JAMB Mathematics Past Questions & Answers - Page 80

∫xⁿdx = \(\frac{x_{n + 1}}{n + 1}\)

∫dx = x + k

where k is constant

∫(x² + 3x − 5)dx

∫x² dx + ∫3xdx − ∫5dx

\(\frac{2_{2 + 1}}{2 + 1}\) + \(\frac{3x^{1 + 1}}{1 + 1}\) − 5x + k

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) − 5x + k

The locus of a point p(x, y) such that pv = pw where v = (1, 1)

and w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance

\(\sqrt{(x − x_1)^2 + (y − y_1)^2}\) = \(\sqrt{(x − x_2)^2 + (y − y_2)^2}\)

\(\sqrt{(x -1)^2 + (y - 1)^2}\) = \(\sqrt{(x - 3)^2 + (y - 5)^2}\)

square both sides
(x - 1)² + (y - 1)² = (x - 3)² + (y - 5)²

x² - 2x + 1 + y2 - 2y + 1 = x² - 6x + 9 + y2 - 10y + 25

x² + y² -2x -2y + 2 = x² + y² - 6x - 10y + 34

Collecting like terms
x² - x² + y² - y² - 2x + 6x -2y + 10y = 34 - 2

4x + 8y = 32

Divide through by 4

x + 2y = 8

Y ∝ \(\frac{1}{2}\)

Y = 6, X = 7

Y = \(\frac{k}{x}\) where k is constant

6 = \(\frac{k}{7}\)

k = 42

MACICITA is an eight letter word = 8!

Since we have repeating letters, we have to divide to remove duplicates accordingly. There are 2A, 2C, 2I

∴ \(\frac{8!}{2! 2! 2!}\)

MATHEMATICS is an eleven-letter word = 11!

There are 2Ms and 2As and 2Ts

Divide the number of repeating letters

= \(\frac{11!}{2!2!2!}\)