2001 - JAMB Mathematics Past Questions and Answers - page 2

m² + n² = 29 .......(1)

m + n = 7 ............(2)

From (2),

m = 7 - n

but m² + n² = 29, substituting;

(7-n)² + n² = 29

49 - 14n + n² + n² = 29

=> 2n² -14n + 20 = 0

Thus n² -7n + 10 = 0

Factorizing;

(n-5)(n-2) = 0

n - 5 = 0, => n = 5

n - 2 = 0, => n = 2.

When n = 5,

m + n = 7, => m = 2,

When n = 2,

m + n = 7, => m = 5.

Thus (m,n) = (5,2) and (2,5)

By definition a*b = a + b + 1.

Let the inverse of the element 2 be x,

Therefore 2*x = -1

i.e. 2 + x + 1 = -1

3 + x = -1

x = -1 - 3

x = -4

1st statement: U₆ = 1/2(U₁₂)

a + (n -1)d = 1/2[a + (n-1)d]

a + 5d = a + 11d

2(a + 5d) = a + 11d

2a + 10d = a + 11d

Solving, => a = d

Hence the first term is equal to the common difference

Given: 4x² - 9y² + 20x + 25

Collect like terms: 4x² + 20x + 25 - 9y²

(2x + 5)(2x + 5) - 9y²

(2x + 5)² - (3y)²

(2x - 3y +5)(2x + 3y + 5)

(r/L) = (θ/360°)

Given θ = 300, and L = 7.2cm,

=> r = (300 x 7.2)/360

r = 6cm

Hint: Make a sketch of the moving points such that the hypotenuse is 8cm and the adjacent x cm.

Cos 30 = x/8

x = 8 cos 30 = 8√3cm

Cos 30 = 5/x

x cos 30 = 5, => x = 5√3

Coordinates of P = -5, 3, 0

Coordinates of Q = 0, 5

Gradient of PQ = (y₂ - y₁) (X₂ - X₁) = (5 - 0)/(0 -5√3)

= 5/5√3 = 1/√3

Equation of PQ = y - y₁ = m (x -x₁)

y - 0 = 1/√3 (x -(-5√3))

Thus: √3y = x + 5√3

Cos θ° = t² + t² -(√3t)² 2 x t x t

= 2t² - 3t² 2t²

= -t²/2t²

= -1/2

Thus θ = cos^-1 (-0.5) = 120°

Hint: If two lines a perpendicular, the gradient of one is equal to minus the reciprocal of the other (P, 4) and (6, -2)

Let the ext. angle = x

Thus int. angle = 2x

But sum of int + ext = 180 (angle of a straight line).

2x + x = 180

3x = 180

x = 180/3 = 60

Each ext angle = 360/n

=> 60 = 360/n

n = 360/60 = 6