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MATHS Case study (4 Marks)

Model Paper 7 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 3 questions.

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Case study (4 Marks)

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3 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
The conjugate of a complex number $z$, is the complex number, obtained by changing the
sign of imaginary part of z . It is denoted by $\bar{z}$.
The modulus (or absolute value) of a complex number, z = a + ib is defined as the non
negative real number
$\sqrt{a^2+b^2}$. It is denoted by |z|. i.e.
$|z|=\sqrt{a^2+b^2}$
Multiplicative inverse of z is $\frac{\bar{z}}{|z|^2}$. It is also called reciprocal of z .
$z \bar{z}=|z|^2$
i. If $f(z)=\frac{7-z}{1-z^2}$, where $z=1+2 i$, then find $|f(z)|$. ( 1 )
ii. Find the value of $(z+3)(\bar{z}+3)$. (1)
iii. If ( $x-i y)(3+5 i)$ is the conjugate of $-6-24 i$, then find the value of $x+y$. (2)
OR
If $z=3+4 i$, then find $\bar{z}$. (2)
Answer
i. Let z = 1 + 2i
$\Rightarrow|z|=\sqrt{1+4}=\sqrt{5}$
Now, $f ( z )=\frac{7-z}{1-z^2}=\frac{7-1-2 i}{1-(1+2 i)^2}$
$\begin{array}{l}=\frac{6-2 i}{1-1-4 i 2-4 i}=\frac{6-2 i}{4-4 i} \\ =\frac{(3-i)(2+2 i)}{(2-2 i
(2+2 i)} \\ =\frac{6-2 i+6 i-2 i^2}{4-4 i^2}=\frac{6+4 i+2}{4+4} \\ =\frac{8+4 i}{8}=1+\frac{1}{2} i\end{array}$
$f(z)=1+\frac{1}{2}$
$\therefore|f( z )|=\sqrt{1+\frac{1}{4}}=\sqrt{\frac{4+1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$
Given that: $( z +3)(\bar{z}+3)$
Let z = x + yi
So $( z +3)(\bar{z}+3)=( x + yi +3)( x - yi +3)$
$\begin{array}{l}=[(x+3)+y i][(x+3)-y i] \\ =(x+3)^2-y^2 i^2 \\ =(x+3)^2+y^2 \\ =|x+3+i y|^2 \\
=|z+3|^2\end{array}$
iii. The conjugate of -6 - 24i is -6 + 24i.
It is given that -6 + 24i = (x – iy) (3 + 5i)
$-6+24 i=3 x+5 x i-3 i y-5 y i^2$
-6 + 24i = (3x + 5y) + i(5x - 3y)
Comparing the real and imaginary parts,
3x + 5y = -6  
5x - 3y = 24
Solving these two equations we get x = 3 and y = -3.
Therefore, x = 3 and y = -3
Then x + y = 3 - 3 = 0
OR
z = 3 + 4i
$\Rightarrow \bar{z}=3-4 i$
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Question 24 Marks
Answer
i. Let $E_1$ and $E_2$ denotes the events that Ankit and Vinod will respectively qualify the exam.
$\begin{array}{l}P\left(E_1 \cup E_2\right)= P \left( E _1\right)+ P \left( E _2\right)- P \left( E _1 \cap E _2\right) \\ =0.05+0.10-0.02=0.13\end{array}$
iii. Let $E_1$ and $E_2$ denotes the events that Ankit and Vinod will respectively qualify the exam.
$\begin{array}{l}=P\left(E_1^{\prime} \cap E_2^{\prime}\right)=P\left(\left(E_1 \cup E_2\right)^{\prime}\right) \\ =1- P \left(E_1 \cup E_2\right)=1-0.13=0.87\end{array}$
OR
Let $E_1$ and $E_2$ denotes the events that Ankit and Vinod will respectively qualify the exam.
The probability that Vinod will not qualify the exam. 
Probability that only one of them will qualify the exam $= P \left(\left( E _1- E _2\right) \cup\left( E _2- E _1\right)\right)$
$\begin{array}{l}=P\left(E_1-E_2\right)+P\left(E_2-E_1\right) \\ =P\left(E_1 \cup E_2\right)-P\left(E_1 \cap E_2\right) \\ =0.13-0.02=0.11\end{array}$
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Question 34 Marks
Representation of a Relation
A relation can be represented algebraically by roster form or by set-builder form and visually it can be represented by an arrow diagram which are given below
i. Roster form In this form, we represent the relation by the set of all ordered pairs belongs to R.
ii. Set-builder form In this form, we represent the relation $R$ from set $A$ to set $B$ as $R=\{(a, b)$ : $a \in A, b \in B$ and the rule which relate the elements of A and B ).
iii. Arrow diagram To represent a relation by an arrow diagram, we draw arrows from first element to second element of all ordered pairs belonging to relation R. 
Questions:
i. If n(A) = 3 and B = {2, 3, 4, 6, 7, 8} then find the number of relations from A to B. (1)
ii. If A = {a, b} and B = {2, 3}, then find the number of relations from A to B. (1)
iii. If A = {a, b} and B = {2, 3}, write the relation in set-builder form. (2) 
OR
Express of R = {(a, b): 2a + b = 5; a, b ∈ W} as the set of ordered pairs (in roster form). (2)
Answer
i. Number of relations $=2^{ mn }$
$=2^{3 \times 6}=2^{18}$
ii. Number of relations $=2^{ mn }$
$=2^{2 \times 2}=2^4=16$
iii. $R=\{(x, y): x \in P, y \in Q$ and $x$ is the square of $y\}$
|OR
Here, W denotes the set of whole numbers.
We have $2 a + b =5$ where $a , b \in W$
$\therefore a=0 \Rightarrow b=5$
$\Rightarrow a=1 \Rightarrow b=5-2=3$
and $a =2 \Rightarrow b=1$
For a > 3, the values of b given by the above relation are not whole numbers.
$\therefore A =\{(0,5),(1,3),(2,1)\}$
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