2006 - JAMB Mathematics Past Questions and Answers - page 2

y = 3x³ - 9x²

dy/dx = 9x² - 18x

As dy/dx = 0

9x² - 18x = 0

9x(x-2) = 0

9x = 0 which implies x = 0

x-2 = 0 implies x = 2

d²y/dx² = 18x - 18

when x = 0

d²y/dx² < 0 ∴ x is is minimum

when x = 2d²y/dx² = 18

∴ > 0 x is minimum

dy/dx = x + cos x

y = ∫(x + cos x)dx

y = ¹/₂x² + sin x + C

y = (cosθ - sinθ)

dy/dx = 2(cosθ - sinθ)(-sinθ - cosθ)

dy/dx = 2(-cosθsinθ - cos²θ + sin²θ)

dy/dx = 2(- cos²θ + sin²θ)

= -2(cos²θ - sin²θ)

= -2(1 - sin²θ)

= -2cos2θ

(\int_{-4}^0(1 - 2x)dx=[x-x^2]_{-4}^0= (0 - 0 + C) - (-4(-4)^2 + C)\

= C - (-4-16+C)\

= C-(-20+C)\

= C+20-C\

= 20)

(\frac{\frac{7}{9}-\frac{2}{3}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}\

=\frac{\frac{7-6}{9}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}\

=\frac{\frac{1}{9}}{\frac{1}{3}+\left(\frac{2}{5}\times \frac{5}{4}\right)}\

=\frac{\frac{1}{9}}{\frac{1}{3}+\frac{1}{2}}\

=\frac{\frac{1}{9}}{\frac{2+3}{6}}\

=\frac{\frac{1}{9}}{\frac{5}{6}}\

=\frac{1}{9}\times\frac{6}{5}\

=\frac{2}{15})

m : n = 13 : 11

m² - n² = (m - n)²

= 13² - 11² : (13 + 11)²

= 169 - 121 : 24²

= 48 : 576

= 1 : 12

(3^{-4}\times 9^2 \times 81^{-1}\

=log_9 (3^{-4}\times 9^2 \times 81^{-1})\

=log_9 \left(\frac{1}{3^4}\times 9^2 \times \frac{1}{81}\right)\

=log_9 \left(\frac{1}{81}\times \frac{81}{1}\times \frac{1}{81}\right)\

=log_9 \frac{1}{81}\

=log_9 \frac{1}{9^2}\

=log_9 9^{-2}\

=-2log_9 9\

-2 \times 1\

=-2)

(K2)₆ * 3₆ = 3₅(K4)₅

k6¹ + 26⁰ + 35⁰ = 35⁰ + K5¹ + 45⁰

6K + 2 = 5K + 4

6K - 5k = 4 - 2

K = 2

Let n(L∩P) = x

n(L) only = 350 - x

n(P) only = 200 - x

∴350 - x + x + 200 - x = 500

550 - x = 500

x = 550 - 500

x = 50 speak both

∴x% = (50/500) * (100/1)

= 10%

N20,000 - N10,000 = N10,000

1^st tax on N5000 at N50 per N1000

= 5*50 = N250

2^nd tax on the remaining N5000 at N55 per N1000 = 5*55 = N275

Total tax paid = N250 + N275

= N525