Mathematics Test

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Mathematics Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Find the vector equation of the line passing through the point with position vector $\hat{i}-2 \hat{j}+5 \hat{k}$ and perpendicular to the plane $\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0$.

A
$\vec{r}=(\hat{i}+2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})$

B
$\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}+3 \hat{j}-\hat{k})$

C
$\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})$

D
$\vec{r}=(\hat{i}+2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}+3 \hat{j}+\hat{k})$

Solution

The vector equation of a line passing through a point with position vector $\overrightarrow{r_{1}}$ and parallel to $\vec{m}$ is given by:

$\vec{r}=\overrightarrow{r_{1}}+\lambda \vec{m}$ where, $\lambda$ is a scalar

Given here:

Equation of plane is $\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0$ and the line is perpendicular to the given plane.

It is given that, the required line is perpendicular to the plane $\vec{r} \cdot(2 \hat{i}-3 \hat{j}-\hat{k})=0$

i.e., The required line is parallel to the vector $\vec{m}=2 \hat{i}-3 \hat{j}-\hat{k}$

The line passes through the point whose position vector is $\hat{i}-2 \hat{j}+5 \hat{k}$ i.e., $\overrightarrow{r_{1}}=\hat{i}-2 \hat{j}+5 \hat{k}$

As we know that, the vector equation of a line passing through a point with position vector $\overrightarrow{r_{1}}$ and parallel to $\vec{m}$ is given by:

$\vec{r}=\overrightarrow{r_{1}}+\lambda \vec{m}$

So, the required equation of line is:

$\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})$
Question 2
1 / -0

What is the probability of solving a given problem if three students $(A, B$ and $C)$, try it independently, with respective probabilities $\frac{4}{7}, \frac{3}{8}$ and $\frac{1}{2}$?

A
$\frac{97}{112}$

B
$\frac{95}{112}$

C
$\frac{97}{111}$

D
$\frac{97}{125}$

Solution

Given,

$P(A)=\frac{4}{7}, P(B)=\frac{3}{8}$ and $P(C)=\frac{1}{2}$

We know that,

$P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)$

$\Rightarrow P(A \cap B)=P(A) \times P(B)$

So, the probability of the question getting solved is

$P(A \cup B \cup C)=\frac{4}{7}+\frac{3}{8}+\frac{1}{2}-\left(\frac{4}{7} \times \frac{3}{8}\right)-\left(\frac{3}{8} \times \frac{1}{2}\right)-\left(\frac{1}{2} \times \frac{4}{7}\right)+\left(\frac{4}{7} \times \frac{3}{8} \times \frac{1}{2}\right)$

$\Rightarrow P(A \cup B \cup C)=\frac{4}{7}+\frac{3}{8}+\frac{1}{2}-\left(\frac{3}{14}\right)-\left(\frac{3}{16}\right)-\left(\frac{2}{7}\right)+\left(\frac{3}{28}\right)$

$\Rightarrow P(A \cup B \cup C)=\frac{81}{56}-\left(\frac{3}{14}\right)-\left(\frac{3}{16}\right)-\left(\frac{2}{7}\right)+\left(\frac{3}{28}\right)$

$\Rightarrow P(A \cup B \cup C)=\frac{81}{56}+\frac{3}{28}-\left(\frac{24+21+32}{112}\right)$

$\Rightarrow P(A \cup B \cup C)=\frac{87}{56}-\frac{77}{112}$

$\Rightarrow P(A \cup B \cup C)=\frac{174-77}{112}$

$\Rightarrow P(A \cup B \cup C)=\frac{97}{112}$
Question 3
1 / -0

If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$, find the angle between $\vec{a}$ and $\vec{b}$.

A
$\frac{π}{2}$

B
$\frac{π}{3}$

C
$\frac{π}{6}$

D
$\frac{π}{4}$

Solution

Consider, the angle between $\vec{a}$ and $\vec{b}$ is $\theta$
Given, $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$
$\Rightarrow \vec{a}+\vec{b}=-\vec{c}$
$\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|-\overrightarrow{\mathrm{c}}|$
Squaring on both side, we get
$\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|-\overrightarrow{\mathrm{c}}|^{2}$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{b}}|^{2}=|-\overrightarrow{\mathrm{c}}|^{2}$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+2 \mathrm{ab} \cos \theta=|-\overrightarrow{\mathrm{c}}|^{2}$
$\left.\Rightarrow(3)\right|^{2}+(5)^{2}+2(3)(5) \cos \theta=(7)^{2}$
$\Rightarrow 30 \cos \theta=15$
$\Rightarrow \cos \theta=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{3}$
Hence, If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0},|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7,$ then the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$
Question 4
1 / -0

Given that, E = {2, 4, 6, 8, 10}, if n represents a member (component) of E, write the set containing all the numbers represented by (n + 1):

A
{5, 7, 9, 11,12}

B
{3, 5, 7, 9, 11}

C
{3, 5, 9, 11,12}

D
{3, 5, 7, 9, 15}

Solution

Given,

E = {2, 4, 6, 8, 10}

Let, A = {x | x = n + 1, n ∈ E}

Thus, x = 3, for 2 ∈ E

X = 5, for 4 ∈ E etc

So, A = {3, 5, 7, 9, 11}
Question 5
1 / -0

If $y=\frac{4+3 x^{3}}{6 x^{4}}$, then find the value of $\frac{d y}{d x}$.

A

$-\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+3\right)$

B

$-\frac{1}{6 x^{4}}\left(\frac{16}{x}+3\right)$

C

$-\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+1\right)$

D

$\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+3\right)$

Solution

Given that:

$y=\frac{4+3 x^{3}}{6 x^{4}}$

$=\frac{4}{6 x^{4}}+\frac{3 x^{3}}{6 x^{4}}$

$=\frac{2}{3 x^{4}}+\frac{1}{2 x}$

$y=\frac{2}{3} x^{-4}+\frac{1}{2} x^{-1}$

As we know that:

If $y=a x^{-n}$

Then,

$ \frac{d y}{d x}=a\left(-n x^{-n-1}\right)$

Therefore,

$\frac{d y}{d x}=\frac{2}{3}\left(-4 x^{-5}\right)+\frac{1}{2}\left(-1 x^{-2}\right)$

$=-\frac{8}{3} x^{-5}-\frac{1}{2} x^{-2}$

$=-\frac{8}{3 x^{5}}-\frac{1}{2 x^{2}}$

$=-\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+3\right)$
Question 6
1 / -0

A man starts repaying his loan as first installment of Rs.1000 . If he increases the installment by Rs. 50 every month. What amount will he pay in the 20th installment ?

A
1900

B
1950

C
2000

D
1850

Solution

Given,

a = 1000

d = 50

To Find: What amount will he pay in the 20^th instalment

T_n = a_n = a + ( n - 1 ) d

Now a₂₀ = a + 19d

= 1000 + 19 × 50

= 1000 + 950

= 1950
Question 7
1 / -0

If the distances of P (x, y) from A (4, 1) and B (-1, 4) are equal, then which of the following is true?

A
3x = 2y

B
5x = 3y

C
x = 2y

D
x = y

Solution

Concept:

The distance 'd' between two points (x₁, y₁) and (x₂, y₂) is obtained by using the Pythagoras' Theorem: d² = (x₁ - x₂)² + (y₁ - y₂)².

Calculation:

Using the formula for the distance between two points:

AP² = (x - 4)² + (y - 1)²

BP² = (x + 1)² + (y - 4)²

Since the distances are equal, we have:

AP² = BP²

⇒ (x - 4)² + (y - 1)² = (x + 1)² + (y - 4)²

⇒ x² - 8x + 16 + y² - 2y + 1 = x² + 2x + 1 + y² - 8y + 16

⇒ 10x = 6y

⇒ 5x = 3y
Question 8
1 / -0

There are $49$ cards in a box numbered from $1$ to $49$. Every card is numbered with only $1$ number. Probability of picking up a card, the number printed on which is a multiple of $5$ but not that of $10$ or $15$ is:

A
$\frac{2}{49}$

B
$\frac{3}{49}$

C
$\frac{4}{49}$

D
$\frac{5}{49}$

Solution

Given:

In a box there are $49$ card from number $1$ to $49$ .

According to the qustion,

Card numbers which are multiples of $5$ are $5,10,15,20,25,30,35,40,45$

Card numbers which are multiples of $10$ or $15$ are $10,15,20,30,40,45$

The latter ones are to be ruled out from the former ones.

So, the only card numbers we are left with are $5,25,35$.

$\therefore$ Probability of picking up a card, the number on which is a multiple of $5$ but not that of $10$ and $15=\frac{3}{49}$
Question 9
1 / -0

$\lim _{x \rightarrow 0}\left\{\frac {\left(a^{x}-b^{x}\right)}{ x}\right\} \text { is equal to }$

A

$\mathrm {log \space a}$

B

$\mathrm {log \space b}$

C

$\log (\frac {a}{b})$

D
Does not exist

Solution

$\lim _{x \rightarrow 0}\left\{\frac {\left(a^{x}-b^{x}\right)}{x}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac {\left(a^{x}-b^{x}-1+1\right)}{x}\right\}$
$=\lim_{x\rightarrow0}\frac {(a^x-1)-(b^x-1)}{x}$
$=\lim_{x\rightarrow0}\frac {(a^x-1)}{x}-\frac {(b^x-1)}{x}$
$=\lim _{x \rightarrow 0}\frac {\left(a^{x}-1\right)}{x}-\lim _{x \rightarrow 0}\frac {\left(b^{x}-1\right)}{x}$
$=\log a-\log b$
$=\log (\frac {a}{b})$
Question 10
1 / -0

A sequence $b_{0}, b_{1}, b_{2} \ldots$ is defined by letting $b_{0}=5$ and $b_{k}=4+b_{k-1}$ for all natural numbers $k$. Show that $b_{n}=5+4 n$ for all:

A
Natural number

B
Real number

C
Complex number

D
Composit number

Solution

Given:

We have, sequence $b_{0}, b_{1}, b_{2} \ldots$ is defined by letting $b_{0}=5$ and $b_{k}=4+b_{k-1}$ for all natural numbers $\mathrm{k}$.

Let $P(n): b_{n}=5+4 n$, for all natural numbers

For $n=1$

$ b_{1}=5+4 \times 1=9$

Also $b_{0}=5$

$\therefore b_{1}=4+b_{0}=4+5=9$

Thus, $P(1)$ is true.

Now, let us assume that $P(n)$ is true for some natural number $n=m$.

$\therefore P(m): b_{m}=5+4 m$.......(1)

Now, to prove that $P(k+1)$ is true, we have to show that:

$P(m+1): b_{m+1}=5+4(m+1)$

$b_{m+1}=4+b_{m+1-1}\left(\right.$ As $\left.b_{k}=4+b_{k-1}\right)$

$=4+b_{m}$

$=4+5+4 m=5+4(m+1) \quad[.......\text{Using}(1)]$

So, $P(m+1)$ is true whenever $P(m)$ is true. So, by the principle of mathematical induction $P(n)$ is true for any natural number $n$.

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Mathematics Test - 16

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Mathematics Test - 16

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Question 1

\(\vec{r}=(\hat{i}+2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})\)

\(\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}+3 \hat{j}-\hat{k})\)

\(\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}-3 \hat{j}-\hat{k})\)

\(\vec{r}=(\hat{i}+2 \hat{j}+5 \hat{k})+\lambda \cdot(2 \hat{i}+3 \hat{j}+\hat{k})\)

Solution

Question 2

\(\frac{97}{112}\)

\(\frac{95}{112}\)

\(\frac{97}{111}\)

\(\frac{97}{125}\)

Solution

Question 3

π2

π3

π6

π4

Solution

Question 4

{5, 7, 9, 11,12}

{3, 5, 7, 9, 11}

{3, 5, 9, 11,12}

{3, 5, 7, 9, 15}

Solution

Question 5

\(-\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+3\right)\)

\(-\frac{1}{6 x^{4}}\left(\frac{16}{x}+3\right)\)

\(-\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+1\right)\)

\(\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+3\right)\)

Solution

Question 6

1900

1950

2000

1850

Solution

Question 7

3x = 2y

5x = 3y

x = 2y

x = y

Solution

Question 8

\(\frac{2}{49}\)

\(\frac{3}{49}\)

\(\frac{4}{49}\)

\(\frac{5}{49}\)

Solution

Question 9

\(\mathrm {log \space a}\)

\(\mathrm {log \space b}\)

\(\log (\frac {a}{b})\)

Does not exist

Solution

Question 10

Natural number

Real number

Complex number

Composit number

Solution

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$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{4}$