2005 - JAMB Mathematics Past Questions and Answers - page 4

y = sin (2x³ + 3x - 4)

let u = 2x³ + 3x - 4

∴du/dx = 6x²

y = sin U

dy/du = cos U

dy/dx = du/dx * dy/dy

∴dy/dx = (6x² + 3) cos U

= (6x² + 3)cos(2x³ + 3x - 4)

Increase in radius dr/dt = 0.5

Area of the circular disc = πr²

Increase in area: dA/dr = 2πr

Rate of increase in area: dA/dt = dA/dr * dr/dt

when r = 6cm

= 2πr * 0.5

= 2 * π * 6 * 0.5

2 * π * 6 * ¹/₂ = 6 π cm²/sec

f(x) = 2 + x - x^2
dy/_dx = 1-2x

As ^dy/_dx = 0

1-2x = 0

2x = 1

x = ¹/₂

At x = ¹/₂

f(x) = 2 + x - x²

= 2 + ¹/₂ -(¹/₂)²

= 2 + ¹/₂ - ¹/₄

= (8+2-1) / 4 = ⁹/₄

Area bounded by

y = x² - x + 3 and y = 3

x² - x + 3 = 3

x² - x = 0

x(x -1) = 0

x= 0 and x = 1

(\int^{1}_{0})(x² - x + 3)dx = [¹/₃x³ - ¹/₂x² + 3x](^1 _0)

(¹/₃(1)³ - ¹/₂(1)² + 3(1) - (¹/₃(0)³ - ¹/₂(0)² + 3(0))

= (¹/₃ - ¹/₂ + 3)- 0

= (\frac{2-3+18}{6})

= ¹⁷/₆

(\int_0^{\frac{\pi}{2}})sin 2x dx = [-¹/₂cos 2x + C](_0^{\frac{\pi}{2}})

=[-¹/₂ cos 2 * ^π/₂ + C] - [-¹/₂ cos 2 * 0]

= [-¹/₂ cos π] - [-¹/₂ cos 0]

= [-¹/₂x - 1] - [-¹/₂ * 1]

= ¹/₂ -(-¹/₂) = ¹/₂ + ¹/₂ = 1

Angle of Excellent

= 360 - (120+80+90)

= 360 - 290

= 70^o

If 360^o represents 36 students

1^o will represent ³⁶/₃₆₀

50^o will represent ³⁶/₃₆₀ * ⁷⁰/₁

= 7

(mean=\frac{60+5x}{18+x}\3.5=\frac{60+5x}{18+x}\frac{7}{2}=\frac{60+5x}{18+x})

7(18+x) = 2(60+5x)

126 + 7x = 120 + 10x

10x - 7x = 126 - 120

3x = 6

x = 2

1.35, 1.25, 1.35, 1.40, 1.35, 1.50, 1.35, 1.50, and 1.20

Modal height = 1.35

Range (r) = 1.50 - 1.20

= 0.30

∴m + 2r = 1.35 + 2(0.30)

= 1.35 + 0.60

= 1.95

(mean(x) = \frac{20t}{5} = 4t\standard\hspace{1mm}deviation=\sqrt{\frac{10t^2}{5}}\sqrt{2}=\sqrt{\frac{10t^2}{5}}\sqrt{2}=\sqrt{2t^2}\2 = 2t^2\t^2=1\t=1)