2021 - JAMB Mathematics Past Questions and Answers - page 1

Given:  \(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) =  \(\frac{1}{(r + 2)}\),

\(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\)  = \(\frac{5}{3}\)

L.C.M. => (2r - 1) (r + 2) 

\(\frac{2(r + 2) - 1(2r - 1)}{(2r - 1) (r + 2)}\) = \(\frac{5}{3}\)

\(\frac{2r + 4 - 2r + 1}{ (2r - 1) (r + 2)}\) = \(\frac{5}{3}\)

3 = (2r - 1) (r + 2) or 2r\(^2\) + 3r - 2 

2r\(^2\) + 3r - 2 - 3 = 0

2r\(^2\) + 3r - 5 = 0

x = 1 or - \(\frac{5}{2}\)

\(V = \frac{2}{3} \pi r^{3}\)

\(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\)

\(\frac{\mathrm d V}{\mathrm d r} (r = 2) = 2\pi (2)^{2}\)

= \(8\pi\)

y=3x\(^2\) + 5x - 3

dy/dx = 6x + 5

as dy/dx = 0

6x + 5 = 0

x = \(\frac{-5}{6}\)

At maximun: 3 \( ^2{\frac{-5}{6}}\)  + 5 \(\frac{-5}{6}\) - 3

3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3

App the L.C.M.: 36

= \(\frac{25 - 50 - 36}{36}\)

= \(\frac{-61}{36}\)

Area of Trapezium = 1/2(sum of parallel sides) x h

Hence,

91 = \(\frac{1}{2}\) (5 + 9)h

91 = 7h
h = \(\frac{91}{7}\)
h = 13cm

Given  2y+8x-17 = 0

Equation of the line: y = mx + c

2y = -8x + 17

y = -4x  + \(\frac{17}{2}\)

The slope, m\(_1\) = 4

For parallel lines, m\(_1\). m\(_2\) = -4

where slope ( -4) = \(\frac{y_2 - y_1}{x_2 - x_1}\) at points (-1, -p) and (-2,2)

-4( \(x_2 - x_1\) ) = \(y_2 - y_1\) 

-4 ( -2 - -1) = 2 - -p

p = 4 - 2 = 2

***

y = 2x\(^2\)(2x - 1)
y = 4x\(^3\) - 2x\(^2\)
dy/dx = 12x\(^2\) - 4x
at x = -1,
dy/dx = 12(-1)\(^2\) - 4(-1)
= 12 + 4 = 16

Mean deviation formula:  (Σ|x - x|)/n
x = 2.5
= (|1 - 2.5| + |2 - 2.5| + |3 - 2.5| + |4 - 2.5|)/4
= 4/4

= 1