JEE Advanced Level Test: Relations and Functions- 2 - JEE MCQ

R is reflexive and transitive but not symmetric 

R is symmetric and transitive but not reflexive 

a ⊥ a is not true. So, R is not reflexive
a ⊥ b and b ⊥ c does not imply a ⊥ c. So, R is not transitive
But, a ⊥ b ⇒ b ⊥ a is always true. 

(i) |a – a| = 0 < 1 is always true
(ii) a R b ⇒ |a – b| < 1 ⇒ |-(a – b)| < ⇒ |b – a| < 1 ⇒ b R a.
(iii) 2R 1 and
But, 2 is not related to 1/2. So, R is not transitive. 

Domain of sin ^-1x is [-1,1]

R = {(a, b): a = b − 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 − 2, (3, 8) ∉ R
And, as 8 ≠ 7 − 2
∴ (8, 7) ∉ R
Now, consider (6, 8).
We have 8 > 6 and also, 6 = 8 − 2.
∴ (6, 8) ∈ R

R is reflexive
∴ (3, 3), (6,6), (9, 9), (12, 12) ∈ R 
again ∴ (6, 12) ∈ R but (12, 6) ∉ R ⇒ R is not symmetric
R is transitive
[∴ (3, 6) ∈ R, (6, 12) ∈ R, (3, 12) ∈ R others are clear 

f(x) = (a – xⁿ)^1/n = y 

The function is injective (one-to-one) if every element of the codomain is mapped to by at most one element of the domain

The function is surjective (onto) if every element of the codomain is mapped to by at least one element of the domain

f(x)=2x 

Domain of f is N 

Codomain of f is N

Every element of N(codomain) is mapped to only one element in N(domain)

i.e. f(x)=f(y)

⟹2x=2y⟹x=y

Hence f is one one function

Every even element of N(codomain) has corresponding value in N(domain).

But, for any odd number in N(codomain) has no corresponding value in Ndomain.

Hence, f is not onto function.

Given, function f : R→R such that f(x) = x², 
Let A and B be two sets of real numbers.
Let x₁, x₂ ∈ A such that f(x₁) = f(x₂).
⇒1 + (x₁)² = 1 + (x₂)²
​⇒ (x₁)² − (x₂)² = 0
⇒(x₁ − x₂)(x₁ + x₂)=0
⇒ x₁ = ± x₂. 
Thus f(x₁) = f(x₂) does not imply that x₁ = x₂.
For instance, f(1) = f(−1) = 1, i.e. , two elements (1, -1) of A have the same image in B. So, f is many-one function.

⇒ x² = x²

∴ f is one-one

Let y∈R and let y=x³. then x=y^⅓ ∈ R

Thus for each y in the codomain R there exists y^1/3 in R such that 

f(y¹³)=(y¹³)³=y

∴ f is onto.

Hence f is one -one onto.

f(x) = sinx
f(x) = sinx is a one-one function.
codomain = range
therefore, f(x) = sinx is an onto function.

cos(2π - θ) = cosq ⇒ f is many-one.
Range (f) = [-1,1] ⊂ R ⇒ f is in to.

Domain = {1, 3, 2}

= cos (log x) cos (cos y)   [cos(log(x/y))+cos(log(xy))] 
= cos (log x) cos (log y)  [cos(log x - log y)+cos(log x +log y)]
= cos (log x) cos (log y) [2 cos(log  x)cos(log y)] = 0

(fof) (x) = f[f(x)] – {(3 – x³)^1/3} = f(y)  where y = (3 – x³)^1/3
= (3 – y³)^1/3 = [3- (3 – x³)]^1/3 = (x³)^1/3 = x 

(fof) (x) = f[f(x)] = f(x² - 3x + 2)² - 3(x² – 3x + 2)
= y² – 3y + 2 = (x² – 3x + 2)² – 3(x² – 3x + 2) + 2 =  (x⁴ – 6x³ + 10x² – 3x)

Since f(2) = f(3) = 1, then inverse does not exists
Inverse exist for (C) f^-1 = {(3, 1), (2, 3), (1, 2)}.

(i) xRx ⇔ 2x² - 3x.x + x²  

∴ R is reflexive
(ii) For x = 1, y= 2; 2x² - 3xy + y² = 0
∴ 1 R 2 but 2.2² - 3.2.1 + 1² = 3 ≠ 0.
So, 2 is not R-related to 1.
∴ R is not symmetric. 

Since, (3, 3), (6, 6), (9, 9), (12, 12) ∈ R ⇒ R is reflexive relation.
Now, (6, 12)∉R but (12, 6) ∉ R  ⇒ R is not a symmetric relation.
Also, (3, 6),(6, 12) ∈ R ⇒ (3, 12) ∈ R
⇒ R is transitive relation.

Clearly, (x, y)R(x, y),  
since xy = yx. This shows that R is reflexive.
Further, (x, y)R(u, v) ⇒ xv = yu ⇒ uy vx and hence (u, v)R(x, y). This shows that R is symmetric.
Similarly, (x, y)R(u, v) and (u, v)R(a, b) ⇒ xv = yu and ub = va ⇒ xb = ya and hence (x, y)R(a, b). Thus, R is transitive. Thus, R is an equivalent relation.

Since, R is a defined as aRb iff |a – b| > 0
For reflexive aRa iff |a – a| > 0
Which is not true, So, R is not reflexive
For symmetric aRb iff |a – b| > 0
Now bRa iff |b – a| > 0
⇒ |a – b| > 0  ⇒ ArB
Thus, R is symmetric.
For transitive aRb iff |a – b| > 0
bRc iff |b – c| > 0
⇒ |a – b + b – c| > 0
⇒ |a – c| > 0 ⇒ | c – a| > 0 ⇒ aRc
∴ R is transitive

If R is an equivalence relation, then R^-1 is also an equivalence relation.

Given, r = {(a, b)| a, b∈R and  as an irrational number
(i) Reflexive  which is irrational number.
(ii) Symmetric
Now,  which is not an irrational.
Also,  which is an irrational.
Which is not symmetric.
(iii) Transitive


∴ It is not transitive.