JAMB Mathematics Past Questions & Answers - Page 202

cos$\theta$ = $\frac{13^2 + 9^2 - 11^2}{2(13)(9)}$

= $\frac{169 + 81 - 21}{26 \times 9}$

cos$\theta$ = $\frac{129}{26 \times 9}$

= $\frac{43}{78}$

f(x - 2) = 4x² + x + 7

x - 2 = 1, x = 3

f(x - 2) = f(1)

= 4(3)² + 3 + 7

= 36 + 10

= 46

2x - 3y = -10; 10x - 6y = -5

2x - 3y = -10 x 2

10x - 6y = -5

4x - 6y = -20 .......(i)

10x - 6y = -5.......(ii)

eqn(ii) - eqn(1)

6x = 15

x = $\frac{15}{6}$

= $\frac{5}{2}$

x = 2$\frac{1}{2}$

Sub. for x in equ.(ii) 10($\frac{5}{2}$) - 6y = -5

6y = 25 + 5 → 30

y = $\frac{30}{6}$

y = 5

$\frac{2}{2r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$

$\frac{2}{2r - 1}$ - $\frac{1}{r + 2}$ = $\frac{5}{3}$

$\frac{2r + 4 - 2r + 1}{2r - 1 (r + 2)}$ = $\frac{5}{3}$

$\frac{5}{(2r + 1)(r + 2)}$ = $\frac{5}{3}$

5(2r - 1)(r + 2) = 15

(10r - 5)(r + 2) = 15

10r2 + 20r - 5r - 10 = 15

10r2 + 15r = 25

10r2 + 15r - 25 = 0

2r2 + 3r - 5 = 0

(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)

(r - 1)(2r + 5) = 0

r = 1 or $\frac{-5}{2}$

2x + y = 4......(i)

x^2 + xy = -12........(ii)

from eqn (i), y = 4 - 2x

= x2 + x(4 - 2x)

= -12

x2 + 4x - 2x2 = -12

4x - x2 = -12

x2 - 4x - 12 = 0

(x - 6)(x + 2) = 0

sub. for x = 6, in eqn (i) y = -8, 8

=(6,-8); (-2, 8)