2001 - JAMB Mathematics Past Questions and Answers - page 3

PQ² = (x₂ - x₁)² + (y₂ - y₁)²

= 12² + 5²

= 144 + 25

= 169

PQ = √169 = 13

But PQ = diameter = 2r, r = PQ/2 = 6.5 units

Hint: Simply make a sketch of the question and find the bearing to be 070°

Capacity = 3080 m³

Base diameter = 14 m, thus radius = 7 m.

Volume of cylinder = capacity of cylinder = πr²h. = 3080

22/7 x 7² x h = 3080

h = 3080/(22x7)

h = 20 m

Since ST = the radius r, => SOT is equilateral.

Each angle = 60°

Length of arc ST = θ/360 x 2πr

= 60/360 x 2 x π x r = πr/3

y = 2kx² + x + 1

Gradient function dy/dx = 4kx + 1

at dy/dx = 9, => 9 = 4kx + 1

=> 8 = 4kx, and k = 8/4x.

At x = 1, k = 8/4 = 2

HINT: Integrate the function easily by letting the value in bracket = letter "U". ie U = 2x-3, and then dx = du/2.

Substitute into the function and work it out.

Hint:

y = 4 - x² and y = 2x + 1

=> 4 - x² = 2x + 1

=> x² + 2x - 3 = 0

(x+3)(x-1) = 0

thus x = -3 or x = 1.

Integrating x² + 2x - 3 from (-3, to 1) w.r.t x will give 31/3 = 10(1/3)

Volume of sphere, V = 4/3 x πr³

Rate of change of V = dv/dr

Thus if V = 4/3 x πr³,

=> dv/dr = 4πr²

At r = 1, Rate = 4 x π x 1 = 4π