Mth202 Guess papers
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Laws of Logic.
1. Cumulative Law
p∧q = q∧ p
p∨q = q∨ p
2. Associative Law
( p∧q)∧r = p∧(q∧r)
( p∨q)∨r = p∨(q∨r)
3. Distributive Law
p∧(q∨r)=(p∧q)∨( p∧r)
p∨(q∧r)=( p∨q)∧(p∨r)
4. Idempotent Law
p∧ p = p
p∨ p = p
5. De-Morgans Law
~ (p∧q)=~ p∨ ~ p
~ (p∨q)=~ p∧~ p
6. Absorption Law
p∨( p∧q)= p
p∧( p∨q)= p
7. Identity Law
p∧t = p
p∨c = p
8. Universal Bond Law
p∨t =t
p∧c =c
9. Negation Law
p∨ ~ p =t (Tautology)
p∧~ p =c (Contradiction)
10. Negation of t and c
~t =c
~ c =t
11. Double Negation Law
~ (~ p)= p
Set Identities.
Let A,B, and C be subsets of a Universal set U. then
1. Idempotent Law:
a. A∪A = A , b. A∩A = A
2. Commutative Law:
a. A∪B = B∪A , b. A∩B = B∩A
3. Associative Law:
a. A∪(B∪C)=(A∪B)∪C ,
b. A∩(B∩C)=(A∩B)∩C
4. Distributive Law:
a. A∪(B∩C)=(A∪B)∩(A∪C) ,
b. A∩(B∪C)=(A∩B)∪(A∩C)
5. Identity Laws:
a. A∪ ∅ = A , b. A∩ ∅ = ∅
c. A∪ U = U , d. A∩ U = A
6. Complement Laws:
a. A∪ Ac = U, b. A∩ Ac = ∅
c. Uc = ∅ , d. ∅c = U
7. Double Complement Law:
a. (Ac)c = A,
8. De-Morgan’s Law:
a. ( A∪B)c = Ac ∩Bc , b. (A∩B)c = Ac ∪Bc
9. Alternate Representation for Set Difference:
A− B = A∩Bc
10. Subset Laws:
a. A∪B ⊆ C iff A ⊆ C and B ⊆ C
c. C ⊆ A∩B iff C ⊆ A and C ⊆ B
11. Absorption Laws:
a. A∪( A∩B) = A, b. A∩( A∪B) = A
Implications(Conditional)& Bi-Conditional
Implication (Conditional) →
1- a. p → q ≡ ~ q → ~ p, b. p → q ≡ ~ p ∨ q
2- if p → q , then its inverse is ~ p→ ~ q
3- if p → q , then its converse is q → p
→ operator is not a commutative coz p → q ≠ q → p
4- if p → q , then its contra positive is ~ q→ ~ p
so p → q ≡ ~ q→ ~ p, so implication is equivalent to its
contra positive.
Bi-Conditional ↔
1- p ↔ q ≡ ( p → q ) ∧ ( q → p )
2- ~ p ↔ q ≡ p ↔ ~ q
Laws of Logic.
1- Commutative Law : p ↔ q ≡ q ↔ p
2- Implication Law: p → q ≡ ~ p ∨ q
≡ ~ ( p ∧ ~ q)
3- Exportation Law: ( p∧ q ) → r ≡ p → ( q→ r )
4- Equivalence: p ↔ q ≡ ( p → q ) ∧ ( q → p )
5- Reductio ad absurdum: p → q ≡ ( p ∧ ~ q)→ c
1.
Solution:
f: R R
g: R R
f(x) = x2 g(x) = 3x + 1
(fog)(x) = f [g(x)]
= f [3x + 1]
= (3x + 1)2
= 9x2 + 6x + 1
(gof)(x) = g [f(x)]
= g [x2]
= 3(x2) + 1
= 3x2 + 1
We observe that fog gof, that is the commutative law does not hold for the composition of
functions.
2.
Solution:
f: R R
f(x) = x2 - 3x + 2
(fof)(x) = f [f(x)]
= f (x2 - 3x + 2)
= (x2 - 3x + 2)2 - 3(x2 - 3x +2) + 2
= x4 + 9x2 + 4 - 6x3 - 12x + 4x2
- 3x2 + 9x - 6 + 2
= x4 - 6x3 + 10x2 - 3x.
3.
f: R R defined by f(x) = x2 + 3x + 1
g: R R defined by g(x) = 2x - 3
i. fog(x) = f [g(x)]
= f [2x - 3]
= (2x - 3)2 + 3(2x - 3) + 1
= 4x2 - 12x + 9 + 6x - 9 + 1
= 4x2 - 6x + 1
ii. gof(x) = g [f(x)]
= g [x2 + 3x + 1]
= 2 (x2 + 3x + 1) - 3
= 2x2 + 6x + 2 - 3
= 2x2 + 6x – 1
iii. fof(x) = f [f(x)]
= f [x2 + 3x + 1]
= (x2 + 3x + 1)2 + 3(x2 + 3x + 1) + 1
= x4 + 11x2 + 1 + 6x3 + 6x + 3x2 + 9x + 3 + 1
= x4 + 6x3 + 14x2 + 15x + 5
iv. gog(x) = g [g(x)]
= g [2x - 3]
= 2 (2x - 3) - 3
= 4x - 6 - 3
= 4x - 9
4.
f: R R defined by f(x) = x + 1
g: R R defined by g(x) = x - 1
fog(x) = f [g(x)]
= f (x - 1)
= (x - 1) + 1
= x
gof (x) = g [f(x)]
= g [x + 1]
= (x + 1) -1
= x
fog = gof = IR
IR is the identity function.
IR(x) = x.
5.
f: N Z0 defined by f(x) = 2x
g: Z0 Q defined by g(x) = 1/x
h: Q R defined by h(x) = 5x + 2
To verify associativity we have to prove that ho(gof) = (hog)of.
Consider
[ho(gof)](x) = h [(gof) (x)]
= h [g (f (x))]
= h [g (2x)]
= h [1/2x]
= 5 * 1/2x + 2
= 5/2x + 2
[(hog)of] (x) = (hog) [f(x)]
= (hog) (2x)
= h [g (2x)]
= h (1/2x)
= 5 * 1/2x + 2
= 5/2x + 2
6.
f: R R is the identity function
f(x) = x
fof (x) = f [f(x)]
= f (x)
= x
(ff) (x) = f (x) * f (x)
= x * x
= x2
What is the smallest integer N such that
a. ⎡N/7⎤ = 5 b. ⎡N/9⎤ = 6
a. N = 7 ⋅ (5 – 1) + 1 = 7 ⋅ 4 + 1 = 29
b. N = 9 ⋅ (6 – 1) + 1 = 9 ⋅ 5 + 1 = 46
EXAMPLE:
Use the Euclidean algorithm to find gcd(330, 156)
1.Divide 330 by 156:
This gives 330 = 156 · 2 + 18
2.Divide 156 by 18:
This gives 156 = 18 · 8 + 12
3.Divide 18 by 12:
This gives 18 = 12 · 1 + 6
4.Divide 12 by 6:
This gives 12 = 6 · 2 + 0
Hence gcd(330, 156) = 6.
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