2011 - JAMB Mathematics Past Questions and Answers - page 3

(\begin{vmatrix} 2 & 3 \ 5 & 3x \end{vmatrix}) = (\begin{vmatrix} 4 & 1 \ 3 & 2x \end{vmatrix})

(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)

6x - 15 = 8x - 3

6x - 8x = 15 - 3

-2x = 12

x = (\frac{12}{-2})

= -6

(\begin{vmatrix} 4 & 2 & -1 \ 2 & 3 & -1 \ -1 & 1 & 3 \end{vmatrix})

4 (\begin{vmatrix} 3 & -1 \ 1 & 3 \end{vmatrix}) -2 (\begin{vmatrix} 2 & -1 \ -1 & 3\end{vmatrix}) -1 (\begin{vmatrix} 2 & 3 \ -1 & 1 \end{vmatrix})

4[(3 x 3) - (-1 x 1)] -2 [(2x 3) - (-1 x -1)] -1 [(2 x 1) - (-1 x 3)]

= 4[9 + 1] -2 [6 - 1] -1 [2 + 3]

= 4(10) - 2(5) - 1(5)

= 40 - 10 - 5

= 25

N = [2 3]

N^-1 = (\frac{adj N}{|N|})

adj N = (\begin{vmatrix} 4 & -3 \ -1 & 2 \end{vmatrix})

|N| = (2 x4) - (1 x 3)

= 8 - 3

=5

N^-1 = (\frac {1}{5}) (\begin{vmatrix} 4 & -3 \ -1 & 2 \end{vmatrix})

Interior angle = (n - 2)180

but, n = 12

= (12 -2)180

= 10 x 180

= 1800

let each interior angle = x

x = (\frac{(n - 2)180}{n})

x = (\frac{1800}{12})

= 150^o

Perimeter of circle = Perimeter of square

28cm = 4L

L = (\frac{28}{4}) = 7cm

Area of square = L²

= 7²

= 49cm²

From Pythagoras theorem

|OA|² = |AN|² + |ON|²

7² = |AN|² + (5)²

49 = |AN|² + 25

|AN|² = 49 - 25 = 24

|AN| = (\sqrt {24})

= (\sqrt {4 \times 6})

= 2√6 cm

|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)

|AN| + |NB| = |AB|

2√6 + 2√6 = |AB|

|AB| = 4√6 cm

Volume of cube = L³

3³ = 27cm³

volume of rectangular tank = L x B X h

= 3 x 4 x 5

= 60cm³

volume of H₂O the tank can now hold

= volume of rectangular tank - volume of cube

= 60 - 27

= 33cm³

P(x, y) Q(8, 6)

midpoint = (5, 8)

x + 8 = 5

(\frac{y + 6}{2}) = 8

x + 8 = 10

x = 10 - 8 = 2

y + 6 = 16

y + 16 - 6 = 10

therefore, P(2, 10)

2y = 5x + 4 (4, 2)

y = (\frac{5x}{2}) + 4 comparing with

y = mx + e

m = (\frac{5}{2})

Since they are perpendicular

m₁m₂ = -1

m₂ = (\frac{-1}{m_1}) = -1

(\frac{5}{2}) = -1 x (\frac{2}{5})

The equator of the line is thus

y = mn + c (4, 2)

2 = -(\frac{2}{5})(4) + c

(\frac{2}{1}) + (\frac{8}{5}) = c

c = (\frac{18}{5})

(\frac{10 + 5}{5}) = c

y = -(\frac{2}{5})x + (\frac{18}{5})

5y = -2x + 18

or 5y + 2x - 18 = 0