2019 - JAMB Mathematics Past Questions and Answers - page 8

y = 8x\(^3\) - 3x\(^2\) + 7x - 1

\(\frac{\mathrm d^2 y}{\mathrm d x^2}  = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)

= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)

\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)

= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)

= \(\sqrt{3^2 + 4^2}\)

= \(\sqrt{9 + 16}\)

= \(\sqrt{25}\)

= 5 units

the equation is 4y = 7x + 3

\(\implies y = \frac{7}{4} x + \frac{3}{4}\)

the slope = coefficient of x = \(\frac{7}{4}\)

the slope of a perpendicular line = \(\frac{-1}{\frac{7}{4}}\)

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

for perpendicular line passes (-3, 1)

\(\therefore\) using the equation of line \(y = mx + b\)

m = slope

b = intercept.

\(y = \frac{-4}{7} x + b\)

we substitute y = 1 and x = -3 in the equation to find the intercept:

\(1 = \frac{-4}{7} (-3) + b\)

\(1 = \frac{12}{7} + b\)

\(b = \frac{-5}{7}\)

\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)

\(7y + 4x + 5 = 0\)

Mean = \(\frac{\sum fx}{\sum f}\)

\(3 = \frac{26 - y}{3y + 2}\)

\(3(3y + 2) = 26 - y\)

\(9y + 6 = 26 - y\)

\(9y + y = 26 - 6\)

\(10y = 20 \implies y = 2\)