Test: Relations & Functions- 1 - JEE MCQ

Let A and B be two sets. Then a relation R from set A to set B is a subset of A × B. Thus, R is a relation from A to B ⇔ R ⊆ A × B.

Here, 0 ≤ x- 3 ≤ 7 - x  
⇒0 ≤ x - 3 and x - 3 ≤ 7 - x
By solvation, we will get 3 ≤ x ≤ 5
So x = 3,4,5 find the values of ^7-x P_{x - 3} by substituting the values of x
at x = 3 ⁴P₀ = 1
at x = 4 ³P₁ = 3 
at x = 5 ²P₂ = 2

To be reflexive, a line must be perpendicular to itself, but which is not true. So, R is not reflexive
For symmetric, if  l₁ R l₂ ⇒ l₁ ⊥ l₂.
⇒  l₂ ⊥ l₁ ⇒ l₁ R l₂ hence symmetric
For transitive,  if l₁ R l₂ and l₂ R l₃
⇒ l₁ R l₂  and l₂ R l₃  does not imply that l₁ ⊥ l₃ hence not transitive.

Because, the element b in the domain A has no image in the co-domain B.

Because, f(- x) = f(x) is the necessary condition for a function to be an even function, which is only satisfied by x²+ sin²x .

Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive.

The relation { } ⊂ A x A on a is surely not reflexive. However, neither symmetry nor transitivity is contradicted. So { } is a transitive and symmetry relation on A.

A relation R on a non empty set A is said to be reflexive if fx Rx for all x ∈ R, Therefore, R is reflexive.

The domain in ordered pair (x,y) is represented by x-coordinate. Therefore, the domain of the given function is given by : {1, 3, 2}.

x - 1 ≥ 0 and 6 – x ≥ 0 ⇒ 1 ≤ x ≤ 6.

As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1, 
⇒ x = 6 ,when y = 2, 
⇒ x = 4, when y = 3, 
⇒ x = 2.
therefore domain is {2, 4, 6}.

The number of onto functions that can be defined from a finite set A containing n elements onto a finite set B containing 2 elements = 2ⁿ − 2.

A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ yRx, for all x , y ∈ R .

We have , 

Therefore, range of f(x) is {-1}.

For even function: f(-x) = f(x) , 
therefore, f(− x)
 = sin (− x)² = sin x² = f(x).

If R is a relation defined by xRy : ifx ⩽ y, then R is reflexive and transitive But, it is not symmetric. Hence, R is not an equivalence relation.

A relation R on a non empty set A is said to be transitive if fxRy and y Rz ⇒ xRz, for all x ∈ R. Here, (1, 2) and (2, 3) belongs to R implies that (1, 3) belongs to R.

A relation R on a non empty set A is said to be transitive if fx Ry and yRz ⇒ x Rz, for all x ∈ R.

We have, f(x) = |x−1|, which always gives non-negative values of f(x) for all x ∈ R.Therefore range of the given function is all non-negative real numbers i.e. [0,∞).

As the denominator of the function  is a modulus function i.e.

Correct Answer :- b

Explanation:- A = {a, b, c} and R = {(a, a), (b, b), (c, c), (b, c)}

Any relation R is reflexive if fx Rx for all x ∈ R. Here ,(a, a), (b, b), (c, c) ∈ R. Therefore , R is reflexive.

For the transitive, in the relation R there should be (a,c)

Hence it is not transitive.

A relation R on a non empty set A is said to be reflexive iff xRx for all x ∈ R . .
A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ y Rx, for all x , y ∈ R .
A relation R on a non empty set A is said to be transitive if fx Ry and y Rz ⇒ x Rz, for all x ∈ R.
An equivalence relation satisfies all these three properties.

Correct answer is D.

A polynomial function has all exponents as integral whole numbers.