2 Marks Questions

Question 12 Marks

An integer is chosen at random from the numbers ranging from 1 to 50. What is the probability that the integer chosen is a multiple of 2 or 3 or 10?

Answer

We have to find the probability that the integer is chosen is a multiple of 2 or 3 or 10
Out of 50 integers an integer can be chosen in ${ }^{50} C _1$ ways.
Total number of elementary events $={ }^{50} C _1=50$
Consider the following events:
A = Getting a multiple of 2, B = Getting a multiple of 3 and, C = Getting a multiple of 10
Clearly, A = {2, 4,..., 50}, B = {3, 6,..., 48}, C = {10, 20,..., 50}
A n B = {6,12,..., 48}, B n C = {30}, A n C = {10, 20,..., 50} and, A n B n C = {30}
$\begin{array}{l}\therefore P(A)=\frac{25}{50}, P(B)=\frac{16}{50}, P(C)=\frac{5}{50}, P(A \cap B)=\frac{8}{50}, P(B \cap C)=\frac{1}{50} \\
P(A \cap C)=\frac{5}{50} \text { and } P(A \cap B \cap C)=\frac{1}{50}\end{array}$
$\begin{array}{l}\text { Required probability }= P ( A \cap B \cap C ) \\
= P ( A )+ P ( B )+ P ( C )- P ( A \cap B )- P ( A \cap C )- P ( B \cap A )+ P ( A \cap B \cap C ) \\
=\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{8}{50}-\frac{1}{50}-\frac{5}{50}+\frac{1}{50}=\frac{33}{50}\end{array}$

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Question 22 Marks

f, g and h are three functions defined from R to R as follows:
i. $f(x)=x^2$
ii. $g(x)=x^2+1$
iii. $h(x)=\sin x$
Then, find the range of each function.

Answer

i. Given: $f: R \rightarrow R$ such that $f(x)=x^2$
since the value of $x$ is squared, $f(x)$ will always be equal or greater than 0
$\therefore$ the range is $[0, \infty)
ii. Given: g: R \rightarrow R$ such that $g(x)=x^2+1$
since, the value of x is squared and also adding with 1, g(x) will always be equal or greater than 1
$\therefore$ Range of $g(x)=[1, \infty)$
iii. Given: $h: R \rightarrow R$ such that $h(x)=\sin x$
We know that, sin (x) always lies between -1 to 1
$\therefore$ Range of $h(x)=(-1,1)$

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Question 32 Marks

Find the domain and range of the function $f(x)=1-|x-2|$

Answer

According to the question, $f(x)=1-|x-2|$
We observe that $f(x)$ is defined for all $x \in R$. Therefore, Domain $(f)=R$
$\therefore 0 \leq|x-2|<\infty$ for all $x \in R$
$\Rightarrow-\infty<-|x-2| \leq 0$ for all $x \in R$
$\Rightarrow-\infty<1-|x-2| \leq 1$ for all $x \in R$
$\Rightarrow-\infty < f(x) \leq 1$ for all $x \in R$
$\therefore$ Range $(f)=(-\infty, 1]$

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Question 42 Marks

In what ratio is the line joining the points (2, 3) and (4, -5) divided by the line passing through the points (6, 8) and (-3, -2).

Answer

Therefore required equation of the line joining the points (6, 8) and (-3, -2) is
$y-8=\frac{-2-8}{-3-6}(x-6)$
$\Rightarrow 10 x-9 y+12=0$
Suppose 10x - 9y + 12 = 0 divide the line joining the points (2, 3) and (4, -5) at point P in the ratio k:1
$\therefore P \equiv\left(\frac{4 k+2}{k+1}, \frac{-5 k+3}{k+1}\right)$
P lies on the line 10x - 9y + 12 = 0
$\therefore 10\left(\frac{4 k+2}{k+1}\right)-9\left(\frac{-5 k+3}{k+1}\right)+12=0$
$\begin{array}{l}\Rightarrow 40 k+20+45 k-27+12 k+12=0 \\
\Rightarrow 97 k+5=0 \\
\Rightarrow k=\frac{-5}{97}\end{array}$

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Question 52 Marks

If $A=\{a, b, c, d, e\}, B=\{a, c, e, g\}$ and $C=\{b, e, f, g\}$, verify that: $(A \cap B) \cap C=A \cap(B \cap C)$

Answer

Suppose $x$ be any element of $(A \cap B) \cap C$
$\begin{array}{l}\Rightarrow x \in(A \cap B) \text { and } x \in C \\
\Rightarrow x \in A \text { and } x \in B \text { and } x \in C \\
\Rightarrow x \in A \text { and } x \in(B \cap C) \\
\Rightarrow x \in A \cap(B \cap C)\end{array}$
$\Rightarrow(A \cap B) \cap C \subset A \cap(B \cap C) \ldots( i )$
Now, suppose $x$ be an element of $A \cap(B \cap C)$
Then, $x \in A$ and $(B \cap C)$
$\begin{array}{l}\Rightarrow x \in A \text { and } x \in B \text { and } x \in C \\ \Rightarrow x \in(A \cap B) \text { and } x \in C \\
\Rightarrow x \in(A \cap B) \cap C \\
\Rightarrow A \cap(B \cap C) \subset(A \cap B) \cap C \ldots \ldots(\text { ii) }\end{array}$
Using (i) and (ii), we have $(A \cap B) \cap C=A \cap(B \cap C)$
Hence, proved.

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Question 62 Marks

Differentiate the function with respect to $x:\left(3 x^2-x+1\right)^4$.

Answer

To Find : $\frac{d}{d x}\left(3 x^2- x +1\right)^4$
Formula used: $\frac{d}{d x}\left(y^n\right)=n y^{ n -1} \times \frac{d y}{d x}$
Let $y=\left(3 x^2-x+1\right)^4$
So, by using the above formula, we have
$\frac{d}{d x}\left(3 x^2- x +1\right)^4=4\left(3 x ^2- x +1\right)^3 \times \frac{d}{d x}\left(3 x ^2- x +1\right)=4\left(3 x ^2- x +1\right)^3(6 x -1)$

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Question 72 Marks

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be a black card (i.e., a club or, a spade).

Answer

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52. Let C be the event ‘card drawn is black card’.
Since total number of black cards = 26
So, $P ( C )=\frac{26}{52}=\frac{1}{2}$
Thus, probability of a black card $=\frac{1}{2}$

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