MTH 202 Assigement Solution BY Rehan Ahmed vuskg

ID

 = Question No 1:


Sol:


P = The number is divisible by 12~P = The number is not divisible by 12 q= The number is divisible by 3 and 4.~q = The number is not divisible by 3 and 4.

Meanings	Symbols	Called
not	~	Tilde
if and only if	↔	Double arrow
Therefore	∨	Vel

p	q	~P	~q	~P↔~q	P ∨ q
T	T	F	F	T	T
T	F	F	T	F	T
F	T	T	F	F	T
F	F	T	T	T	F

Test the truth table is valid or invalid:An argument is valid if the conclusion is true when all the premises are true. An argument is invalid if the conclusion is false when all the premises are true.
Premises           Premises         Premises   conclusion↓          ↓         ↓        ↓

~P↔~q	P ∨ q	p	q
T	T	T	T
F	T	T	F
F	T	F	T

T

F

F

F

←   Critical Row




In the First critical row, the conclusion is True when the premises ~P↔~q, P ∨ q , p are true. Therefore, the argument is valid.

Question No 2:


Sol:

f (x) =
x + 1
x + 2
a)  Find Domain and range of f
Find Domain?



Set the denominator in
x + 1
x + 2
equal to 0 to Find where the expression is undefined.X + 2 ¹ 0
Subtract 2 from both sides of the equation

Domain
X ¹ -2


R-{-2}
or(−∞,−2)∪(−2,∞)
Find Range?Let:
x + 1
y =   x + 2x+1 = xy+2y x –xy = 2y -1 x(1-y) = 2y-1 2 y -1x = 1 - y
So the   1-y  ¹ 0
Y  ¹ 1
R-{-1}or(−∞,1)∪(1,∞)


b) Determine whether
1) f is injective
2) f is surjective
3) f is bijective

1) f is injective  

f (x1 ) =x1  + 1 =x1 + 2
f (x2 )x2  + 1x2  + 2

(x1 +1)(x2 + 2) = (x2 +1)(x1 + 2)x1 x2 + 2x1 + x2 + 2 = x2 x1 + 2x2 + x1 + 2 2x1 + x2 = 2x2 + x12x1 - x1 = 2x2 - x2x1 = x2So, the function is injective / one to one

2) f is surjective  
y Î R, x Î R

x + 1
x + 2 = yx + 1 = y(x+2) x = xy + 2y - 1 x - xy = 2y - 1 x(1 - y) = 2y – 1x = 2 y - 1
1 - y
So, the function is not surjective.

3) f is bijective
A function f: X→Y that is both one-to-one (injective) and onto (surjective) is called a bijective function

So, the Function is not bijective