## 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
C
D
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
B
C
D
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
B
C
10
D
Explanation.
Mode = L1 + ($$\frac{D_1}{D_1 + D_2}$$)C

D1 = frequency of modal class - frequency of the class before it

D1 = 5 - 2 = 3

D2 = frequency of modal class - frequency of the class that offers it

D2 = 5 - 3 = 2

L1 = lower class boundary of the modal class

L1 = 5 - 5

C is the class width = 8 - 5.5 = 3

Mode = L1 + ($$\frac{D_1}{D_1 + D_2}$$)C

= 5.5 + $$\frac{3}{2 + 3}$$C

= 5.5 + $$\frac{3}{5}$$ x 3

= 5.5 + $$\frac{9}{5}$$

= 5.5 + 1.8

= 7.3 $$\approx$$ = 7

Question 38
A
$$\frac{1}{3}$$
B
-$$\frac{1}{3}$$
C
D
-1
Explanation.
y = x3 + x2 - x + 1

$$\frac{dy}{dx}$$ = $$\frac{d(x^3)}{dx}$$ + $$\frac{d(x^2)}{dx}$$ - $$\frac{d(x)}{dx}$$ + $$\frac{d(1)}{dx}$$

$$\frac{dy}{dx}$$ = 3x2 + 2x - 1 = 0

$$\frac{dy}{dx}$$ = 3x2 + 2x - 1

At the maximum point $$\frac{dy}{dx}$$ = 0

3x2 + 2x - 1 = 0

(3x2 + 3x) - (x - 1) = 0

3x(x + 1) -1(x + 1) = 0

(3x - 1)(x + 1) = 0

therefore x = $$\frac{1}{3}$$ or -1

For the maximum point

$$\frac{d^2y}{dx^2}$$ < 0

$$\frac{d^2y}{dx^2}$$ 6x + 2

when x = $$\frac{1}{3}$$

$$\frac{dx^2}{dx^2}$$ = 6($$\frac{1}{3}$$) + 2

= 2 + 2 = 4

$$\frac{d^2y}{dx^2}$$ > o which is the minimum point

when x = -1

$$\frac{d^2y}{dx^2}$$ = 6(-1) + 2

= -6 + 2 = -4

-4 < 0

therefore, $$\frac{d^2y}{dx^2}$$ < 0

the maximum point is -1

Question 39
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 40
A
B
C
D
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 = $$\sqrt\frac{\sum f(x - \bar{x})^2}{\sum f}$$

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

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

= $$\sqrt {6}$$

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