Mathematics Test

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Mathematics Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Evaluate the integral $\int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x$.

A
$\frac{\pi^{2}}{32}$

B
$\frac{\pi^{2}}{30}$

C
$\frac{\pi^{3}}{20}$

D
$\frac{\pi^{2}}{50}$

Solution

Given, $\int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x$.

Let $t=\tan ^{-1} x$, $ \quad \ldots \ldots(i)$

Differentiating with respect to $x$,

Then $\frac{dt}{dx}=\frac{1}{1+x^{2}}$

$d t=\frac{1}{1+x^{2}} d x$

The new limits to $eq^{n}(i)$,

When $x=0$, $t=0$

And when $x=1, t=\frac{\pi}{4}$

Therefore,

$\int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x$ $=\int_{0}^{\frac{\pi}{4}} t d t$

$=\left[\frac{t^{2}}{2}\right]_{0}^{\frac{\pi}{4}}$

Put the value of limit,

$=\frac{1}{2}\left[\frac{\pi^{2}}{16}-0\right]$

$=\frac{\pi^{2}}{32}$
Question 2
1 / -0

The value of $(\sqrt{-6} \times \sqrt{-6})$ is:

A
6

B
-6

C
6i

D
-6i

Solution

Let;
$x=(\sqrt{-6} \times \sqrt{-6})$
We know that:
${i}=\sqrt{(-1)}$
${i}^{2}=-1$
$x=(\sqrt{6} \times i \times \sqrt{6} \times i)$
$x=6 i^{2}$
$x=-6$
Question 3
1 / -0

If $A=\left\{x \in R : x^{2}=2\right\}$ and $B=\left\{y \in R : y^{2}-5 y+6=0\right\}.$ Find $n(A \times B)$.

A
2

B
4

C
6

D
8

Solution

Given:

$A=\left\{x \in R: x^{2}=2\right\}$

And, $B=\left\{y \in R: y^{2}-5 y+6=0\right\}$

Then,

${x}^{2}-2=0 \quad$ (Given)

$\Rightarrow {x}=\pm \sqrt{2}$

$\Rightarrow {A}=\{\sqrt{2},-\sqrt{2}\}$

$\therefore {n}({A})=2$.

Similarly,

${y}^{2}-5 {y}+6=0 \quad$ (Given)

$\Rightarrow y^{2}-3 y-2 y+6=0$

$\Rightarrow(y-3)(y-2)=0$

$\Rightarrow {y}=2,3$

$\Rightarrow {B}=\{2,3\}$

$\therefore {n}({B})=2$.

As we know that, if $n(A)=p, n(B)=q$, then:

$n(A \times B)=n(A) \times n(B)=p \times q$.

$\Rightarrow {n}({A} \times {B})={n}({A}) \times {n}({B})$

$=2 \times 2$

$=4$
Question 4
1 / -0

Let $\mathrm{A}=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]$. If $\mathrm{M}$ and $\mathrm{N}$ are two matrices given by $\mathrm{M}=\sum_{\mathrm{k}=1}^{10} \mathrm{~A}^{2 \mathrm{k}}$ and $\mathrm{N}=\sum_{\mathrm{k}=1}^{10} \mathrm{~A}^{2 \mathrm{k}-1}$ then $\mathrm{MN}^2$ is :

A
a non-identity symmetric matrix

B
a skew-symmetric matrix

C
neither symmetric nor skew-symmetric matrix

D
an identity matrix

Solution

$A=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]$

$A^2=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}-4 & 0 \\ 0 & -4\end{array}\right]=-4 I$

$ \mathrm{M}=\mathrm{A}^2+\mathrm{A}^4+\mathrm{A}^6+\ldots .+\mathrm{A}^{20} \\$

$ =-4 \mathrm{I}+16 \mathrm{I}-64 \mathrm{I}+\ldots . \text { upto } 10 \text { terms } \\$

$ =-\mathrm{I}[4-16+64 \ldots+\text { upto } 10 \text { terms }] \\$

$=-\mathrm{I} .4\left[\frac{(-4)^{10}-1}{-4-1}\right]=\frac{4}{5}\left(2^{20}-1\right) \mathrm{I} \\$

$ \mathrm{N}=\mathrm{A}^1+\mathrm{A}^3+\mathrm{A}^5+\ldots .+\mathrm{A}^{19} \\$

$ =\mathrm{A}-4 \mathrm{~A}+16 \mathrm{~A}+\ldots . \text { upto } 10 \text { terms } \\$

$ =\mathrm{A}\left(\frac{(-4)^{10}-1}{-4-1}\right)=-\left(\frac{2^{20}-1}{5}\right) \mathrm{A} \\$

$ \mathrm{N}^2=\frac{\left(2^{20}-1\right)^2}{2^5} \mathrm{~A}^2=\frac{-4}{24}\left(2^{20}-1\right)^2 \mathrm{I} \\$

$ \mathrm{MN}^2=\frac{-16}{125}\left(2^{20}-1\right)^3 \mathrm{I}=\mathrm{KI} \quad(\mathrm{K} \neq \pm 1) \\$

$ \left(\mathrm{MN}^2\right)^{\mathrm{T}}=(\mathrm{KI})^{\mathrm{T}}=\mathrm{KI} \\$

$ \therefore \mathrm{A} \text { is correct }$
Question 5
1 / -0

A plane passes through the points $\mathrm{A}(1,2,3), \mathrm{B}(2,3,1)$ and $\mathrm{C}(2,4,2)$. If 0 is the origin and $\mathrm{P}$ is $(2,-1,1)$, then the projection of OP on this plane is of length.

A
$\sqrt{\frac{2}{3}}$

B
$\sqrt{\frac{2}{11}}$

C
$\sqrt{\frac{2}{7}}$

D
$\sqrt{\frac{2}{5}}$

Solution

Refer diagram, the normal vector be $\mathrm{n}$ and it is perpendicular to both

$\mathrm{AB}$ and $\mathrm{AC}$

$\mathrm{AB} \times \mathrm{AC}=\mathrm{n}$

$\begin{aligned} & \text { Now, } A(1,2,3), B(2,3,1) \text { and } C(2,4,2) \\ & \text { Then, } A B=(2-1) \hat{i}+(3-2) \hat{j}+(1-3) \hat{k} \\ & =\hat{i}+\hat{j}-2 \hat{k} \\ & -1) \hat{i}+(4-2) \hat{j}+(2-3) \hat{k} \\ & \begin{array}{l}A C=(2-1) \hat{i}+(4 \\ =\hat{i}+2 \hat{j}-\hat{k}\end{array}\end{aligned}$

Now, $A B \times A C=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \hat{j} & \mathrm{k} & \\ 1 & 1 & -2 \\ 1 & 2 & -1\end{array}\right|$

$\begin{aligned} & =\hat{i}(-1+4)-\hat{j}(-1+2)+\hat{k}(2-1) \\ & =3 \hat{i}-\hat{j}+\hat{k} \\ n & =3 \hat{i}-\hat{j}+\hat{k}\end{aligned}$

Let P be any point on normal vector and O be origin. Then refer the diagram, projection of OP on plane have length OM .

$\begin{aligned} & \text { OP }=2 \hat{i}-\hat{j}+\hat{k} \text { and } n=3 \hat{i}-\hat{j}+\hat{k} \\ & \text { OP. } n=|O P||n| \cos \theta \\ & (6+1+1)=\sqrt{4+1+1}(\sqrt{9+1+1}) \cos \theta \\ & 8=\sqrt{6} \sqrt{11} \cos \theta \Rightarrow \cos \theta=\frac{8}{\sqrt{66}} \\ & \text { Again, } \sin \theta=\frac{|\mathrm{OM}|}{|\mathrm{OP}|}, \text { gives }|\mathrm{OM}|=\sin \theta \cdot|\mathrm{OP}| \\ & \Rightarrow|\mathrm{OM}|=\sqrt{1-\cos ^2 \theta|\mathrm{OP}|} \\ & \left.=\sqrt{1-\frac{64}{66}} \sqrt{4+1+1} \quad \text { (use } \cos \theta=8 \sqrt{66}\right) \\ & =\sqrt{\frac{2}{66}} \cdot \sqrt{6} \\ & \therefore|O M|=\sqrt{\frac{2}{11}}\end{aligned}$
Question 6
1 / -0

In any group, the number of improper subgroups is:

A
2

B
3

C
Depends on the group

D
1

Solution

If G is a group, then the subgroup consisting of G itself is the improper subgroup of G. All other subgroups are proper subgroups

In any group, the number of improper subgroups is 2.
Question 7
1 / -0

In linear programming, lack of points for a solution set is said to have:

A
No feasible solution

B
A feasible solution

C
Single point method

D
Infinte point method

Solution

If there is no point in the feasible set, there is no feasible solution of the linear programming model.

In linear programming, lack of points for a solution set is said to have no feasible solution
Question 8
1 / -0

Find the area of the region (in square unit) bounded by the curve $y=x-2$ and $x=0$ to $x=4$.

A
2 sq. unit

B
4 sq. unit

C
3 sq. unit

D
6 sq. unit

Solution

Given curve and lines are:

$y=x-2$ and $x=0$ to $x=4$

In figure $\triangle \mathrm{ABC}$ and $\triangle \mathrm{AOD}$ is similar.

So, Area of the region $=2 \times($ Area of $\triangle \mathrm{ABC})$

For the area of $\triangle A B C$

We know that:

$=\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}}\left|\mathrm{y}_{2}-\mathrm{y}_{1}\right| \mathrm{dx}$

$=\int_{2}^{4}(\mathrm{x}-2) \mathrm{dx}$

$=\left[\frac{\mathrm{x}^{2}}{2}-2 \mathrm{x}\right]_{2}^{4}$

$=\frac{4^{2}}{2}-2(4)-\left(\frac{2^{2}}{2}-2 \times 2\right)$

$=8-8-2+4$

$=2 $ sq. unit

So,

Area of the region $=2 \times($ Area of $\triangle \mathrm{ABC})$

$=2 \times 2$

$=4 $ sq. unit

Question 9

1 / -0

Let $R$ be a relation on a set $A$ such that $R=R^{-1}$, then $R$ is:

Reflexive

Symmetric

Transitive

None of these

Solution

Let ${R}$ be a relation on ${Z}$, and let ${x}, {y}, {z} \in {Z}$.

Reflexive	$xRx$
Symmetric	$xRy$ implies $yRx$
Transitive	$xRy$ and $yRz$ implies $xRz$

Let $(a, b) \in R$
Then $(a, b) \in R \Rightarrow(b, a) \in R^{-1}$
$(b, a) \in R\left[\right.$ Because $\left.R=R^{-1}\right]$
$(a, b) \in R \Rightarrow(b, a) \in R$, for all $(a, b) \in A$
$a R b \Rightarrow b R a$
$\therefore {R}$ is symmetric

Question 10
1 / -0

Find the values of ${k}$ so the line $\frac{2 {x}-2}{2 {k}}=\frac{4-{y}}{3}=\frac{{z}+2}{-1}$ and $\frac{{x}-5}{1}=\frac{{y}}{{k}}=\frac{{z}+6}{4}$ are at right angles

A
$0$

B
$-2$

C
$2$

D
$1$

Solution

Given lines are $\frac{2 {x}-2}{2 {k}}=\frac{4-{y}}{3}=\frac{{z}+2}{-1}$ and $\frac{{x}-5}{1}=\frac{{y}}{{k}}=\frac{{z}+6}{4}$

Write the above equation of a line in the standard form of lines

$\Rightarrow \frac{2(x-1)}{2 k}=\frac{-(y-4)}{3}=\frac{z+2}{-1} \Leftrightarrow \frac{(x-1)}{k}=\frac{y-4}{-3}=\frac{z+2}{-1}$

So, the direction ratio of the first line is $({k},-3,-1)$

$\frac{{x}-5}{1}=\frac{{y}}{{k}}=\frac{{z}+6}{4}$

So, direction ratio of second line is $(1,{k}, 4)$

Lines are perpendicular,

$\therefore({k} \times 1)+(-3 \times {k})+(-1 \times 4)=0$

$\Rightarrow k-3 k-4=0$

$\Rightarrow-2 k-4=0$

$\therefore{k}=-2$

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Reflexive	\(xRx\)
Symmetric	\(xRy\) implies \(yRx\)
Transitive	\(xRy\) and \(yRz\) implies \(xRz\)

Mathematics Test - 48

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Mathematics Test - 48

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SHARING IS CARING

Question 1

\(\frac{\pi^{2}}{32}\)

\(\frac{\pi^{2}}{30}\)

\(\frac{\pi^{3}}{20}\)

\(\frac{\pi^{2}}{50}\)

Solution

Question 2

6

-6

6i

-6i

Solution

Question 3

2

4

6

8

Solution

Question 4

a non-identity symmetric matrix

a skew-symmetric matrix

neither symmetric nor skew-symmetric matrix

an identity matrix

Solution

Question 5

23

211

27

25

Solution

Question 6

2

3

Depends on the group

1

Solution

Question 7

No feasible solution

A feasible solution

Single point method

Infinte point method

Solution

Question 8

2 sq. unit

4 sq. unit

3 sq. unit

6 sq. unit

Solution

Question 9

Reflexive

Symmetric

Transitive

None of these

Solution

Question 10

\(0\)

\(-2\)

\(2\)

\(1\)

Solution

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$\sqrt{\frac{2}{3}}$

$\sqrt{\frac{2}{11}}$

$\sqrt{\frac{2}{7}}$

$\sqrt{\frac{2}{5}}$