# Mathematics 2011 | Study Mode

Question 31
A
$$\frac{2}{3}$$
B
$$\frac{3}{5}$$
C
$$\frac{1}{5}$$
D
$$\frac{4}{5}$$
##### Explanation
tan$$\theta$$ = $$\frac{3}{4}$$

from Pythagoras tippet, the hypotenus is T

i.e. 3, 4, 5.

then sin $$\theta$$ = $$\frac{3}{5}$$ and cos$$\theta$$ = $$\frac{4}{3}$$

cos$$\theta$$ - sin$$\theta$$

$$\frac{4}{5}$$ - $$\frac{3}{5}$$ = $$\frac{1}{5}$$

Question 32
A
195o
B
135o
C
225o
D
045o
##### Explanation
tan$$\theta$$ = $$\frac{100}{100}$$ = 1

$$\theta$$ = tan-1(1) = 45o

The bearing of x from z is N45oE or 135o

Question 33
A
12x + 1
B
6x + 5
C
6x + 1
D
12x + 5
##### Explanation
(2x + 1)(3x + 1) IS

2x + 1 $$\frac{d(3x + 1)}{d}$$ + (3x + 1) $$\frac{d(2x + 1)}{d}$$

2x + 1 (3) + (3x + 1) (2)

6x + 3 + 6x + 2 = 12x + 5

Question 34
A
33m
B
5
C
2
D
6
##### Explanation
$$\int^{1}_{0}$$(3 - 2x)dx

[3x - x2]o

[3(1) - (1)2] - [3(0) - (0)2]

(3 - 1) - (0 - 0) = 2 - 0

= 2

Question 35
A
$$\frac{3}{4}$$ sin 4x + k
B
-$$\frac{1}{4}$$ sin 4x + k
C
-$$\frac{3}{4}$$ sin 4x + k
D
$$\frac{1}{4}$$ sin 4x + k
##### Explanation
$$\int^{1}_{0}$$ cos4 x dx

let u = 4x

$$\frac{dy}{dx}$$ = 4

dx = $$\frac{dy}{4}$$

$$\int^{1}_{0}$$cos u. $$\frac{dy}{4}$$ = $$\frac{1}{4}$$$$\int$$cos u du

= $$\frac{1}{4}$$ sin u + k

= $$\frac{1}{4}$$ sin4x + k

Question 36
A
7
B
6
C
8
D
5
##### Explanation
Let the numbers be a, a + 1, a + 2, a + 3

a + a + 1 + a + 2 + a + 3 = 34

4a = 34 - 6

4a = 28

a = $$\frac{28}{4}$$

= 7

The least of these numbers is a = 7

Question 37
A
(8,5)
B
(3, 5)
C
(5 , 8)
D
(5 , 3)
##### Explanation
Median = $$\frac{\sum fx}{\sum f}$$

$$\begin{array}{c|c} No & 0 & 1 & 2 & 3 & 4 & 5 \\ F & 1 & 4 & 3 & 8 & 2 & 5 \\ fx & 0 & 4 & 6 & 24 & 8 & 25 \end{array}$$

$$\sum fx$$ = 0 + 4 + 6 + 24 + 8 + 25 = 67

$$\sum f$$ = 23

Median = $$\frac{\sum fx}{\sum f}$$ = $$\frac{67}{23}$$ = 2.913

= $$\approx$$ 3

Range = 5 - 0 = 5

(3, 5)

Question 38
A
5
B
6
C
7
D
2
##### Explanation
$$\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \\ x & 4 & 7 & 10 \\ f & 2 & 2 & 2 \\ f - x & 8 & 14 & 20 \\ |x - \bar{x}|^2 & 9 & 0 & 9 \\ |x - \bar{x}|^2 & 18 0 & 18 \end{array}$$

$$\bar{x}$$ = $$\frac {\sum fx}{\sum f}$$

= $$\frac {8 + 14 + 20}{2 + 2 + 2}$$

= $$\frac{42}{6}$$

$$\bar{x}$$ = 7

S.D = $$\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}$$

= $$\sqrt\frac{18 + 0 + 18}{6}$$

= $$\sqrt\frac{36}{6}$$

= $$\sqrt {6}$$

Question 39
A
6!
B
7!
C
5!
D
8!
##### Explanation
ELATION
Since there are 7 letters. The first letter can be arranged in 7 ways, , the second letter in 6 ways, the third letter in 5 ways, the 4th letter in four ways, the 3rd letter in three ways, the 2nd letter in 2 ways and the last in one way.

therefore, 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! ways

Question 40
A
24
B
60
C
12
D
120
##### Explanation
The first person will sit down and the remaining will join.

i.e. (n - 1)!

= (5 - 1)! = 4!

= 24 ways

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