Case study (4 Marks)

Question 14 Marks

Two complex numbers $Z _1= a + ib$ and $Z _2= c + id$ are said to be equal, if $a = c$ and $b = d$.
$i.$ If $(x+i y)(2-3 i)=4+i$ then find the value of $(x, y)$.
$ii.$ If $\frac{(1+i)^2}{2-i}=x+i y$, then find the value of $x+y. (1)$
$iii.$ If $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$, then find the values of $a$ and $b. (2)$
$OR$
If $(a-2,2 b+1)=(b-1, a+2)$, then find the real values of $a$ and $b. (2)$

Answer

Given, $(a - 2, 2b + 1) = (b - 1, a + 2)$
Comparing $x$ coordinates of both the sides, we get,
$a-2=b-1$
$\therefore a-b=1 \ldots(1)$
Comparing $y$ coordinates of both the sides, we get,
$2 b+1=a+2$
$\therefore a-2 b=-1 \ldots(2)$
Subtracting equation $(2)$ from $(1),$ we get,
$(a-a)+(-b-(-2 b))=1-(-1)$
$\therefore(-b+2 b)=1+1$
$\therefore b=2$
Put this value in equation $(1),$ we get,
$a-2=1$
$\therefore a=3$

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Question 24 Marks

On her vacation, Priyanka visits four cities. Delhi, Lucknow, Agra, Meerut in a random order.

$i$. What is the probability that she visits Delhi before Lucknow? $(1)$
$ii$. What is the probability she visit Delhi before Lucknow and Lucknow before Agra? $(1)$
$iii.$ What is the probability she visits Delhi first and Lucknow last? $(2)$
OR
What is the probability she visits Delhi either first or second? $(2)$

Answer

$\text { i. }(x+i y)(2-3 i)=4+i$
$ 2 x-(3 x) i+(2 y) i-3 y i^2=4+i$
$2 x+3 y+(2 y-3 x) i=4+i$
Comparing the real imaginary parts,
$2 x+3 y=4 \ldots \text { (i) }$
$2 y-3 x=1 \ldots \text { (ii) }$
Solving eq $(i)$ eq $(ii), \ 4x + 6y = 8$
$-9 x+6 y=3$
$13 x=5 $
$\Rightarrow x=\frac{5}{13}$
$y=\frac{14}{13}$
$\therefore(x, y)=\left(\frac{5}{13}, \frac{14}{13}\right)$
$\text { ii. } x+i y=\frac{(1+i)^2}{2-i}$
$x+i y=\frac{(1+i)^2}{2-i}$
$=\frac{1+2 i+i^2}{2-i}=\frac{2 i}{2-i}$
$=\frac{2 i(2+i)}{(2-i)(2+i)}=\frac{4 i+2 i^2}{4-i^2}$
$=\frac{4 i-2}{4+1}=\frac{-2}{5}+\frac{4 i}{5}$
$\Rightarrow x=\frac{-2}{5}, y=\frac{4}{5} $
$\Rightarrow x+y=\frac{-2}{5}+\frac{4}{5}=\frac{2}{5}$
$iii$. We have $\left(\frac{1-i}{1+i}\right)^{100}=a+b i$
$\Rightarrow\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^{100}=a+b i$
$\Rightarrow\left(\frac{1+i^2-2 i}{1-i^2}\right)^{100}=a+b i$
$\Rightarrow\left(\frac{1-1-2 i}{1+1}\right)^{100}=a+b i$
$\Rightarrow\left(\frac{-2 i}{2}\right)^{100}=a+b i$
$\Rightarrow(-i)^{100}=a+b i$
$\Rightarrow i^{100}=a+b i$
$\Rightarrow\left(i^4\right)^{25}=a+b i$
$\Rightarrow(1)^{25}=a+b i$
$\Rightarrow 1=a+b i$
$\Rightarrow 1+0 i=a+b i$
Comparing the real and imaginary parts,
We have $a = 1, b = 0$
Hence $(a, b) = (1, 0)$

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Question 34 Marks

Method to Find the Sets When Cartesian Product is Given
For finding these two sets, we write first element of each ordered pair in first set say $A$ and corresponding second element in second set $B\ ($say$)$.
Number of Elements in Cartesian Product of Two Sets
If there are $p$ elements in set $A$ and $q$ elements in set $B$, then there will be pq elements in $A \times B$ i.e. if $n(A)=p$ and $n(B)=q$, then $n(A \times B)=p q$.
$i$. The Cartesian product $A \times A$ has $9$ elements among which are found $(-1,0)$ and $(0,1)$. Find the set $A$ and the remaining elements of $A \times A. (1)$
$ii. A$ and $B$ are two sets given in such a way that $A \times B$ contains $6$ elements. If three elements of $A \times B$ are $(1, 3 ), (2,5)$ and $(3, 3),$ then find the remaining elements of $A \times B. (1)$
$iii$. If the set $A$ has $3$ elements and set $B$ has $4$ elements, then find the number of elements in $A \times B$. $(2)$
OR
If $A \times B=\{(a, 1),(b, 3),(a, 3),(b, 1),(a, 2),(b, 2)\}$. Find $A$ and $B .(2)$

Answer

$i.$ Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4 !$ i.e. $24$
$\therefore A=\{a, b\}$ and $B=\{1,2,3\}$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
Let $E _1$ be the event that Priyanka visits $A$ before $B$ .
Then,
$E_1=\{A B C D, A B D C, A C B D, A C D B, A D B C, A D C B,$
$ C A B D, C A D B, C D A B, D A B C, D A C B, D C A B\}$
$\Rightarrow n \left( E _1\right)=12$
$\therefore P ($ she visits $A$ before $B )= P \left( E _1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}$
$ii$. Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4!$ i.e. $24$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
$ E _1=\{ ABCD , ABDC , ACBD , ACDB , ADBC , ADCB ,$
$ CABD , CADB , CDAB , DABC , DACB , DCAB \}$
$\Rightarrow n \left( E _1\right)=12$
$\therefore P ($she visits $A$ before $ B )=P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}$
$iii$. Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4!$ i.e. $24$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
Let $E _3$ be the event that she visits $A$ first and $B$ last.
Then,
$E_3=\{ ACDB , ADCB \}$
$n\left(E_3\right)=2$
$\because P($ she visits $A$ first and $B$ last $)=P\left(E_3\right)$
$=\frac{n\left(E_3\right)}{n(S)}=\frac{2}{24}=\frac{1}{12}$
OR
Let the Priyanka visits four cities Delhi, Lucknow, Agra, Meerut are respectively $A, B, C$ and $D$.
Number of way's in which Priyanka can visit four cities $A, B, C$ and $D$ is $4!$ i.e. $24$
$\therefore n ( S )=24$
Clearly, sample space for this experiment is
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
Let $E _4$ be the event that she visits $A$ either first or second.
Then, $E_4=\{A B C D, A B D C, A C B D, A C D B, A D B C, A D C B,$
$ B A C D, B A D C, C A B D, C A D B, D A B C, D A C B\}$
$\Rightarrow n\left(E_4\right)=12$
Hence $, P ($she visits $A$ either first or second$)$
$=P\left(E_4\right)=\frac{n\left(E_4\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}$

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Maths Case study (4 Marks)

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