Mathematics 2011 | Study Mode

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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 = \(\frac{4}{18}\)

= \(\frac{2}{9}\)

Correct option: C
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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 \(\theta\)

Correct option: A
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Question 43
A
{a, b, d, e}
B
{b, d}
C
{a, e}
D
{c}
Explanation

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


Correct option: D
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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

Correct option: D
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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

Correct option: D
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Question 46
A
30%
B
25%
C
35%
D
20%
Explanation
\(\frac{70}{360} \times 36 = \frac{70}{10}\)

= 7

Correct option: D
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Question 47
A
180
B
135
C
210
D
105
Explanation

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


Correct option: A

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