JAMB Mathematics Past Questions & Answers

1,641.

Find the inverse $\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}$

$\begin{vmatrix} 2 & -\frac{3}{2} \\ -3 & -\frac{5}{2} \end{vmatrix}$

$\begin{vmatrix} 2 & -\frac{3}{2} \\ -3 & \frac{5}{2} \end{vmatrix}$

$\begin{vmatrix} 2 & \frac{3}{2} \\ -3 & \frac{5}{2} \end{vmatrix}$

Correct answer is B

Let A = $\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}$

Then |A| = $\begin{pmatrix} 5 & 3 \\ 6 & 4 \end{pmatrix}$ = 20 - 18 = 2

Hence A-1 = $\frac{1}{|A|}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}$

= $\frac{1}{2}\begin{pmatrix} 4 & -3 \\ -6 & 5 \end{pmatrix}$

= $\begin{pmatrix} 4 \times 1/2 & -3 \times 1/2 \\ -6 \times 1/2 & 5 \times 1/2 \end{pmatrix}$

= $\begin{pmatrix} 2 & -\frac{3}{2} \\ -3 & \frac{5}{2} \end{pmatrix}$

1,642.

If P = $\begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix}$ and Q = $\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}$ , find 2P + Q

$\begin{vmatrix} 7 & 7 \\ 14 & 8 \end{vmatrix}$

$\begin{vmatrix} 14 & 8 \\ 7 & 7 \end{vmatrix}$

$\begin{vmatrix} 7 & 7 \\ 8 & 14 \end{vmatrix}$

$\begin{vmatrix} 8 & 14 \\ 7 & 7 \end{vmatrix}$

View answer & explanation

Correct answer is B

2P + Q = 2 $\begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix}$ + $\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}$

= $\begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix}$ + $\begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}$

= $\begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix}$

1,643.

If a binary operation * is defined by x * y = x + 2y, find 2 * (3 * 4)

View answer & explanation

Correct answer is A

x * y = x + 2y (given)

3 * 4 = 3 + 2(4) = 11

Hence, 2 * (3 * 4) = 2 * 11

= 2 + 2(11)

= 2 + 22

= 24

1,644.

The n^th term of the progression $\frac{4}{2}$ , $\frac{7}{3}$ , $\frac{10}{4}$ , $\frac{13}{5}$ is ...

$\frac{1 - 3n}{n + 1}$

$\frac{3n + 1}{n + 1}$

$\frac{3n + 1}{n - 1}$

$\frac{3n - 1}{n + 1}$

View answer & explanation

Correct answer is B

Using T_n = $\frac{3n + 1}{n + 1}$ ,

T₁ = $\frac{3(1) + 1}{(1) + 1} = \frac{4}{2}$

T₂ = $\frac{3(2) + 1}{(2) + 1} = \frac{7}{3}$

T₃ = $\frac{3(3) + 1}{(3) + 1} = \frac{10}{4}$

1,645.

Solve for x: |x - 2| < 3

x < 5

-2 < x < 3

-1 < x < 5

x < 1

View answer & explanation

Correct answer is C

|x - 2| < 3 implies -(x - 2) < 3 .... or .... +(x - 2) < 3 -x + 2 < 3 .... or .... x - 2 < 3 -x < 3 - 2 .... or .... x < 3 + 2 x > -1 .... or .... x < 5 combining the two inequalities results, we get; -1 < x < 5