2003 - JAMB Mathematics Past Questions and Answers - page 1

L ∝ 1/W

L = K/W

K = WL

= 1.5 * 20

30

3x - 5y + 5 = 0 → eqn1

4x - 7y + 8 = 0 → eqn2

eqn1 * 4; 12x - 20y + 20 = 0 → eqn3

eqn2 * 3; 12x - 21y + 24 = 0 → eqn4

eqn3 - eqn4 = y - 4 = 0

∴ y = 4

From eqn1,

3x - 5y + 5 = 0

3x - 5(4) + 5 = 0

3x - 20 + 5 = 0

3x - 15 = 0

3x = 15

x = 5

x and y = 5, 4 respectively

(r=\frac{n}{n-2}\hspace{1mm}and\hspace{1mm}r=\frac{n+3}{n}\∴\frac{n}{n-2}=\frac{n+3}{n}\n^2 = n^2 +3n - 2n-6\0=n-6\∴n=6\But\hspace{1mm}r = \frac{n}{n-2}\r=\frac{6}{6-2}\frac{6}{4}=\frac{3}{2})

Lines bounding Δ OPQ

OQ; y - x = 0

y - x ≥ 0

PQ; x + 1 = 0

x + 1 ≥ = 0

PO; y + x = 0

y + x ≤ 0

∴ x + 1 ≥ 0, y + x ≤ 0, y - x ≥ 0

4abx - 2axy - 12b²x + 6bxy = (4abx - 2axy) - (12b²x - 6bxy)

= 2ax(2b - y) -6bx(2b - y)

= (2ax - 6bx)(2b - y)

= 2x(a - 3b)(2b - y)

(S_n = 252, a = -16\hspace{1mm}and\hspace{1mm}l = 72\S_n = \frac{n}{2}(-16+72)\252 = \frac{n}{2}(-16+72)\n=\frac{504}{56}\n=9)

Area of Trapezium = 1/2(sum of parralel sides) * ht

21 = 1/2(5 + 9)h

42 = 14h

h = 42/14

h = 3cm

y = 2x² + 5x - 3

dy/dx = 4x + 5

= 4 + 5

= 9

(\int^{2} {3}(x^2 - 2x)dx=\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{2}{3}\left[\frac{x^3}{3}-x^2 + C\right ]^{2}_{3}\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\9-9-\left[\frac{8}{3}-4 \right ]=\frac{-8}{3}+4=\frac{4}{3})