Mathematics 2011 | Study Mode

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Question 31
A
\(\frac{2}{3}\) 
B
\(\frac{3}{5}\) 
C
\(\frac{1}{5}\) 
D
\(\frac{4}{5}\) 
Explanation.
Share Answer 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\(\theta\)

\(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)

Correct Option:
C
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Question 32
A
195o 
B
135o 
C
225o 
D
045o 
Explanation.
Share Answer tan\(\theta\) = \(\frac{100}{100}\) = 1

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

The bearing of x from z is N45oE or 135o

Correct Option:
B
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Question 33
A
12x + 1 
B
6x + 5 
C
6x + 1 
D
12x + 5 
Explanation.
Share Answer (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

Correct Option:
D
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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.
Share Answer \(\int^{1}_{0}\) 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

Correct Option:
D
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Question 36
A
B
C
D
Explanation.
Share Answer 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

Correct Option:
A
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Question 37
A
B
C
10 
D
Explanation.
Share Answer 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 38
A
\(\frac{1}{3}\) 
B
-\(\frac{1}{3}\) 
C
D
-1 
Explanation.
Share Answer 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 39
A
(8,5) 
B
(3, 5) 
C
(5 , 8) 
D
(5 , 3) 
Explanation.
Share Answer 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)

Correct Option:
B
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Question 40
A
B
C
D
Explanation.
Share Answer \(\begin{array}{c|c}Class Interval & 3 - 3 & 6 - 8 & 9 - 11 \\ x & 4 & 7 & 10 \\ f & 2 & 2 & 2 \\ f - x & 8 & 14 & 20 \\ |x - \bar{x}|^2 & 9 & 0 & 9 \\ |x - \bar{x}|^2 & 18 0 & 18 \end{array}\)

\(\bar{x}\) = \(\frac {\sum fx}{\sum f}\)

= \(\frac {8 + 14 + 20}{2 + 2 + 2}\)

= \(\frac{42}{6}\)

\(\bar{x}\) = 7

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

= \(\sqrt\frac{18 + 0 + 18}{6}\)

= \(\sqrt\frac{36}{6}\)

= \(\sqrt {6}\)

Correct Option:
B
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