2 Marks Questions

Question 12 Marks

Is $B =\left\{ x : x ^2+2 x +1=0, x \in N \right\}$ a singleton set?

Answer

We have, $B=\left\{x: x^2+2 x+1=0, x \in N\right]$
Now, $x^2+2 x+1=0$
$\Rightarrow(x+1)^2=0$
$\Rightarrow x=-1$ which is not a natural number.
Thus. $B =\{ \}=\phi$
Hence, B is not a singleton set.

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Question 22 Marks

If A and B are two events associated with a random experiment for which P(A) = 0.60, P(A or B) = 0.85 and P(A and B) = 0.42, find P(B).

Answer

We have given that: P(A) = 0.60, P(A or B) = 0.85 and P(A and B) = 0.42
To find : $P ( B )$
Formula used: $P ( A$ or B $)= P ( A )+ P ( B )- P ( A$ and B $)$
Substituting the values in the above formula we get
$\begin{array}{l}0.85=0.60+P(B)-0.42 \\ 0.85=0.18+P(B) \\ 0.85-0.18=P(B) \\ 0.67=P(B) \\ P(B)=0.67\end{array}$

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Question 32 Marks

Find a point on the y-axis which is equidistant from A (-4, 3) and B(5, 2).

Answer

Let the point on the y-axis be P(0, y)
Here, it is given that P is equidistant from A(-4, 3) and B(5, 2).
i.e., PA = PB
$\Rightarrow \sqrt{(-4-0)^2+(3-y)^2}=\sqrt{(5-0)^2+(2-y)^2}$
Squaring both sides, we obtain
$\begin{array}{l}\Rightarrow(-4-0)^2+(3-y)^2=(5-0)^2+(2-y)^2 \\ \Rightarrow 16+9-6 y+y^2=25+4-4 y+y^2 \\ \Rightarrow 25-6 y=29-4 y\end{array}$
$\begin{array}{l}\Rightarrow 2 y =-4 \\ \Rightarrow y =-2\end{array}$
Thus, the required point on the y-axis is (0, -2).

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Question 42 Marks

Differentiate: $\left(x^2-4 x+5\right)\left(x^3-2\right)$.

Answer

To find: Differentiation of $\left(x^2-4 x+5\right)\left(x^3-2\right)$
Formula used: (i) (uv)′ = u′v + uv′ (Using Leibnitz or product rule)
(ii) $\frac{d x^n}{d x}=n x^{n-1}$
Let $u =\left( x ^2-4 x +5\right)$ and $v =\left( x ^3-2\right)$
$\begin{aligned} u^{\prime} & =\frac{d u}{d x}=\frac{d\left(x^2-4 x+5\right)}{d x}=2 x-4 \\ v^{\prime} & =\frac{d v}{d x}=\frac{d\left(x^3-2\right)}{d x}=3 x^2\end{aligned}$
Put the above obtained values in the formula:-
(uv)′ = u′v + uv′
$\left[\left(x^2-4 x+5\right)\left(x^3-2\right)\right]^{\prime}=(2 x-4) \times\left(x^3-2\right)+\left(x^2-4 x+5\right) \times\left(3 x^2\right)$
$\begin{array}{l}=2 x^4-4 x-4 x^3+8+3 x^4-12 x^3+15 x^2 \\ =5 x^4-16 x^3+15 x^2-4 x+8 .\end{array}$

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Question 52 Marks

Let R be a relation from N to N defined by $R =\left\{( a , b ): a , b \in N\right.$ and $\left.a = b ^2\right\}$. Are the following true?
(i) $( a , a ) \in R$ for all $a \in N$
(ii) $( a , b ) \in R$ implies $( b , a ) \in R$
(iii) (a, b) $\in R$, (b, c) $\in R$ implies (a, c) $\in R$
Justify your answer in each case.

Answer

Here $R =\left\{( a , b ): a , b \in N\right.$ and $\left.a = b ^2\right\}$

(i) No $(3,3) \in R$ because $3 \neq 3^2$

(ii) No. $(9,3) \in R$ but $(3,9) \in R$

(iii) No. $(81,9) \in R(9,3) \in R$ but $(81,3) \neq R$

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Question 62 Marks

If $\frac{5}{14}$ is the probability of occurrence of an event, find
i. the odds in favour of its occurrence.
ii. the odds against its occurrence.

Answer

i. We know that,
If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$
Given, probability $=\frac{5}{14}$
We know, probability of an event to occur $=\frac{a}{a+b}$
Here,a = 5 and a + b = 14 i.e. b = 9
So, $\frac{a}{a+b}=\frac{5}{14}$
odds in favor of its occurrence = a : b = 5 : 9
Conclusion: Odds in favor of its occurrence is 5: 9
ii. As we solved in part (i), a = 5 and b = 9
Also, we know, odds against its occurrence is b: a = 9: 5
Conclusion: Odds against its occurrence is 9 : 5

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Question 72 Marks

Let $R=\left\{\left(x, x^2\right): x\right.$ is a prime number less than 10$\}$
i. Write R in roster form.
ii. Find dom (R) and range (R).

Answer

i. Here we have, $\left\{\left( x , x ^2\right)\right.$ : x is a prime number less than 10$\}$.
Roster form of $R =\{(1,1),(2,4),(3,9),(5,25),(7,49)\}$
ii. The domain of R is the set of first co-ordinates of R
Domain of R = {1, 2, 3, 5, 7}
The domain of R is the set of first co-ordinates of R
Range(R) = {1, 4, 9, 25, 49}

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