Case study (4 Marks)

Question 14 Marks

Arka bought two cages of birds: Cage $-I $ contains $5$ parrots and $1$ owl and Cage $-II$ contains $6$ parrots. One day Arka forgot to lock both cages and two birds flew from Cage $-I$ to Cage $-II$ $($simultaneously$)$. Then two birds flew back from cage $-II$ to cage $-I \ ($simultaneously$)$.
Assume that all the birds have equal chances of flying.
On the basis of the above information, answer the following questions: $-$
$(i)$ When two birds flew from Cage $-I$ to Cage $-II$ and two birds flew back from Cage $-II$ to Cage $-I$ then find the probability that the owl is still in Cage $-I.$
$(ii)$ When two birds flew from Cage $-I$ to Cage $-II$ and two birds flew back from Cage $-II$ to Cage-I, the owl is still seen in Cage $-I,$ what is the probability that one parrot and the owl flew from Cage $-I$ to Cage $-II$?

Answer

Let $E_1$ be the event that the parrot and the owl fly from cage $-I$
$E_2$ be the event that two parrots fly from cage $-I$
$E_3$ be the event that the owl is still in cage $-I$
$E_3^c$ be the event that the owl is not in cage $-I$
$ n\left(E_1 \cap E_3\right)=\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_1} \times 1_{C_1}\right)=35$
$n\left(E_1 \cap E_3^c\right)=\left(5_{C_1} \times 1_{C_1}\right)\left(7_{C_2}\right)=105 $
$ n\left(E_2 \cap E_3\right)=\left(5_{C_2}\right)\left(8_{C_2}\right)=280$
$n\left(E_2 \cap E_3^c\right)=0 $
$n ( S )=35+105+280+0=420 \ ($Total sample points$)$
$(i)$ Probability that the owl is still in Cage $-I = P \left(E_3\right)= P \left(E_1 \cap E_3\right)+ P \left(E_2 \cap E_3\right)$
$=\frac{35+280}{420}=\frac{315}{420}=\frac{3}{4}$
$(ii)$ The probability that one parrot and the owl flew from Cage $-I$ to Cage $-II$ given
that the owl is still in cage $-I$ is $P\left(\frac{E_1}{E_8}\right)$
$ P\left(\frac{E_1}{E_3}\right)=\frac{P\left(E_1 \cap E_3\right)}{P\left(E_3\right)}=\frac{P\left(E_1 \cap E_3\right)}{P\left(E_1 \cap E_3\right)+P\left(E_2 \cap E_3\right)} \ ($by Baye's Theorem$)$
$=\frac{\frac{35}{420}}{\frac{315}{420}}=\frac{1}{9} $

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Question 24 Marks

Ramesh, the owner of a sweet selling shop, purchased some rectangular card board sheets of dimension 25 cm by 40 cm to make container packets without top. Let x cm be the length of the side of the square to be cut out from each corner to give that sheet the shape of the container by folding up the flaps.
Based on the above information answer the following questions.
(i) Express the volume (V) of each container as function of x only.
(ii) Find $\frac{d V}{d x}$
(iii) (a) for what value of x, the volume of each container is maximum ?
OR
(iii) (b) Check whether V has a point of inflection at x $x=\frac{65}{6}$ or not ?

Answer

(i) $V=(40-2 x)(25-2 x) x cm^3$
(ii) $\frac{d V}{d x}=4(3 x-50)(x-5)$
(iii)For extreme values $\frac{d V}{d x}=4(3 x-50)(x-5)=0$
$\begin{array}{c} \Rightarrow x=\frac{50}{3} \text { or } x=5 \\ \frac{d^2 V}{d x^2}=24 x-260 \\ \therefore \frac{d^2 V}{d x^2} \text { at } x=5 \text { is }-140<0 \end{array}$
$\therefore V$ is max when $x=5$
(iii) OR
For extreme values $\frac{d V}{d x}=4\left(3 x^2-65 x+250\right)$
$\frac{d^2 V}{d x^2}=4(6 x-65)$
$\frac{d V}{d x}$ at $x=\frac{65}{6}$ exists and $\frac{d^2 V}{d x^2}$ at $x=\frac{65}{6}$ is 0.
$\frac{d^2 V}{d x^2}$ at $x=\left(\frac{65}{6}\right)^{-}$is negative and $\frac{d^2 V}{d x^2}$ at $x=\left(\frac{65}{6}\right)^{+}$is positive
$\therefore x=\frac{65}{6}$ is a point of inflection.

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Question 34 Marks

An organization conducted bike race under $2$ different categories$-$boys and girls. In all, there were $250$ participants. Among all of them finally three from Category $1$ and two from Category $2$ were selected for the final race. Ravi forms two sets $B$ and $G$ with these participants for his college project.
Let $B =\left\{b_1, b_2, b_3\right\}, G =\left\{g_1, g_2\right\}$ where $B$ represents the set of boys selected and $G$ the set of girls who were selected for the final race.
Ravi decides to explore these sets for various types of relations and functions.
On the basis of the above information, answer the following questions:
$(i)$ Ravi wishes to form all the relations possible from $B$ to $G$. How many such relations are possible?
$(ii)$ Among these relations, how many are functions from $B$ to $G$ ?
$OR$
$(iii) (b)$ If the track of the final race $($for the biker $b_1)$ follows the curve
$x^2=4 y ;($ where $0 \leq x \leq 20 \sqrt{2} \ 0 \leq y \leq 200)$,then state whether the track represents a one$-$one and onto function or not. $($Justify$).$

Answer

$(i)$ Number of relations is equal to the number of subsets of the set $B \times G=2^{n(B \times G)}$
$=2^{n(B) \times n(G)}=2^{3 \times 2}=2^6$ ( Where $n (A)$ denotes the number of the elements in the finite set $A$ )
$(ii)$ Number of functions $=$
$\text { (number of elements in the co-domain) }{ }^{\text {number of elements in the domain }}=(n(G))^{n(B)}=2^3$
$(iii) (a)$ Number of one-one functions $=0$, as number of elements in the co$-$domain $(G)$ is less than the number of elements in the domain $(B)$.
number of onto functions $=2^3-2=6$
Total number of functions $=$
$\text { (number of elementsin the co-domain) }{ }^{\text {number of elements in the domain }}=(n(G))^{n(B)}=2^3$
Out of which there are two possibilities in which all the elements in the domain set $(B)$ is mapped to the one element the co-domain set $(G)$

So, number of onto functions $=2^{ 3 }-2=6$
OR
$(iii) (b)$ One$-$one and onto function
$x^2=4 y . \text { let } y=f(x)=\frac{x^2}{4}$
Let $x_1, x_2 \in[0,20 \sqrt{2}]$ such that $f\left(x_1\right)=f\left(x_2\right) $
$\Rightarrow \frac{x_1{ }^2}{4}=\frac{x_1{ }^2}{4}$
$\Rightarrow x_1^2=x_2^2 $
$\Rightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)=0 $
$\Rightarrow x_1=x_2\text { as } x_1, x_2 \in[0,20 \sqrt{2}]$
$\therefore f$ is one$-$one function
Now, $0 \leq y \leq 200$ hence the value of $y$ is nonnegative
and $f(2 \sqrt{y})=y$
$\therefore$ for any arbitrary $y \in[0,200]$, the pre image of $y$ existsin $[0,20 \sqrt{2}]$ hence $f$ is ontofunction.

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MATHS Case study (4 Marks)

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