2 Marks Questions

Question 12 Marks

Find the coordinates of the point which divides the join of $A(-5, 11)$ and $B(4, -7)$ in the ratio $2 : 7.$

Answer

Let $P(x, y)$ be the point that divides the join of $A(-5, 11)$ and $B(4, -7)$ in the ratio $2 : 7$
We know that: If $m_1: m_2$ is the ratio in which the join of two points is divided by another point $(x, y)$,
then $x =\frac{m_1 x_2+m_2 x_1}{m_1+m_2}$
$y =\frac{m_1 y_2+m_2 y_1}{m_1+m_2}$
Here, $x_1=-5, x_2=4, y_1=11$ and $y_2=-7$
Substituting,we get
$x=\frac{2 \times 4+7 \times-5}{2+7}$
$x=\frac{8-35}{9}$
$x=\frac{-27}{9}$
$\Rightarrow x=-3$
$y=\frac{2 \times-7+7 \times 11}{2+7}$
$y=\frac{-14+77}{9}$
$y=\frac{63}{9}$
$\Rightarrow y=8$
Thus, the coordinates of the point which divided the join of $A(-5, 11)$ and $B(4, -7)$ in the ratio $2 : 7$ is $(-3, 8).$

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Question 22 Marks

Is $A =\{ x : x \in N , 1< x \leq 2\}$ null set?

Answer

We know that,Natural numbers = 1, 2, 3, 4, 5, 6, 7,..
Natural number greater than 1 (1 < x) = 2, 3, 4, 5....
Natural number less than or equal to $2(x \leq 2)=2$
$\Rightarrow$ one element in this set
$\therefore$ It is not a null set.

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Question 32 Marks

If $A$ and $B$ are two events associated with a random experiment such that $P ( A )=0.25, P ( B )=0.4$ and $P ( A$ or $B )= 0.5 ,$ find the values of
$i. P ( A$ and B $)$
$ii. P ( A$ and $\bar{B})$

Answer

$i.$ It is given that
$: P ( A )=0.25, P ( A$ or B $)=0.5$ and $P ( B )=0.4$
To find $: P(A$ and $B)$
Formula used : $P(A$ or $B) = P(A) + P(B) - P(A$ and $B)$
Substituting the value in the above formula we get,
$0.5 = 0.25 + 0.4 - P(A$ and $B)$
$0.5 = 0.65 - P(A$ and $B)$
$P(A$ and $B) = 0.65 - 0.5$
$P(A$ and $B) = 0.15$
$ii.$ Given : $P(A) = 0.25, P(A$ and $B) = 0.15 ($ from part $(i))$
To find: $P ( A$ and $\bar{B})$
To find : $P ( A$ and $\bar{B})$
Formula used : $P ( A$ and $\vec{B})= P ( A )- P ( A$ and $B )$
Substituting the value in the above formula we get,
$ P ( A \text { and } \bar{B})=0.25-0.15$
$P ( A \text { and } \bar{B})=0.10$
$P ( A \text { and } \bar{B})=0.10$

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Question 42 Marks

Two dice are thrown simultaneously. Find the probability of getting a total of at least $10.$

Answer

We know that in a single throw of two dice, the total number of possible outcomes is $(6 \times 6)=36$.
Let $S$ be the sample space of the event and is given by
$n(S) = 36.$
Let $E_5=$ event of getting a total of at least $10$ . Then,
$E_5=$ event of getting a total of $10$ or $11$ or $12=((4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}$.
$\Rightarrow n \left( E _5\right)=6$
$\therefore P \left( E _3\right)=\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

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Question 52 Marks

Evaluate: $\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}$.

Answer

We have: $\underset{{\theta \rightarrow 0}}{\lim}\left[\frac{1-\cos (4 \theta)}{1-\cos 6 \theta}\right]$
$\underset{{\theta \rightarrow 0}}{\lim}\left[\frac{2 \sin ^2 2 \theta}{2 \sin ^2 3 \theta}\right]\left\{\because 1-\cos A =2 \sin ^2\left(\frac{A}{2}\right)\right\}$
$\underset{{\theta \rightarrow 0}}{\lim}\left[\frac{\sin ^2 2 \theta}{(2 \theta)^2} \times \frac{(2 \theta)^2}{\frac{\sin ^2 3 \theta}{(33)^2} \times(3 \theta)^2}\right]$
$\underset{{\theta \rightarrow 0}}{\lim}\left[\left(\frac{\sin 2 \theta}{2 \theta}\right)^2 \times\left(\frac{3 \theta}{\sin 3 \theta}\right)^2 \times \frac{4}{9}\right]$
$=\frac{4}{9}\left[\because \lim _{X \rightarrow 0} \frac{\sin x}{x}=1\right]$

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Question 62 Marks

Find the domain of $f(x)=\frac{1}{x+2}$.

Answer

Here $f(x)=\frac{1}{x+2}$
f(x) assume real values for all real values of x except for x + 2 = 0 i.e. x = - 2.
Thus domain of $f(x)=R-\{-2\}$.

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Question 72 Marks

If $A$ and $B$ are any two non-empty sets, then prove that $A \times B=B \times A \Leftrightarrow A=B$

Answer

First, let $A = B$. Then, we have to prove that $A \times B=B \times A$
Now, $A = B$
$\Rightarrow A \times B=A \times A \text { and } B \times A=A \times A$
$\Rightarrow A \times B=B \times A$
Conversely, let $A \times B=B \times A$. Then, we have to prove that $A = B$.
Let x be an arbitrary element of A. Then,
$x \in A$
$\Rightarrow (x, b) \in A \times B \text { for all } b \in B .$
$\Rightarrow (x, b) \in B \times A$
$\Rightarrow x \in B$
$\therefore A \subseteq B$
Let y be an arbitrary element of A. Then,
$y \in B$
$\Rightarrow (a, y) \in A \times B \text { for all } a \in A$
$\Rightarrow (a, y) \in B \times A$
$\Rightarrow y \in A$
$\therefore B \subseteq A$
Hence, A = B.
Hence, $A \times B=B \times A \Leftrightarrow A=B$

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