2021 - JAMB Mathematics Past Questions and Answers - page 1
Solve the following equation: \(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
( -1,\(\frac{5}{2}\) )
( 1, - \(\frac{5}{2}\) )
( \(\frac{5}{2}\), 1 )
(2,1)
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 - \(\frac{5}{2}\)
In how many ways can 2 students be selected from a group of 5 students for a debating competition?
20 ways
10 ways
15 ways
25 ways
\(\hspace{1mm} ^{5}C_{2}\hspace{1mm}ways\hspace{1mm}=\frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}=\frac{5\times4\times3!}{3!\times2\times1}=10\hspace{1mm}ways\)
Find the rate of change of volume V of a hemisphere with respect to its radius r, when r = 2
8π
2π
16π
4π
\(V = \frac{2}{3} \pi r^{3}\)
\(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\)
\(\frac{\mathrm d V}{\mathrm d r} (r = 2) = 2\pi (2)^{2}\)
= \(8\pi\)
Find the maximum value of y=3x\(^2\) + 5x - 3
6
0
2
No correct option
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
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
A trapezium has two parallel sides of lengths 5cm and 9cm. If the area is 91cm\(^2\), what is the distance between the parallel sides?
13 cm
12 cm
8 cm
9 cm
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
Find the value of p if the line which passes through (-1, -p) and (-2,2) is parallel to the line 2y+8x-17 = 0
\(\frac{-2}{7}\)
\(\frac{7}{6}\)
\(\frac{-6}{7}\)
2
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
The ratio of the length of two similar rectangular blocks is 2:3. If the volume of the larger block is 351cm\(^3\), find the volume of the other block.
234.00 cm3
166.00 cm3
526.50 cm3
687cm3
Let x = total volume, 2 : 3 = 2 + 3 = 5
\(\frac{3}{5}\)x = 351
x = \(\frac{351 \times 5}{3}\)
= 585
Volume of the smaller block = \(\frac{2}{5}\) x 585
= 234.00cm\(^3\)
Find the derivative of the function y = 2x\(^2\)(2x - 1) at the point x = -1
-4
16
18
-8
***
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
Correct 241.34(3 x 10\(^{-3}\))\(^2\) to 4 significant figures
0.0014
0.001448
0.0022
0.002172
(3 x 10-\(^3\))\(^2\)
= 3\(^2\)x\(^2\)
= \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\)
x\(^2\) = \(\frac{1}{x^3}\)
= 24.34 x 3\(^2\) x \(\frac{1}{10^6}\)
= \(\frac{2172.06}{10^6}\)
= 0.00217206
= 0.002172
Find the mean deviation of 1, 2, 3 and 4
1.0
1.5
2.0
2.5
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