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.