1999 - JAMB Mathematics Past Questions and Answers - page 4

y + x - x² ≥ 0

y = x³ - x

on x axis, y = 0

∴x² - x = 0

x(x² - 1) = 0

x = 0 or x² - 1 = 0

x² = 1

x = +/-1

∴ x = -1, 0 and 1 which are the roots of the equation y + x - x² ≥ 0

Also y - x ≤ 0

=> y ≤ 0 + x

∴ the region where y ≤ 0

XY² = XY² + Y² - 2XY.YZ cos 120

XZ² = 2² + 1² - 2 x 2 1 cos 120

= 4 + 1 -2 x 2 x 1 x -cos(180 - 120)

= 4 + 1 + 4 cos 60

= 5 + 4 x ¹/₂

= 5 + 2

= 7

XY = √7 m

PQRS is a cyclic quad
^{^}P = 180 - 110 (opp ∠s of a cyclic quad)
^{^}P = 70

In ΔPTS, ^{^}S = ^{^}T (base ∠s of Isc Δ)

∴^{^}T = (180-70) / 2

= 110/2

= 55^o

But ^{^}T x(corr ∠)

∴x =55

∠EFH = ∠EGH(∠s in same segment)

= 34^o

∠HEG = ∠HFG(∠s in same segment)

= X

also ∠HFG = ∠EHF (alternate ∠s)

But ∠EHG + ∠EFG =180(opp sof a cyclic quad)

42 + x + 34 + x = 180

2x + 76 = 180

2x = 180 - 76

2x = 104

x = 52^o

6² = 9 x χ

36 = 9χ
χ = ³⁶/₉
χ = 4

ΔABE and ΔACD are similar

6x = 3(x+4)

6x = 3x + 12

3x = 12

x = 4

In ΔACD; L² = (4+4)² + 6²

= 8² + 6²

= 64 + 36

= 100
L =

√ 100 

= 10cm

In ABE, AE² = 4² + 3²

= 16 + 9

= 25

AE = 5cm

∴L = 10 - 5

= 5cm

(\int_0 ^4 x^2 dx = \left[\frac{1}{3}x^2+C\right]_0 ^4\

=\frac{1}{3}\times 4^3 - \frac{1}{3}\times 0^3\

=\frac{64}{3}-0\

=\frac{64}{3})sq units

% of those above 15 but less than 21 years

age = ²⁵/₅₀ x ¹⁰⁰/₁

= 50%

< EFH = < EGH = 34^o (angles n the same segment)

< GHF = < GEF = 42^o(angles in the same segment)

< FOG = 42 + 34 = 76(exterior angle)

< FOG = < EOH = 76(vertically opposite angle)

< EDO = 90^o, < DOE = (\frac{76}{2}) = 38^o

< HEG = 90^o - 38^o = 52^o