Mathematics Test

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Mathematics Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

In a room, there are eight couples. Out of them if 4 people are selected at random, the probability that they may be couples is?

A

$\frac{{ }^{8} c_{4}}{{ }^{16} c_{4}}$

B

$\frac{{ }^{8} c_{2}}{{ }^{16} c_{8}}$

C

$\frac{{ }^{8} \mathrm{c}_{2}}{{ }^{8} \mathrm{C}_{4}}$

D

$\frac{{ }^{8} \mathrm{C}_{2}}{{ }^{16} \mathrm{C}_{4}}$

Solution

Given:

In a room, there are eight couples.

$\Rightarrow$ Eight couples $=16$ peoples

We have to select four peoples out of 16 peoples.

$\Rightarrow$ Total possible cases $={ }^{16} \mathrm{C}_{4}$

Now, we have to select four people- they may be couples

So, we have to select two couples from eight couples.

$\Rightarrow$ Favourable cases $={ }^{8} \mathrm{C}_{2}$

So, required Probability $=\frac{{ }^{8} \mathrm{c}_{2}}{{ }^{16} \mathrm{c}_{4}}$
Question 2
1 / -0

If the angles of a triangle ABC are in AP and b : c = $\sqrt{3} : \sqrt{2}$ ,then what is the measure of angle A?

A
30°

B
45°

C
60°

D
75°

Solution

We know that if A, B and C are angles and a, b and c are the sides of a ΔABC, then:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Let us consider Δ ABC with A, B and C as its angles and a, b and c as its sides.

It is given that angles are in AP i.e A, B and C are in AP.

⇒ 2 × B = A + C

As we know that, A + B + C =180°

⇒ 3 × B = 180° ----(∵ 2 × B = A + C)

⇒ B = 60°

As we know that, if A, B and C are angles and a, b and c are the sides of a Δ ABC, then:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

$\Rightarrow \frac{b}{c} = \frac{\sin B}{\sin C}$

As it is given that b : c = $\sqrt{3} : \sqrt{2}$ and sin C = sin 60° = $\sqrt{3} : \sqrt{2}$ .

$\Rightarrow \frac{\sqrt{3}}{\sqrt{2}} = \frac{\frac{\sqrt{3}}{2}}{\sin C} \Rightarrow \sin C = \frac{1}{\sqrt{2}}$

⇒ C = 45°

∵ A + B + C = 180° ⇒ A + 60° + 45° = 180°

⇒ A = 75°.
Question 3
1 / -0

Find the value of$\left|\begin{array}{ccc}a+x & y & z \\ x & a+y & z \\ x & y & a+z\end{array}\right|$.

A
$a^{3}(a-x+y+z)$

B
$a^{2}(a+x-y+z)$

C
$a^{3}(a+x+y+z)$

D
$a^{2}(a+x+y+z)$

Solution

Given,

$\left|\begin{array}{ccc}a+x & y & z \\ x & a+y & z \\ x & y & a+z\end{array}\right|$

Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}$, we get

$=\left|\begin{array}{ccc}a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z\end{array}\right|$

Taking $a+x+y+z$ common from $C_{1}$, we get

$=(a+x+y+z)\left|\begin{array}{ccc}1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z\end{array}\right|$

Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$, we get

$=(a+x+y+z)\left|\begin{array}{ccc}1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right|$

$=(a+x+y+z) [1(a^2-0)-y(0-0)+z(0-0)]$

$=a^{2}(a+x+y+z)$
Question 4
1 / -0

The expected value of the number of heads obtained when three fair coins are tossed simultaneously is:

A
$1$

B
$1.5$

C
$0$

D
$-1$

Solution

If three fair coins are tossed then the different values of the number of heads we can get are 0,1,2 or 3 heads and the total number of different outcomes possible is $2 \times 2 \times 2=8$
$
\mathrm{p}(0)=\frac{{ }^{3} \mathrm{C}_{0}}{8}=\frac{1}{8}
$
$\mathrm{p}(1)=\frac{{ }^{3} \mathrm{C}_{1}}{8}=\frac{3}{8}$
$\mathrm{p}(2)=\frac{{ }^{3} \mathrm{C}_{2}}{8}=\frac{3}{8}$
$
\mathrm{p}(3)=\frac{{ }^{3} \mathrm{C}_{3}}{8}=\frac{1}{8}
$
$\therefore \mathrm{E}($ Head $)=\Sigma($ Value $\times$ Probability)
$
\begin{array}{l}
=\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(2 \times \frac{3}{8}\right)+\left(3 \times \frac{1}{8}\right) \\
=\frac{12}{8}=1.5
\end{array}
$
Question 5
1 / -0

What is the minimum value of $\frac{a^{2}}{\cos ^{2} x}+\frac{b^{2}}{\sin ^{2} x}$ where $a>0$ and $b>0$

A
(a + b)²

B
(a - b)²

C
a² + b²

D
|a² + b²|

Solution

The given equation is

$\frac{a^{2}}{\cos ^{2} x}+\frac{b^{2}}{\sin ^{2} x}=a^{2} \sec ^{2} x+b^{2} \operatorname{cosec}^{2} x$

As we know that, minimum value of $m \sec ^{2} \theta+n \operatorname{cosec}^{2} \theta=(\sqrt{m}+\sqrt{n})^{2}$

Here, $m=a^{2}$ and $n=b^{2}$.

So, the minimum value of $\frac{a^{2}}{\cos ^{2} x}+\frac{b^{2}}{\sin ^{2} x}=a^{2} \sec ^{2} x+b^{2} \operatorname{cosec}^{2} x=(a+b)^{2}$
Question 6
1 / -0

What is the general solution of the differential equation $x^{2} d y+y^{2} d x=0$?

A
$x+y=C$

B
$x y=c$

C
$\frac{x}{y}=c$

D
$C(x+y)=x y$

Solution

Given:

$x^{2} d y+y^{2} d x=0$

$\Rightarrow \frac{d y}{y^{2}}=-\frac{d x}{x^{2}}$

Now, by integrating both the sides we get

$\Rightarrow \int y^{-2} d y+\int x^{-2} d x=0$

$\Rightarrow-\frac{1}{y}-\frac{1}{x}=-C_{1}$

$\Rightarrow \frac{1}{y}+\frac{1}{x}=C_{1}$

$\Rightarrow x+y=C_{1}(x y)$

$\Rightarrow C(x+y)=x y,$ where $C=\frac{1}{c_{1}}$
Question 7
1 / -0

Consider a binary operation $^{*}$ on $\mathrm{N}$ defined as $a^{*} b=a^{3}+b^{3},$ Choose the correct answer.

A
Is * both associative and commutative?

B
Is * commutative but not associative?

C
Is * associative but not commutative?

D
Is * neither commutative nor associative?

Solution

On $\mathrm{N},$ the operation $^{*}$ is defined as $a^{*} b=a^{3}+b^{3}$.
For, $a, b, \in \mathbf{N},$ we have
$a^{*} b=a^{3}+b^{3}=b^{3}+a^{3}=b^{*} a \quad$[Addition is commutative in $\left.\mathrm{N}\right]$
Therefore, the operation $^{*}$ is commutative.
It can be observed that:
$(1 ^* 2)^{*} 3=\left(1^{3}+2^{3}\right)^{*} 3=(1+8) ^{*} 3=9 ^{*} 3=9^{3}+3^{3}=729+27=756$ and
$1^{*}\left(2^{*} 3\right)=1^{*}\left(2^{3}+3^{3}\right)=1^{*}(8+27)=1^{*} 35=1^{3}+35^{3}=1+42875=42876$ $\therefore(1 ^* 2) ^* 3 \neq 1^{*}(2 ^{*} 3),$ where $1,2,3 \in \mathrm{N}$
Therefore, the operation * is not associative.
Thus, the operation $^{*}$ is commutative, but not associative.
Question 8
1 / -0

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|,$ then which one of the following is correct?

A
$|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|$

B
$\overrightarrow{\mathrm{a}}$ is parallel to $\overrightarrow{\mathrm{b}}$

C
$\overrightarrow{\mathrm{a}}$ is perpendicular to $\overrightarrow{\mathrm{b}}$

D
$\overrightarrow{\mathrm{a}}$ is unit vector

Solution

Given:

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$

Squaring both sides,

$|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}$

Now we have

$|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^2=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2-4\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2$

$\Rightarrow-4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$

$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$

$\therefore$ Vector a is perpendicular to b.
Question 9
1 / -0

If $A=\left[\begin{array}{cc}2 & 3 \\ -1 & 2\end{array}\right]=\frac{1}{2}(P+Q)$ where $P$ is symmetric and $Q$ is skew symmetric matrix then $P$ and $Q$ are:

A
$P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]$ and $Q=\left[\begin{array}{ll}0 & 4 \\ 4 & 0\end{array}\right]$

B
$P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]$ and $Q=\left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right]$

C
$P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]$ and $Q=\left[\begin{array}{cc}0 & -4 \\ -4 & 0\end{array}\right]$

D
None of these

Solution

Given,
$A=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]=\frac{1}{2}(P+Q)$
Where $P$ is symmetric and $Q$ is a skew-symmetric matrix.
As we know,
Any square matrix can be be expressed as the sum of the symmetric and skew-symmetric matrix. i.e If $A$ is a square matrix then $A$ can be expressed as where $A + A ^{\prime}$ is symmetric and $A - A ^{\prime}$ is skew-symmetric matrix.
On comparing $A=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]=\frac{1}{2}(P+Q)$ with $A=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)$ we get,
$ P=A+A^{\prime}$ and $Q=A-A^{\prime}$
As, $A=\left[\begin{array}{cc}2 & 3 \\ -1 & 2\end{array}\right]$
$\therefore A^{\prime}=\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$
So, $\begin{aligned}P=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & -1 \\
3 & 2
\end{array}\right]\end{aligned}$
$\begin{aligned}\Rightarrow P=\left[\begin{array}{ll}
4 & 2 \\
2 & 4
\end{array}\right]\end{aligned}$
Similarly,
$\begin{aligned}
&Q=\left[\begin{array}{cc}
2 & 3 \\
-1 & 2
\end{array}\right]-\left[\begin{array}{cc}
2 & -1 \\
3 & 2
\end{array}\right] \end{aligned}$
$\begin{aligned} \Rightarrow Q=\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]
\end{aligned}$
So,
$\begin{aligned}
&P=\left[\begin{array}{ll}
4 & 2 \\
2 & 4
\end{array}\right]\end{aligned}$ and $\begin{aligned}Q=\left[\begin{array}{cc}
0 & 4 \\
-4 & 0
\end{array}\right]
\end{aligned}$
Question 10
1 / -0

Find the differentiation of the following function $\sec ^{-1} \tan x .$

A
$\frac{1}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}$

B
$\frac{1}{\sin x \sqrt{\sin ^{2} x+\cos ^{2} x}}$

C
$\frac{2}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}$

D
None of the above

Solution

Given term $\sec ^{-1} \tan x$

Let us consider $f(x)=\sec ^{-1} \tan x$

As we know that the differentiation of $\sec ^{-1} x$ is given as

$\frac{d\left(\sec ^{-1} x\right)}{d x}=\frac{1}{x \sqrt{x^{2}-1}}$

In order to solve the problem we will use the chain rule.

According to the chain rule, we have:

$\frac{d p}{d q}=\frac{d p}{d u} \cdot \frac{d u}{d q}$

Using the chain rule let us proceed

So, by using the formula, we get

$\frac{d f(x)}{d x}=\frac{d\left(\sec ^{-1} \tan x\right)}{d x}$

$=\frac{1}{\tan x \sqrt{\tan ^{2} x-1}} \times \frac{d \tan x}{d x}$

$\frac{1}{\tan x \sqrt{\tan ^{2} x-1}} \times \sec ^{2} x \quad \quad\left[\because \frac{d \tan x}{d x}=\sec ^{2} x\right]$

$=\frac{1}{\frac{\sin x}{\cos x} \sqrt{\frac{\sin ^{2} x}{\cos ^{2} x}-1}} \times \frac{1}{\cos ^{2} x}$

$=\frac{1}{\frac{\sin x}{\cos x} \sqrt{\frac{\sin ^{2} x-\cos ^{2} x}{\cos ^{2} x}}} \times \frac{1}{\cos ^{2} x}$

By simplifying above equation, we get

$=\frac{1}{\frac{\sin x}{\cos x} \frac{\sqrt{\sin ^{2} x-\cos ^{2} x}}{\cos x}} \times \frac{1}{\cos ^{2} x}$

$=\frac{1}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}$

Thus, the differentiation of $\sec ^{-1} \tan x$ is $\frac{1}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 22

Result

Mathematics Test - 22

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SHARING IS CARING

Question 1

\(\frac{{ }^{8} c_{4}}{{ }^{16} c_{4}}\)

\(\frac{{ }^{8} c_{2}}{{ }^{16} c_{8}}\)

\(\frac{{ }^{8} \mathrm{c}_{2}}{{ }^{8} \mathrm{C}_{4}}\)

\(\frac{{ }^{8} \mathrm{C}_{2}}{{ }^{16} \mathrm{C}_{4}}\)

Solution

Question 2

30°

45°

60°

75°

Solution

Question 3

\(a^{3}(a-x+y+z)\)

\(a^{2}(a+x-y+z)\)

\(a^{3}(a+x+y+z)\)

\(a^{2}(a+x+y+z)\)

Solution

Question 4

\(1\)

\(1.5\)

\(0\)

\(-1\)

Solution

Question 5

(a + b)2

(a - b)2

a2 + b2

|a2 + b2|

Solution

Question 6

\(x+y=C\)

\(x y=c\)

\(\frac{x}{y}=c\)

\(C(x+y)=x y\)

Solution

Question 7

Is * both associative and commutative?

Is * commutative but not associative?

Is * associative but not commutative?

Is * neither commutative nor associative?

Solution

Question 8

\(|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|\)

\(\overrightarrow{\mathrm{a}}\) is parallel to \(\overrightarrow{\mathrm{b}}\)

\(\overrightarrow{\mathrm{a}}\) is perpendicular to \(\overrightarrow{\mathrm{b}}\)

\(\overrightarrow{\mathrm{a}}\) is unit vector

Solution

Question 9

\(P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]\) and \(Q=\left[\begin{array}{ll}0 & 4 \\ 4 & 0\end{array}\right]\)

\(P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]\) and \(Q=\left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right]\)

\(P=\left[\begin{array}{ll}4 & 2 \\ 2 & 4\end{array}\right]\) and \(Q=\left[\begin{array}{cc}0 & -4 \\ -4 & 0\end{array}\right]\)

None of these

Solution

Question 10

\(\frac{1}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}\)

\(\frac{1}{\sin x \sqrt{\sin ^{2} x+\cos ^{2} x}}\)

\(\frac{2}{\sin x \sqrt{\sin ^{2} x-\cos ^{2} x}}\)

None of the above

Solution

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(a + b)²

(a - b)²

a² + b²

|a² + b²|