2017 - JAMB Mathematics Past Questions and Answers - page 1

T = {evenn numbers from 1 to 12}

N = {common factors of 6,8 and 12}

Find T ∩ N

T = {2, 4, 6, 8, 10, 12}

N = {2}

T ∩ N = {2} i.e value common to T & N

2, 1, (\frac{1}{2}), (\frac{1}{4}).....

There are 4 terms in the series

Therefore the next number will be the 5th term

T_n = ar(^{n − 1}) (formular for geometric series)

a = first term = 2

r = common rate = (\frac{\text{next term}}{\text{previous term}}) = (\frac{1}{2})

n = number of terms

T₅ = 5th term = ?

T₅ = ar(^{5 - 1})

= ar(^4)

= 2 × (ar(^{n − 1}))⁴

= 2 × (\frac{1}{16})

= (\frac{1}{8})

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

E₁ = {3, 6, 9, 12, 15, 18}

E₂ = {4, 8, 12, 16, 20}

Probability of E₂ = (\frac{5}{20}) i.e (\frac{\text{Total number in}E_2}{\text{Entire number in set}})

Probability of set E₂ = 1 − (\frac{5}{20})

= (\frac{15}{20})

= (\frac{3}{4})

Curved surface area of cylinder = 2πrh

110 = 2 × (\frac{22}{7}) × r × 5

r = (\frac{110 \times 7}{44 \times 5})

= 3.5cm

Y = Px² + q

Y = 2x² - 1

Px² + q = 2x² - 1

Px² = 2x² - 1 - q

p = (\frac{2x^2 - 1 - q}{x^2})

at x = 2

P = (\frac{2(2)^2 - 1 - q}{2^2})

= (\frac{2(4) - 1 -q}{4})

= (\frac{8 - 1 - q}{4})

P = (\frac{7 - q}{4})

hypotenuse

sin = (\frac{1}{2})

(\sin45 = \frac{1}{\sqrt{2}})

= (\frac{2}{2})

∴ (sin45 + sin30)

= (\frac{1}{\sqrt{2}} + \frac{1}{2})

= (\frac{\sqrt{2}}{2}) + (\frac{1}{2})

= (\frac{\sqrt{2} + 1}{2})

= (\frac{1 + \sqrt{2}}{2})

y = xsinx

(\frac{dy}{dx}) = (1 \sin x + x \cos x)

= (\sin x + x \cos x)

At x = (\frac{\pi}{2})

= sin(\frac{\pi}{r}) + (\frac{\pi}{2} \cos {\frac{\pi}{2}})

= 1 + (\frac{\pi}{2}) × 10

= 1

t ∝ h, t = 20, h

t = ? h = 60

t = kh where k is constant

20 = 50k

k = (\frac{20}{50})

k = (\frac{2}{5})

when h = 60, t = ?

t = (\frac{2}{5}) × 60

t = 24^oC

M = (\frac{\sqrt{SL}}{T}),

make T subject of formula square both sides

M² = (\frac{N^2SL}{T})

TM² = N²SL

T = (\frac{N^2SL}{M^2})