Mathematics 2012 | Study Mode

Question 21
A
3
B
4
C
5
D
7
Explanation
$$\begin{vmatrix} 5 & 3 \\ x & 2 \end{vmatrix}$$ = $$\begin{vmatrix} 3 & 5 \\ 4 & 5 \end{vmatrix}$$

10 - 3x = 15 - 20

-3x = 15 - 20 - 10

-3x = -15

x = 5

Question 22
A
-1
B
O
C
1
D
2
Explanation
Recall that a unit matrice is a diagonal matrix in which the elements in the leading diagonal is unity. Therefore,

I3 = $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

I3 = $$+1\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}$$

I3 = +1(1 - 0) - 0(0 - 0) + 0(0 - 0)

= 1(1)

= 1

Question 23
A
24o
B
30o
C
33o
D
36o
Explanation

Explanation not provided, please join discuss, your contributions are welcomed

Question 24
A
3.92cm3
B
2.13cm3
C
1.97cm3
D
1.62cm3
Explanation
Volume of cuboid = L x b x h

= 0.76cm x 2.6cm x 0.82cm

= 1.62cm3

Question 25
A
line parallel to 7x + 13y + 8 = 0
B
circle
C
semicircle
D
bisector of the line 7x + 13y + 8 = 0
Explanation

Explanation not provided, please join discuss, your contributions are welcomed

Question 26
A
$$\frac{1}{2}$$
B
$$\frac{2}{5}$$
C
$$-\frac{4}{7}$$
D
$$-\frac{2}{3}$$
Explanation
PQ = $$\frac{y_1 - y_0}{x_1 - x_0}$$ = $$\frac{-3 - (-7)}{-2 - 5}$$ = $$\frac{-3 + 7}{-2 - 5}$$ = $$\frac{4}{-7}$$

Question 27
A
$$\sqrt{13}$$
B
$$3\sqrt{2}$$
C
$$\sqrt{26}$$
D
$$10\sqrt{5}$$
Explanation
P1 (4, 3), P2 (x, y)

y = 2x + 4 .....(1)

y = 7 - x .....(2)

Substitute (2) in (1)

7 - x = 2x + 4

7 - 4 = 2x + x

3 = 3x

x = 1

Substitute in eqn (2)

y = 7 - x

y = 7 - 1

y = 6

P2 (1, 6)

Distance between 2 points is given as

D = $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

D = $$\sqrt{(1 - 4)^2 + (6 - 3)^2}$$

D = $$\sqrt{(-3)^2 + (3)^2}$$

D = $$\sqrt{9 + 9}$$

D = $$\sqrt{18}$$

D = $$\sqrt{9 \times 2}$$

D = $$3\sqrt{2}$$

Question 28
A
y = 2x - 3
B
y = 2x + 5
C
y = 3x - 2
D
y = 2x + 1
Explanation
$$\frac{y - y_1}{x + x_1}$$ = $$\frac{y_2 - y_1}{x - x_1}$$

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

= $$\frac{y - 1}{x + 2}$$ = $$\frac{3}{\frac{3}{2}}$$

y = 2x + 5

Question 29
A
$$\frac{1}{2}$$
B
$$\frac{\sqrt{2}}{2}$$
C
$$-\frac{\sqrt{2}}{2}$$
D
$$-\frac{1}{2}$$
Explanation

$$\theta$$ = 135o

Cos 135o = Cos(90 + 45)o

= cos90ocos45o - sin90osin45o

= 0cos45o - (1 x $$\frac{\sqrt{2}}{2}$$)

= $$-\frac{\sqrt{2}}{2}$$

Question 30
A
2x - $$\frac{1}{x^2}$$
B
2x + x2
C
2x - x2
D
2x + $$\frac{1}{x^2}$$
Explanation
y = x2 - $$\frac{1}{x}$$

y = x2 - x-1

$$\frac{\delta y}{\delta x}$$ = 2x + x-2

$$\frac{\delta y}{\delta x}$$ = 2x + $$\frac{1}{x^2}$$

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