Mathematics Test

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Mathematics Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Let $R$ be a relation defined as $x R y$ if and only if $2 x+3 y=20$, where $x, y \in N$. How many elements of the form $(x, y)$ are there in $R$ ?

A
$2$

B
$3$

C
$4$

D
$6$

Solution

Given:

$2 x+3 y=20$

$y=\frac{20-2x}{3}$

Let the value of $x$ be $1$

$y=\frac{20-2\times(1)}{3}$

$y=\frac{20-2}{3}$

$y=\frac{18}{3}$

$y=6$

$(x, y)=(1, 6)$

Let the value of $x$ be $2$

$y=\frac{20-2\times(2)}{3}$

$y=\frac{20-4}{3}$

$y=\frac{16}{3}$

Let the value of $x$ be $3$

$y=\frac{20-2\times(3)}{3}$

$y=\frac{20-6}{3}$

$y=\frac{14}{3}$

Let the value of $x$ be $4$

$y=\frac{20-2\times(4)}{3}$

$y=\frac{20-8}{3}$

$y=\frac{12}{3}$

$y=4$

$(x, y)=(4, 4)$

Let the value of $x$ be $5$

$y=\frac{20-2\times(5)}{3}$

$y=\frac{20-10}{3}$

$y=\frac{10}{3}$

Let the value of $x$ be $6$

$y=\frac{20-2\times(6)}{3}$

$y=\frac{20-12}{3}$

$y=\frac{8}{3}$

Let the value of $x$ be $7$

$y=\frac{20-2\times(7)}{3}$

$y=\frac{20-14}{3}$

$y=\frac{6}{3}$

$y=2$

$(x, y)=(7, 2)$

$R=\{(1,6),(4,4),(7,2)\}$

There are $3$ elements in the form $(x, y)$ are there in $R$.
Question 2
1 / -0

Let $\mathrm{P}$ be a square matrix such that $\mathrm{P}^2=\mathrm{I}-\mathrm{P}$. For $\alpha, \beta, \gamma, \delta \in \mathrm{N}$, if $\mathrm{P}^\alpha+\mathrm{P}^\beta=\gamma \mathrm{I}-29 \mathrm{P}$ and $\mathrm{P}^\alpha-\mathrm{P}^\beta=\delta \mathrm{I}-13 \mathrm{P}$, then $\alpha+\beta+\gamma-\delta$ is equal to :

A
40

B
22

C
24

D
18

Solution

$ \mathrm{P}^2=\mathrm{I}-\mathrm{P} \\$

$ \mathrm{P}^\alpha+\mathrm{P}^\beta=\gamma \mathrm{I}-29 \mathrm{P} \\$

$ \mathrm{P}^\alpha-\mathrm{P}^\beta=\delta \mathrm{I}-13 \mathrm{P} \\$

$ \mathrm{P}^4=(\mathrm{I}-\mathrm{P})^2=\mathrm{I}+\mathrm{P}^2-2 \mathrm{P} \\$

$ \mathrm{P}^4=\mathrm{I}+\mathrm{I}-\mathrm{P}-2 \mathrm{P}=2 \mathrm{I}-3 \mathrm{P} \\$

$ \mathrm{P}^8=\left(\mathrm{P}^4\right)^2=(2 \mathrm{I}-3 \mathrm{P})^2=4 \mathrm{I}+9 \mathrm{P}^2-12 \mathrm{P} \\$

$ =4 \mathrm{I}+9(\mathrm{I}-\mathrm{P})-12 \mathrm{P} \\$

$ \mathrm{P}^8=13 \mathrm{I}-21 \mathrm{P} \ldots(1) \\$

$ \mathrm{P}^6=\mathrm{P}^4 \cdot \mathrm{P}^2=(2 \mathrm{I}-3 \mathrm{P})(\mathrm{I}-\mathrm{P}) \\$

$ =2 \mathrm{I}-5 \mathrm{P}+3 \mathrm{P}^2 \\$

$ =2 \mathrm{I}-5 \mathrm{P}+3(\mathrm{I}-\mathrm{P}) \\$

$ =5 \mathrm{I}-8 \mathrm{P} \ldots(2)$

$ (1)+(2) \\$

$P^8+P^6=18 I-29 P$

(1) $-(2)$

$\mathrm{P}^8-\mathrm{P}^6=8 \mathrm{I}-13 \mathrm{P}$

From (A) $\alpha=8, \beta=6$

$\gamma=18 \\$

$\delta=8 \\$

$ \alpha+\beta+\gamma-\delta=32-8=24$
Question 3
1 / -0

If $\underset{{{{x} \rightarrow \infty}}}{\lim}\left(\frac{{x}^{2}+{x}+1}{{x}+1}-{px}-{q}\right)=-3$, then ${p}$ and ${q}$ is:

A
p = 1, q = 3

B
p = 2, q = 3

C
p = 3, q = 1

D
p = 3, q = 2

Solution

Given:

$\underset{{{{x} \rightarrow \infty}}}{\lim}\left(\frac{{x}^{2}+{x}+1}{{x}+1}-{px}-{q}\right)=-3$

On simplifying, we get:

$\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{x^{2}+x+1-p x^{2}-p x-q x-q}{x+1}\right)=-3$

$\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{x^{2}(1-p)+x(1-p-q)+1-q}{x+1}\right)=-3$

As we can see limit gives finite value. So, this is possible only when the coefficient of higher degree term will be zero.

Therefore,

$(1-p)=0$

Or, $ {p}=1$

$\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{{x}(1-{p}-{q})+1-{q}}{{x}+1}\right)=-3$ ....(1)

Dividing and multiplying by$x$ in both numerator and denominator, we get:

$\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{{x}\left[(1-{p}-{q})+\frac{(1-{q})}{{x}}\right]}{{x}\left[1+\frac{1}{{x}}\right]}\right)=-3$

$\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{\left[(1-{p}-{q})+\frac{(1-{q})}{{x}}\right]}{\left[1+\frac{1}{{x}}\right]}\right)=-3$

$\Rightarrow\left(\frac{[(1-p-q)+0]}{[1+0]}\right)=-3$

$\Rightarrow(1-p-q)=-3$

$\Rightarrow(1-1-q)=-3 \quad(\because p=1)$

$\therefore q=3$
Question 4
1 / -0

Consider the following statements:

1. The null set is a subset of every set.

2. Every set is a subset of itself.

3. If a set has $10$ elements, then its power set will have $1024$ elements.

Which of the above statements are correct?

A
1 and 2 only

B
2 and 3 only

C
1 and 3 only

D
1, 2 and 3

Solution

1. The null set is a subset of every set - The intersection of two sets is a subset of each of the original sets.

So if $\{ \}$ is the empty set and $A$ is any set then $\{ \}$ intersect $A$ is $\{ \}$ which means $\{ \}$ is a subset of $A$ and $\{ \}$ is a subset of $\{ \}$.

You can prove it by contradiction. Let's say that you have the empty set $\{ \}$ and a set $A$.

2. Every set is a subset of itself.

When we say that a set A is a subset of B, we mean that every element in A is also in B. A Venn diagram shows this nicely:

$(A\subseteq A)$

This also means that every set, A, is a subset of itself because every element of A is obviously an element of A.

3. If $n=10$ then $2^{10}=1024$

So, all three statements are true.
Question 5
1 / -0

If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $I_1=\int_a^b x \sin \left(4 x-x^2\right) d x, I_2=\int_a^b \sin \left(4 x-x^2\right) d x$, then $36 \frac{I_1}{I_2}$ is equal to :

A
72

B
88

C
80

D
66

Solution

$\begin{aligned} & \text { Equation of CE } \\ & \begin{array}{l}y-1=-(x-3) \\ x+y=4\end{array}\end{aligned}$

orthocentre lies on the line $x+y=4$

$ \text { so, } \underset{a}{b}+b=4 \\$

$ I_1=\int_a^b x \sin (x(4-x)) d x \ldots \ldots \ldots \ldots . \text { (i) }$

Using king rule

$I_1=\int_{\mathrm{a}}^{\mathrm{h}}(4-\mathrm{x}) \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx}$

$ \text { (i) }+ \text { (ii) } \\$

$ 2 I_1=\int_{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \\$

$ 2 \mathrm{I}_1=4 \mathrm{I}_2 \\$

$ \mathrm{I}_1=2 \mathrm{I}_2 \\$

$ \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \\$

$ \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72$
Question 6
1 / -0

If ${ }^{2 n} P_{3}=100^{n} P_{2}$, then the value of $n$ is:

A
1

B
50

C
13

D
4

Solution

Given:

${ }^{2 n} P_{3}=100^{n} P_{2}$

We know that:

$P_{r}=\frac{n !}{(n-r) !}$

$n !=1 \times 2 \times 3 \times \ldots \times n$

$0 !=1$

Therefore,

${ }^{2 n} P_{3}=100^{n} P_{2}$

$\Rightarrow \frac{2 n !}{(2 n-3) !}=100 \frac{n !}{(n-2) !}$

Expanding the expression on both sides of the equation, we get:

$\Rightarrow(2 n)(2 n-1)(2 n-2)=100(n)(n-1)$

$\Rightarrow 2 n-1=25, n \neq 1$

$\Rightarrow n=13$
Question 7
1 / -0

If $\mathrm{A}=\left[\begin{array}{ll}x & 2 \\ 4 & 3\end{array}\right]$ and $\mathrm{A}^{-1}=\left[\begin{array}{cc}\frac{1}{8} & \frac{-1}{12} \\ \frac{-1}{6} & \frac{4}{9}\end{array}\right]$, then find the value of ${x}$ ?

A
$\frac{28}{3}$

B
$\frac{32}{3}$

C
$\frac{34}{3}$

D
$10$

Solution

$A \times A^{-1}=1$, where $I$ is an identity matrix

$|A|=\frac{1}{\left|A^{-1}\right|}$

Given,

$\mathrm{A}=\left[\begin{array}{ll}x & 2 \\ 4 & 3\end{array}\right]$ and $\mathrm{A}^{-1}=\left[\begin{array}{cc}\frac{1}{8} & \frac{-1}{12} \\ \frac{-1}{6} & \frac{4}{9}\end{array}\right]$

$\left|\mathrm{A}^{-1}\right|=\frac{4}{72}-\frac{1}{72}=\frac{3}{72}=\frac{1}{24}$

$|\mathrm{~A}|=\frac{1}{\left|\mathrm{~A}^{-1}\right|}=24$

$\Rightarrow 3 {x}-8=24$

$\therefore {x}=\frac{32}{3}$
Question 8
1 / -0

The Cartesian equations of a line are $5 x-3=3 y+4=2 z-3$. Direction ratios of the line are:

A
$\left(\frac{1}{3},-\frac{1}{3},1\right)$

B
$\left(\frac{1}{5},\frac{1}{3},\frac{1}{2}\right)$

C
$(2,3,1)$

D
None of these

Solution

The given equation of line is:

$5 x-3=3 y+4=2 z-3$

$\Rightarrow 5\left(x-\frac{3}{5}\right)=3\left(y+\frac{4}{3}\right)=2\left(z-\frac{3}{2}\right)$

$\Rightarrow \frac{x-\frac{3}{5}}{\frac{1}{5}}=\frac{y+\frac{4}{3}}{\frac{1}{3}}=\frac{z-\frac{3}{2}}{\frac{1}{2}}$

So, the equation has Direction Ratios as $\left(\frac{1}{5},\frac{1}{3},\frac{1}{2}\right)$.
Question 9
1 / -0

Evaluate the following expression: $\sin x+\sin (x-\pi)+\sin (x+\pi)$

A
$-\sin x$

B
$\sin x$

C
$\sec x$

D
$\cos x$

Solution

Given,

$\sin x+\sin (x-\pi)+\sin (x+\pi)$

We know that,

$\sin (\pi-x)=\sin x$

$\sin (\pi+x)=-\sin x$

$\sin (-x)=-\sin x $

Therefore,

$=\sin x-\sin (\pi-x)-\sin (\pi+x)$

$=\sin x-\sin (\pi-x)-\sin x$

$=-\sin (\pi-x)$

$=-\sin x$
Question 10
1 / -0

A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan^–¹(1/2) . Water is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10m; is:

A
$\frac{1}{15 π}$

B
$\frac{1}{10 π}$

C
$\frac{2}{π}$

D
$\frac{1}{5 π}$

Solution

Given that water is poured into the tank at a constant rate of $5 \mathrm{~m}$ minute.

$\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=5 \mathrm{~m}^3 / \mathrm{min}$

Volume of the tank is,

$\mathrm{V}=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h}, \ldots . . . .(\mathrm{i})$

where $\mathrm{r}$ is radius and $\mathrm{h}$ is height at any time.

By the diagram,

$\tan \theta=\frac{\mathrm{r}}{\mathrm{h}}=\frac{1}{2} \\$

$ \Rightarrow \mathrm{h}=2 \mathrm{r} \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{2 \mathrm{dr}}{\mathrm{dt}} \ldots . . . \text { (ii) }$

Differentiate eq. (i) w.r.t. ' $t$ ', we get

$\frac{d V}{d t}=\frac{1}{3}\left(\pi 2 r \frac{d r}{d t} h+\pi r^2 \frac{d h}{d t}\right)$

Putting $\mathrm{h}=10, \mathrm{r}=5$ and $\frac{\mathrm{dV}}{\mathrm{dt}}=5$ in the above equation.

$5=\frac{75 \pi}{3} \frac{\mathrm{dh}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dh}}{\mathrm{dt}}=\frac{1}{5 \pi} \mathrm{m} / \min$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 34

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Mathematics Test - 34

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SHARING IS CARING

Question 1

\(2\)

\(3\)

\(4\)

\(6\)

Solution

Question 2

40

22

24

18

Solution

Question 3

p = 1, q = 3

p = 2, q = 3

p = 3, q = 1

p = 3, q = 2

Solution

Question 4

1 and 2 only

2 and 3 only

1 and 3 only

1, 2 and 3

Solution

Question 5

72

88

80

66

Solution

Question 6

1

50

13

4

Solution

Question 7

\(\frac{28}{3}\)

\(\frac{32}{3}\)

\(\frac{34}{3}\)

\(10\)

Solution

Question 8

\(\left(\frac{1}{3},-\frac{1}{3},1\right)\)

\(\left(\frac{1}{5},\frac{1}{3},\frac{1}{2}\right)\)

\((2,3,1)\)

None of these

Solution

Question 9

\(-\sin x\)

\(\sin x\)

\(\sec x\)

\(\cos x\)

Solution

Question 10

115π

110π

2π

15π

Solution

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$\frac{1}{15 π}$

$\frac{1}{10 π}$

$\frac{2}{π}$

$\frac{1}{5 π}$