JAMB Mathematics Past Questions & Answers

1,671.

If y = 4x³ - 2x² + x, find $\frac{\delta y}{\delta x}$

8x² - 2x + 1

8x² - 4x + 1

12x² - 2x + 1

12x² - 4x + 1

View answer & explanation

Correct answer is D

If y = 4x³ - 2x² + x, then;

$\frac{\delta y}{\delta x}$ = 3(4x²) - 2(2x) + 1

= 12x² - 4x + 1

1,672.

If $\sin\theta = \frac{12}{13}$ , find the value of $1 + \cos\theta$

$\frac{25}{13}$

$\frac{18}{13}$

$\frac{8}{13}$

$\frac{5}{13}$

View answer & explanation

Correct answer is B

No explanation has been provided for this answer.

1,673.

Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.

$\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}$

$\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}$

$\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}$

$\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}$

View answer & explanation

Correct answer is A

y - 4x + 3 = 0

When y = 0, 0 - 4x + 3 = 0

Then -4x = -3

x = 3/4

So the line cuts the x-axis at point (3/4, 0).

When x = 0, y - 4(0) + 3 = 0

Then y + 3 = 0

y = -3

So the line cuts the y-axis at the point (0, -3)

Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

$[\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]$

$[\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]$

$[\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]$

$[\frac{3}{8}, \frac{-3}{2}]$

1,674.

The gradient of a line joining (x,4) and (1,2) is $\frac{1}{2}$ . Find the value of x

-3

-5

View answer & explanation

Correct answer is A

$\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1}$

$\frac{1}{2} = \frac{2 - 4}{1 - x}$

1 - x = 2(2 - 4)

1 - x = 4 - 8

1 - x = -4

-x = -4 - 1

x = 5

1,675.

Find the mid point of S(-5, 4) and T(-3, -2)

-4, 2

4, -2

-4, 1

4, -1

View answer & explanation

Correct answer is C

Mid point of S(-5, 4) and T(-3, -2) is

$[\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + 2)]$

$[\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]$

$[\frac{1}{2}(-8), \frac{1}{2}(2)]$

$[-4, 1]$