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
k*6¹ + 2*6⁰ + 3*5⁰ = 3*5⁰ + K*5¹ + 4*5⁰
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