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Case study (4 Marks)

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3 questions · timed · auto-graded

Question 14 Marks
Answer
$i =|Z|=2 \cdot \sqrt{2}$
x = 2, y = -2 
$\begin{array}{l}\cos \theta=\frac{z}{r}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}} \\ \sin \theta=\frac{y}{r}=\frac{-2}{2 \sqrt{2}}=\frac{-1}{\sqrt{2}}\end{array}$
$\operatorname{Arg}(Z)=\frac{-\pi}{4}$
ii. $\bar{z} \bar{z}=|z|^2=(2 \sqrt{2})^2=8$
iii. $|Z|=\sqrt{2^2+(-2)^2}$
$=\sqrt{8}=2 \sqrt{2}$
OR
Real part of 2 - 2i = 2
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Question 24 Marks
Answer
i. Total marbles = 4 + 5 + 3 = 12
Required probability $=\frac{{ }^4 C_2}{{ }^{12} C_2}=\frac{\frac{4 \times 3}{2 \times 1}}{\frac{12 \times 11}{2 \times 1}}=\frac{1}{11}$
ii. Total marbles = 4 + 5 + 3 = 12
Required probability $=\frac{{ }^3 C_3}{{ }^{12} C_3}=\frac{1}{\frac{12 \times 11 \times 10}{3 \times 2}}=\frac{1}{220}$
iii. Total marbles = 4 + 5 + 3 = 12
Required probability $=\frac{{ }^7 C_2}{{ }^{12} C_2}=\frac{\frac{7 \times 6}{2 \times 1}}{\frac{12 \times 11}{2 \times 1}}=\frac{21}{66}=\frac{7}{22}$
OR
Total marbles = 4 + 5 + 3 = 12
Required probability = 1 - P (None is blue)
$\begin{array}{l}=1-\frac{{ }^7 C_3}{{ }^{12} C_3} \\ =1-\frac{\frac{7 \times 6 \times 5}{3 \times 2}}{\frac{13 \times 11 \times 10}{3 \times 2}} \\ =1-\frac{7}{44}=\frac{37}{44}\end{array}$

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Question 34 Marks
Answer
i. Number of functions from $A$ to $B$ are $n(B)^{n(A)}=q^p$
ii. Number of relations from A to B is $2^{ n (A) n(B)}=2^{\text {pA }}$.
iii. Figures A and B show relations. Figure C shows a function but not a relation.
OR
x is a factor of y .
$1,2,4$ and 8 are factors of 8 .
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Case study (4 Marks) - MATHS STD 11 Science Questions - Vidyadip