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MATHS Case study (4 Marks)

Model Paper 8 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 3 questions.

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Case study (4 Marks)

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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,
$\begin{array}{l}a-2=b-1 \\ \therefore a-b=1 \ldots(1)\end{array}$
Comparing y coordinates of both the sides, we get,
$\begin{array}{l}2 b+1=a+2 \\ \therefore a-2 b=-1 \ldots(2)\end{array}$
Subtracting equation (2) from (1), we get, 
$\begin{array}{l}(a-a)+(-b-(-2 b))=1-(-1) \\ \therefore(-b+2 b)=1+1 \\ \therefore
b=2\end{array}$
Put this value in equation (1), we get,
$\begin{array}{l}a-2=1 \\ \therefore a=3\end{array}$
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Question 24 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 
$S =\left\{\begin{array}{l}A B C D, A B D C, A C B D, A C D B, A D B C, A D C B \\ B A C D, B A D C, B C A D, B C D A, B D A C, B D C A \\ C A B D, C A D B, C B A D, C B D A, C D A B, C D B A, \\ D A B C, D A C B, D C A B, D C B A, D B A C, D B C A\end{array}\right\}$
$\begin{array}{l} E _1=\{ ABCD , ABDC , ACBD , ACDB , ADBC , ADCB , CABD , CADB , CDAB , DABC , DACB , DCAB \} \\ \Rightarrow n \left( E _1\right)=12 \\ \therefore P (\text { she visits A before } B )=P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{12}{24}=\frac{1}{2}\end{array}$
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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Question 34 Marks
Answer
$\begin{array}{l}\text { i. }(x+i y)(2-3 i)=4+i \\ \quad 2 x-(3 x) i+(2 y) i-3 y i^2=4+i \\ 2 x+3 y+(2 y-3 x) i=4+i\end{array}$
Comparing the real & imaginary parts, 
$\begin{array}{l}2 x+3 y=4 \ldots \text { (i) } \\ 2 y-3 x=1 \ldots \text { (ii) }\end{array}$
Solving eq (i) & eq (ii), 4x + 6y = 8 
$\begin{array}{l}-9 x+6 y=3 \\ 13 x=5 \Rightarrow x=\frac{5}{13} \\ y=\frac{14}{13} \\ \therefore(x, y)=\left(\frac{5}{13}, \frac{14}{13}\right)\end{array}$
$\begin{array}{l}\text { ii. } x+i y=\frac{(1+i)^2}{2-i} \\ x+i y=\frac{(1+i)^2}{2-i}=\frac{1+2 i+i^2}{2-i}=\frac{2 i}{2-i}=\frac{2 i(2+i)}{(2-i)(2+i)}=\frac{4 i+2 i^2}{4-i^2} \\ =\frac{4 i-2}{4+1}=\frac{-2}{5}+\frac{4 i}{5} \\ \Rightarrow x=\frac{-2}{5}, y=\frac{4}{5} \Rightarrow x+y=\frac{-2}{5}+\frac{4}{5}=\frac{2}{5}\end{array}$
$\begin{array}{l}\text { iii. We have }\left(\frac{1-i}{1+i}\right)^{100}=a+b i \\ \Rightarrow\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^{100}=a+b i \\ \Rightarrow\left(\frac{1+i^2-2 i}{1-i^2}\right)^{100}=a+b i \\ \Rightarrow\left(\frac{1-1-2 i}{1+1}\right)^{100}=a+b i \\ \Rightarrow\left(\frac{-2 i}{2}\right)^{100}=a+b i \\ \Rightarrow(-i)^{100}=a+b i \\ \Rightarrow i^{100}=a+b i \\ \Rightarrow\left(i^4\right)^{25}=a+b i \\ \Rightarrow(1)^{25}=a+b i \\ \Rightarrow 1=a+b i \\ \Rightarrow 1+0 i=a+b i\end{array}$
Comparing the real and imaginary parts, 
We have a = 1, b = 0
Hence (a, b) = (1, 0) 
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Case study (4 Marks) - MATHS STD 11 Science Questions - Vidyadip