Question

Set theory and functions doubts

Answer 1

q3,q6 in one marks
q1,q2 in 4 marks

Answer 2

@ganeshie8

Answer 3

we may use below for q3 :
if (x, y) is a point on \(f(x)\),
then (y, x) will be a point on \(f^{-1}(x)\)

Answer 4

nahi samaj aya :/ very weak in this
how to solve for R?

Answer 5

\(R = {(1, 2), (2, 3), (3, 1)} \)

swap x and y values to get its inverse :

\(R^{-1} = {(2, 1), (3, 2), (1, 3)} \)

Answer 6

to A ka use kya hua matlab?

Answer 7

acha :/ bakwas q

Answer 8

next!

Answer 9

the values used in this relation are taken from set A

Answer 10

yeah got it :|

Answer 11

q6 is equally trivial

Answer 12

just write out all the pairs (x, y) such that "y is a multiple of x"

Answer 13

for example :
{(2, 4), (2, 6)}

Answer 14

they have to be in the relation cuz, 4 = 2*2, and 6 = 2*3

Answer 15

you can find all other ordered pairs

Answer 16

okay! got it :D 4 marks one?

Answer 17

lets work Q2 first

Answer 18

knw how to interpret AxA ?

Answer 19

(1,2) x (3,4) = {(1,3),(1,4),(2,3),(2,4)} ?

Answer 20

yep ! but u need to use flower brackets for sets : {}

Answer 21

{1,2} x {3,4} = {(1,3),(1,4),(2,3),(2,4)} ?

Answer 22

ah yeah..:P

Answer 23

since (1, 2) and (3, 4) are two elements in set AxA,

the x coordinates, 1, 3 must belong to A
the y coordinates, 2, 4 must belong to A

Answer 24

so 1, 2, 3, 4 must be elements in A

Answer 25

and we're given that AxA has 16 elements - that means A must has sqrt(16) =4 elements.

so 1, 2, 3, 4 are ALL the elements in A :

A = {1, 2, 3, 4}

Answer 26

sab theek hai par B intersection C is phi :/

Answer 27

you mean Q1 ?

Answer 28

yeah in Q1..but okay 2 ke baad hi dekhenge :|

Answer 29

Q2 katam hua na ?

Answer 30

yes!

Answer 31

bakwas q again :P

Answer 32

A = {1, 2, 3, 4}, A = {1, 2, 3, 4}

you can write out AxA

Answer 33

yeah got it..

Answer 34

haha set theory appears bakwas, but its applications in number theory and other branches are very powerful

Answer 35

okay lets see Q1

Answer 36

parA - just show that both sides yield an empty set

Answer 37

partB - find AxC, and BxD
then show that one guy is a subset of the other

Answer 38

got it :O

Answer 39

q11 -> 1,2

Answer 40

range of f(x) = domain of "x = f(y)"

solve for x first

Answer 41

\(\large y = \dfrac{x^2+2x+3}{x}\)

Answer 42

\(\large xy = x^2 + 2x+3\)

\(\large x^2+(2-y)x+3 = 0\)

Answer 43

\(\large x = \dfrac{-(2-y) \pm \sqrt{(2-y)^2 - 4(1)(3)}}{2(1)}\)

Answer 44

ye solve hoga? O.o

Answer 45

clearly, x produces some value in reals ONLY when D > 0

Answer 46

D >= 0 *

Answer 47

\(\large (2-y)^2 - 4(1)(3) \ge 0\)

Answer 48

\(\large (2-y)^2 \ge 12\)

Answer 49

y^2-4y-8>=0

Answer 50

\(\large 2-y \ge \sqrt{12}\) or \(\large 2-y \le -\sqrt{12}\)

Answer 51

\(\large y \le 2 - \sqrt{12}\) or \(\large y \ge 2 + \sqrt{12}\)

Answer 52

thats the range ^^

Answer 53

nice :O

Answer 54

let me try part 2

Answer 55

okie

Answer 56

\[\Huge (- \infty , \frac{17}{8}]\]
ye hoga?

Answer 57

http://www.wolframalpha.com/input/?i=range+%28x%5E2-2%29%2F%28x%5E2-3%29

Answer 58

y <= 2/3 or y>1

Answer 59

:/

Answer 60

\[\LARGE yx^2-3y=x^2-2\]
\[\LARGE x^2(y-1)-3y+2=0\]
D>=0
\[\LARGE 9 -4(2)(y-1) \geq 0\]
8y<= 17
y<=17/8 :(

Answer 61

haha the equation is a quadratic in x, not y

Answer 62

\[\LARGE x^2(y-1)-3y+2=0\]

\(\large a = y-1\) x^2 coefficient
\(\large b = 0\) x corfficient
\(\large c = -3y+2\) constant term

Answer 63

D >= 0

\[\large 0 - 4(y-1)(-3y+2) \ge 0\]

Answer 64

\[\large 4(y-1)(-3y+2) \le 0\]

Answer 65

okay..agaya

Answer 66

thanks..now u can fly away..ill tag if any more problems :P

Answer 67

okay have fun :)

Answer 68

hogaye saare :D thanks!