^{2} x + \cos^{2} x = 1$).

= $\frac{2 \sin^{2} x}{\sin^{2} x}$

= 2.