JAMB Mathematics Past Questions & Answers - Page 81

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 = $\frac{7 - q}{4}$

Curved surface area of cylinder = 2πrh

110 = 2 × $\frac{22}{7}$ × r × 5

r = $\frac{110 \times 7}{44 \times 5}$

= 3.5cm

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 − $\frac{5}{20}$

= $\frac{15}{20}$

= $\frac{3}{4}$

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 × $\frac{1}{16}$

= $\frac{1}{8}$

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