2017 - JAMB Mathematics Past Questions and Answers - page 1

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