Mathematics Test

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Mathematics Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The modulus-amplitude form of $\sqrt{3}+i$, where $i=\sqrt{-1}$ is:

A
$2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$

B
$2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$

C
$4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$

D
$4\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$

Solution

Modulus amplitude of form:
$(\sqrt{3}+i)$
$z=r(\cos \theta+i \sin \theta)$
$r \cos \theta=\sqrt{3}, r \sin \theta=1$
$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=4$
$\Rightarrow r^{2}=4$
$\Rightarrow r=2$
$2 \cos \theta=\sqrt{3}$ and $2 \sin \theta=1$
$\Rightarrow \cos \theta=\frac{\sqrt{3}}{2}$ and $\sin \theta=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{6}$ and $\theta=\frac{\pi}{6}$
$\therefore z=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$
Question 2
1 / -0

The smallest positive integer $n$ for which $(1+i)^{2 n}=(1-i)^{2 n}$ is

A
2

B
16

C
12

D
None of these

Solution

$(1+i)^{2 n}=(1-i)^{2 n}$

$\left[\frac{(1+i)}{(1-i)}\right]^{2 n}=1$

Multiply and divide by conjugate, we get
$\left[\frac{(1+i)}{(1-i)} \frac{(1+i)}{(1+i)}\right]^{2 n}=1$
$\left[\frac{(1+i)^{2}}{1-(i)^{2}}\right]^{2n}$
$\left[\frac{1+i^{2}+2 i}{1-(-1)}\right]^{2n}=1 \quad$ where $\Rightarrow i^{2}=-1$
$\left[\frac{1-1+2 i}{1+1}\right]^{2n}=1$
$\left[\frac{2 i}{2}\right]^{2n}=i^{4}~~$ [where $ i^{4}=1$]
$[i]^{2n}=i^{4}$
On comparing
$2 n=4$
$n=2$
Question 3
1 / -0

If there are n elements in a set A then the elements present in power set are?

A
2ⁿ - 1

B
2ⁿ

C
n

D
None of these

Solution

It is given that there are n elements in a set A. So,

n(A) = n

Elements present in power set-

∴ n [P(A)] = 2ⁿ
Question 4
1 / -0

There are 20 cricket players, out of which 5 players can bowl. In how many ways can a team of 11 players be selected so, to include 4 bowlers?

A

${ }^{5} {C}_{4} \times{ }^{15} {C}_{7}$

B

${ }^{5} {C}_{3} \times{ }^{15} {C}_{8}$

C

${ }^{5} {C}_{5} \times{ }^{15} {C}_{6}$

D

${ }^{5} {C}_{2} \times{ }^{15} {C}_{9}$

Solution

Given that:

There are 20 cricket players, out of which 5 players can bowl.

We have to make a team of 11 players so to include 4 bowlers.

So, we select 4 bowlers out of 5 players and the remaining 7 players must be selected from 15 players, i.e.,

Total ways $={ }^{5} {C}_{4} \times{ }^{15} {C}_{7}$
Question 5
1 / -0

What is the eccentricity of a rectangular hyperbola?

A
$1$

B
$2$

C
$\sqrt{2}$

D
$12$

Solution

In a rectangular hyperbola:

$ \Rightarrow\mathrm{a}=\mathrm{b}$

Eccentricity:

$\Rightarrow e=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}$

Simplifying, we have:

$\Rightarrow e=\frac{\sqrt{2 a^{2}}}{a}$

$\Rightarrow e=\frac{\sqrt{2} a}{a}$

Canceling the common terms in the numerator and denominator, we have:

$\Rightarrow e=\sqrt{2}$
Question 6
1 / -0

Find the value of x where $^{x} P_{3} =^{(2 x - 1)} P_{2}, x \geq 3, x \in N$ .

A
3

B
1

C
5

D
7

Solution

The given equation is,

$^{x} P_{3} =^{(2 x - 1)} P_{2}$

$\Rightarrow \frac{x!}{(x - 3)!} = \frac{(2 x - 1)!}{(2 x - 1 - 2)!}$

$\Rightarrow \frac{x!}{(x - 3)!} = \frac{(2 x - 1)!}{(2 x - 3)!}$

$\Rightarrow \frac{x (x - 1) (x - 2) (x - 3)!}{(x - 3)!} = \frac{(2 x - 1) (2 x - 2) (2 x - 3)!}{(2 x - 3)!}$

$\Rightarrow x (x - 1) (x - 2) = (2 x - 1) (2 x - 2)$

$\Rightarrow x^{3} - 3 x^{2} + 2 x = 4 x^{2} - 6 x + 2$

$\Rightarrow x^{3} - 7 x^{2} + 8 x - 2 = 0$

$\Rightarrow (x - 1) (x^{2} + 6 x + 2) = 0$

Which gives $x = 1$ and:

$ x^{2}+6 x+2=0$

Using the Quadratic Formula where,

$a=1, b=6$, and $c=2$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$x=\frac{-6 \pm \sqrt{6^{2}-4(1)(2)}}{2(1)}$

$x=\frac{-6 \pm \sqrt{36-8}}{2}$

$x=\frac{-6 \pm \sqrt{28}}{2}$

$x=\frac{-6 \pm 2 \sqrt{7}}{2}$

$x=\frac{-6}{2} \pm \frac{2 \sqrt{7}}{2}$

$x=-3 \pm \sqrt{7}$

$x=-0.354249$

$x=-5.64575$

Other imaginary roots of $x$ which are impossible because $x \geq 3$ and $x \in N$ .
Question 7
1 / -0

A bag contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. The probability of getting the balls of different colours is:

A
$\frac{28}{121}$

B
$\frac{56}{121}$

C
$\frac{1}{2}$

D
None of these

Solution

There are a total of 7 red + 4 blue $=11$ balls.

Probability of drawing 1 red ball $=\frac{{ }^{7} \mathrm{C}_{1}}{11 \mathrm{C}_{1}}=\frac{7}{11}$

Probability of drawing 1 blue ball $=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{11} \mathrm{C}_{1}}=\frac{4}{11}$

Probability of drawing $(1$ red $)$ and $(1$ blue $)$ ball $=\frac{7}{11} \times \frac{4}{11}=\frac{28}{121}$

Similarly, Probability of drawing $(1$ blue $)$ and $(1$ red $)$ ball $=\frac{4}{11} \times \frac{7}{11}=\frac{28}{121}$

Probability of getting the balls of different colors $=\frac{28}{121}+\frac{28}{121}=\frac{56}{121}$
Question 8
1 / -0

Consider the following in respect of the function f(x) = |x - 3| :

1. f(x) is continuous at x = 3

2. f(x) is differentiable at x = 0.

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

$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 3^{-}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 3^{-}}-(\mathrm{x}-3)=0$

$\mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 3^{+}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 3^{+}}(\mathrm{x}-3)=0$

$f(x=3)=0$

$\therefore \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=3$

$\mathrm{LHD}=\lim _{\mathrm{h} \rightarrow 0^{-}} \frac{\mathrm{f}(0-0)-\mathrm{f}(0)}{-\mathrm{h}}$

$=\lim _{\mathrm{h} \rightarrow 0^{-}} \frac{f(0)-f(0)}{-h}$

$=0$

$\mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0^{+}} \frac{\mathrm{f}(\mathrm{a}+\mathrm{h})-\mathrm{f}(\mathrm{a})}{\mathrm{h}}$

$=0$

$\mathrm{LHD}=\mathrm{RHD},$ so $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$
Question 9
1 / -0

In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the students selected has opted neither for NCC nor for NSS is:

A
1/3

B
1/6

C
2/3

D
5/6

Solution

Given:

Total number of students = 60

$n (N C C) = 40$

$n (N S S) = 30$

$n (N C C \cap N S S) = 20$

We have to find the probability that the student selected has opted for neither NCC nor NSS.

Consider,

The probability that the student selected has neither for NCC nor NSS.

$= \frac{n (E)}{n (S)} = \frac{10}{60} = \frac{1}{6}$
Question 10
1 / -0

What is the degree of the differential equation $\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+2\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)-\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0 ?$

A
6

B
3

C
2

D
1

Solution

Given:

$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+2\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)-\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$

For the given differential equation the highest order derivative is 3.

Now, the power of the highest order derivative is 1.

We know that, the degree of a differential equation is the power of the highest derivative.

So, the degree of the differential equation is 1.

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Mathematics Test - 21

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Mathematics Test - 21

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Question 1

\(2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\)

\(2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\)

\(4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\)

\(4\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\)

Solution

Question 2

2

16

12

None of these

Solution

Question 3

2n - 1

2n

n

None of these

Solution

Question 4

\({ }^{5} {C}_{4} \times{ }^{15} {C}_{7}\)

\({ }^{5} {C}_{3} \times{ }^{15} {C}_{8}\)

\({ }^{5} {C}_{5} \times{ }^{15} {C}_{6}\)

\({ }^{5} {C}_{2} \times{ }^{15} {C}_{9}\)

Solution

Question 5

\(1\)

\(2\)

\(\sqrt{2}\)

\(12\)

Solution

Question 6

3

1

5

7

Solution

Question 7

\(\frac{28}{121}\)

\(\frac{56}{121}\)

\(\frac{1}{2}\)

None of these

Solution

Question 8

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 9

1/3

1/6

2/3

5/6

Solution

Question 10

6

3

2

1

Solution

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