# Mathematics 2011 | Study Mode

Question 41
A
$$\frac{2}{3}$$
B
$$\frac{1}{3}$$
C
$$\frac{2}{9}$$
D
$$\frac{7}{9}$$
##### Explanation
Prime numbers = (43,47,53,59)

N = (43, 44, 45,..., 60)

The universal set contains 18 numbers.

The prime numbers between 43 and 60 are 4

Probability of picking a prime number = $$\frac{4}{18}$$

= $$\frac{2}{9}$$

Question 42
A
sec2 $$\theta$$
B
tan $$\theta$$ cosec $$\theta$$
C
cosec $$\theta$$sec $$\theta$$
D
cosec2$$\theta$$
##### Explanation
$$\frac {\sin\theta}{\cos\theta}$$

$$\frac{\cos \theta {\frac{d(\sin \theta)}{d \theta}} - \sin \theta {\frac{d(\cos \theta)}{d \theta}}}{\cos^2 \theta}$$

$$\frac{\cos \theta. \cos \theta - \sin \theta (-\sin \theta)}{cos^2\theta}$$

$$\frac{cos^2\theta + \sin^2 \theta}{cos^2\theta}$$

Recall that sin2 $$\theta$$ + cos2 $$\theta$$ = 1

$$\frac{1}{\cos^2\theta}$$ = sec2 $$\theta$$

Question 43
A
{a, b, d, e}
B
{b, d}
C
{a, e}
D
{c}

Question 44
A
9
B
8
C
10
D
7
##### 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 45
A
$$\frac{1}{3}$$
B
-$$\frac{1}{3}$$
C
1
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 46
A
30%
B
25%
C
35%
D
20%
##### Explanation
$$\frac{70}{360} \times 36 = \frac{70}{10}$$

= 7

Question 47
A
180
B
135
C
210
D
105