Mathematics Test - 12

Calculation:

Given: f(x) = px + q and g(x) = mx + n

f (g(x)) = g (f(x))

⇒ f (mx + n) = g (px + q)

⇒ p (mx + n) + q = m (px + q) + n

⇒ pmx + pn + q = pmx + mq + n

⇒ pn + q = mq + n

⇒ f (n) = g (q)

∴ Option 3 is correct answer.

Concept:

Concept:

For any two non-empty sets A and B, we have:

I. A X B = {(a, b) | a ∈ A and b ∈ B}

II. B X A = {(b, a) | a ∈ A and b ∈ B}

III. Any two ordered pairs (a, b) = (c, d) if and only if a = c and b = d.

Calculation:

Given: A = {X: X ∈ N, –1 < X < 8} and B = {Y: Y ∈ N, –1 < Y < 1}.

Consider, A = {X: X ∈ N, –1 < X < 8} = {1, 2, 3, 4, 5, 6, 7}

Now, B = {Y: Y ∈ N, –1 < Y < 1} = Ⴔ

Hence, we get

⟹ A × B = {1, 2, 3, 4, 5, 6, 7} × Ⴔ

⟹ A × B = Ⴔ

Hence, A × B = Ⴔ.

Concept:

One -One Function / Injective Function:

A function f : A → B is said to be an one - one function, if different elements in A have different f - images in B.

i.e if f (x₁) = f (x₂) ⇒ x₁ = x₂, ∀ x₁, x₂ ∈ A

Onto Function / Surjective Function:

A function f : A → B is said to be an onto function if each element in B has at least one pre-image in A.

Bijective Function:

A function f : A → B is said to be a bijective function if it is both one -one and onto function.

Calculation:

Given: f : N → N and is defined as f (x) = x + 1

One - One:

Let f (x₁) = f (x₂)

⇒ x₁ + 1 = x₂ + 1

⇒ x₁ = x₂

Hence, the given function f is one - one.

Onto:

Let y = f(x) = x + 1

⇒ x = y -1

But If y = 1 ∈ N then x = 0 ∉ N.

Hence, all the elements of co - domain i.e N does not has a pre - image in the domain i.e N.

Concept:

An operation * on a non-empty set S, is said to be a binary operation if it satisfies the closure property.

Closure Property:

Let S be a non-empty set and a, b ∈ S, if a * b ∈ S for all a, b ∈ S then S is said to be closed with respect to operation *.

Calculation:

Statement 1: Subtraction on the set of all natural numbers is a binary operation.

Let a = 1, b = 2 ∈ N and here the operation is '-'

⇒ a - b = 1 - 2 = - 1 ∉ N

So, we can say that N is not closed with respect to the operation '-'

Hence, statement 1 is false.

Statement 2: Subtraction on the set of all integers is a binary operation.

Let a, b ∈ Z and here the operation is '-'

As we know that the resultant of subtraction of two integers is again an integer.

i.e a - b ∈ Z ∀ a, b ∈ Z.

So, Z is closed with respect to the operation '-'

Hence, statement 2 is true.

So, option B is correct answer.

Concept:

Inverse Relation:

Let R be a relation form A to B. Then the inverse relation, denote by R^-1, from the set B to A, is defined as, R^-1 = {(b, a) : (a, b) ∈ R}.

Calculation:

Given: A = {2, 4, 5}, B = {5, 9, 11} and R is the relation from A to B such that a R b ⇔ b = 2a + 1.

So, when a = 2 ⇒ b = 2a + 1 = 5 ∈ B ⇒ (2, 5) ∈ R

When a = 4 then b = 2a + 1 = 9 ∈ B ⇒ (4, 9) ∈ R

When a = 5 then b = 2a + 1 = 11 ∈ B ⇒ (5, 11) ∈ R

⇒ R = {(2, 5), (4, 9), (5, 11)}

As we know that if R is a relation form A to B. Then the inverse relation, denote by R^-1, from the set B to A, is defined as, R^-1 = {(b, a) : (a, b) ∈ R}.

⇒ R^-1 = {(5, 2), (9, 4), (11, 5)}

Concept:

Let * be a binary operation on a non-empty set S. If there exists an element e in S such that a * e = e * a = a ∀ a ∈ S. Then the element e is said to be an identity element of S with respect to *.

Calculation:

Given: A = {1, - 1} and * is a operation on A such that a * b = ab ∀ a, b ∈ A.

Let e be the identity element of A with respect to *.

As we know that if e is an identity element of a non-empty set S with respect to a binary operation * then a * e = e * a = a ∀ a ∈ S.

Let a ∈ A and because e is the identity element of A with respect to given operation *.

.e a * e = a =  e * a ∀ a ∈ A.

According to the definition of *, we have

⇒ a * e = ae = a

⇒ e = 1 ∈ A

Hence, 1 is the identity element of A with respect to given operation *.

Concept:

Smallest integer function:(Ceiling function)

The function f (x) = [x] is called the smallest integer function and the value of [x] will always be lesser than or equal to the inner function.

⇒ [x] ≥ x.

If f :A → B and g : C → D. Then (fog) (x) will exist if and only if the co-domain of g = domain of f i.e D = A and (gof) (x) will exist if and only if co-domain of f = domain of g i.e B = C.

Here, the codomain of f is Z and the domain of g is also Z so  (gof) (x) is defined.

The graph of the smallest integer function,

Calculation:

Given: f(x) = [x], where [.] denotes smallest integer function and g(x) = 5x 

Here, we have to find the value of g o f(1/2)

⇒ g o f(1/2) = g( f(1/2))

∵ f(x) = [x] so, f(1/2) =  [1/2] = [0.5] = 1

⇒ g o f(1/2) = g(1)

∵ g(x) = 5^x so, g(1) = 5¹ = 5

Hence, g o f(1/2) = 5

Concept:

The function f(x) is said to be one to one function if 

f(a) = f(b) ⇒ a = b,   for every a, b 

Calculations:

The function f(x) is said to be one to one function if 

f(a) = f(b) ⇒ a = b,   for every a, b 

consider the co-domain {1, 2, 3}.

'1' can be related to any of the 5 numbers.
'2' can be associated with any of the other four numbers as '1' has already been associated to one of them.
leaving us with 4 options to associate 2.

finally '3' can be associated with the remaining 3 numbers.
Hence total number of ways of doing so =5 × 4 × 3
= 60

Hence, The number of one-to-one functions from {1, 2, 3} to {1, 2, 3, 4, 5} is 60

Concept:

Definition of Equivalence relation

• A relation is an equivalence relation if & only if it is:
• Reflexive property: aRa
• Symmetric property: if a R b, then b R a
• Transitive property: if a R b and b R c, then a R c

Calculation:

Given that,

S = persons living in Delhi x, y ∈ S.

x R y if x & y born on same day.

⇒ First we check reflexive property so,

x R x or y R y both will be reflective.

So reflexive property true for Relation.

⇒ Now symmetric property,

Given that x R y born on same day so it’s reverse y R x also true y and x also born on same day.

Symmetric property also true for relation.

⇒ For transitive property,

Given that x R y born on same day

Let’s take z who also born on same day of y so y R z ∈ S

So we can say that x R z ∈ S because given that x R y and y R z ∈ S.

Transitive property also true for relation.

Concept:

One–One Function / Injective Function:
A function f: A → B is said to be a one–one function, if different elements in A have different images or associated with different elements in B i.e if

f (x1) = f (x2) ⇒  x1 = x2, ∀ x1, x2 ∈ A.

Into Function:

Any function f: A → B is said to be into function if there exist at least one element in B which does not has a pre-image in A, then the function f is said to be into function.

i.e If Range of function f ⊂ Co-domain of function f, then f is into

Many-one Function:

Any function f: A → B is said to be many-one, if two (or more than two) distinct elements in A have same images in B.

Onto Function / Surjective Function :

Any function f: A → B is said to be onto if every element in B has atleast one pre-image in A.

i.e If Range of function f = Co-domain of function f, then f is onto

Calculation:

Given: f: R → R is a function such that f(x) = [x], where [.] denotes greatest integer function

Let x₁ = 1. 2 and x₂ = 1.3

Now according to the definition of f(x), we get

⇒ f(x₁) = [1.2] = 1 and f(x₂) = [1.3] = 1

As we can see that, f(x1) = f(x2) but x1 ≠ x₂

So, the given function is many-one function.

Clearly, we know that there is no real number x such that f(x) = [x] = 1.1

So, we have atleast one element in the codomain of the given function which does not has a pre-image in the given domain.

So, the given function is an into function.

Hence, the given function is a many-one and into function.