Mathematics Test

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Mathematics Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

Let X denote the number of scores which exceed 4 in 18, tosses of a symmetrical die. Consider the following statements:

1. The arithmetic mean of X is 6.

2. The standard deviation of X is 2.

Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Statement I :

n(X) = 2

p = $\frac{n (X)}{N (S)} = \frac{2}{6} = \frac{1}{3}$

q = 1 - p = 1- $\frac{1}{3} = \frac{2}{3}$

arithmetric mean of X = np

= 18 × $\frac{1}{3} = 6$

Statement 2 : Standard deviation of

X = $\sqrt{v a r i a n c e o f X}$

= $\sqrt{18 \times \frac{1}{3} \times \frac{2}{3}} = \sqrt{4} = 2$

Hence, statements 1 and 2 both are correct.
Question 2
1 / -0

Consider the following statement in respectdifferential equationa $\frac{a^{2} y}{a x^{2}} + \cos (\frac{a y}{a x}) = 0$

1. The degree of the differential equation is not defined.

2. The order of the differential equation is 2.

Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Statement I : Differential equation is not a polynomial equation in its derivatives. So, its degree is not defined.

Statement II : The highest order derivative in the given polynomial is 2.
Question 3
1 / -0

Given that C(n, r) : C(n, r + 1) = 1 : 2 and C(n, r + 1) : C(n, r + 2) = 2 : 3.

What is n equal to?

A
11

B
12

C
13

D
14

Solution

$\frac{n_{C_{r}}}{n_{C_{r + 1}}} = \frac{1}{2}$

$\frac{n! (r + 1)! (n - r - 1)!}{r! (n - r)! n!} = \frac{1}{2}$

$\frac{r + 1}{n - r} = \frac{1}{2} \Rightarrow 3 r - n + 2 = 0$

$\frac{n_{C_{r + 1}}}{n_{C_{r + 2}}} = \frac{2}{3}$

$\frac{n! (r + 2)! (n - r - 2)!}{(r + 1)! (n - r - 1)! n!} = \frac{2}{3}$

$\frac{r + 2}{n - r - 1} = \frac{2}{3} \Rightarrow 5 r - 2 n + 8 = 0$

$n = 14, r = 4$
Question 4
1 / -0

A
abc

B
a+b+c

C
-(a+b+c)

D
-abc

Solution
Question 5
1 / -0

Let X be the set of all citizens of India. Elements x, y in X are said to be related if the difference of their age is 5 years. Which one of the following is correct?

A
The relation is an equivalence relation on X.

B
The relation is symmetric but neither reflexive nor transitive.

C
The relation is reflexive but neither, symmetric nor transitive

D
None of these

Solution

X = Set of all citizens of India

$R = {(x, y) : x, y \in X, |x - y| = 5}$

$|x - x| = 0 \neq 5$

(R is not reflexive)

xRy $\Rightarrow |x - y| = 5$

$\Rightarrow |y - x| = 5$

(R is symmetric)

xRy $\Rightarrow |x - y| = 5$

$y R z \Rightarrow |y - x| = 5$

But $|x - z| \neq 5$

(R is not transitive)
Question 6
1 / -0

The sum of the series formed by the sequence 3, √3, 1 …….. upto infinity is:

A
$\frac{3 \sqrt{3} (\sqrt{3 + 1})}{2}$

B
$\frac{\sqrt[3]{3 (\sqrt{3 - 1})}}{2}$

C
$\frac{3 (\sqrt{3} + 1)}{2}$

D
$\frac{3 (\sqrt{3} - 1)}{2}$

Solution

$3, \sqrt{3}, 1, \frac{1}{\sqrt{3}}, . . . . . . . . ., \infty$

This is a Geometric Progression with a = 3, r = $\frac{1}{\sqrt{3}}$

$S_{\infty} = \frac{a}{1 - r} = \frac{3}{1 - \frac{1}{\sqrt{3}}}$

$\frac{3 \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{3 \sqrt{3} (\sqrt{3} + 1)}{2}$
Question 7
1 / -0

The determinant of an odd order skew symmetric matrix is always:

A
Zero

B
One

C
Negative

D
None of these

Solution

We know, that elements of principle diagonals skew symmertic matrix are all zero.
Question 8
1 / -0

Evaluate- $\tan^{- 1} (\frac{3}{4}) + \tan^{- 1} (\frac{1}{7})$ .

A
$\frac{π}{4}$

B
$\frac{π}{2}$

C
$\frac{2 π}{3}$

D
None of these

Solution

Given:

$\tan^{- 1} (\frac{3}{4}) + \tan^{- 1} (\frac{1}{7})$

$\tan^{- 1} (\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \times \frac{1}{7}})$

$\tan^{- 1} (\frac{\frac{25}{28}}{\frac{25}{28}})$

$\tan^{- 1} (1)$

$\frac{π}{4}$
Question 9
1 / -0

A straight line passes through (1, –2, 3) and perpendicular to the plane 2x + 3y – z = 7.

What are the direction ratios of normal to plane?

A
<2, 3, –1>

B
<2, 3, 1>

C
<–1, 2, 3>

D
None of these

Solution

Direction rations of normal to plane 2x + 3y - z = 7 is <2, 3, –1>
Question 10
1 / -0

The variance of number $x^{1}, x^{2}, x^{3}, . . . . . . . . . . x^{n}$ , is V. Consider the following statement:

1.If every x_iis increased by 2, the variance of the new set of numbers is V.

2. If the numbers x_iis squared, the variance of the set is V²

Which of the following statement is/are correct ?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

I: Variance is not dependent on change of origin.

Therefore, if every xi is increased by 2, the variance of the new set of numbers is not changed.

II: Variance is dependent on change of scale.

If the number xi is squared the variance of the new set is V²

V_xt V_xt = V. V = V²

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Mathematics Test - 7

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Mathematics Test - 7

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Question 1

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 2

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 3

11

12

13

14

Solution

Question 4

abc

a+b+c

-(a+b+c)

-abc

Solution

Question 5

The relation is an equivalence relation on X.

The relation is symmetric but neither reflexive nor transitive.

The relation is reflexive but neither, symmetric nor transitive

None of these

Solution

Question 6

333+12

33-132

33+12

33-12

Solution

Question 7

Zero

One

Negative

None of these

Solution

Question 8

π4

π2

2π3

None of these

Solution

Question 9

<2, 3, –1>

<2, 3, 1>

<–1, 2, 3>

None of these

Solution

Question 10

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

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$\frac{3 \sqrt{3} (\sqrt{3 + 1})}{2}$

$\frac{\sqrt[3]{3 (\sqrt{3 - 1})}}{2}$

$\frac{3 (\sqrt{3} + 1)}{2}$

$\frac{3 (\sqrt{3} - 1)}{2}$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{2 π}{3}$