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Let A = {1, 2, 3} and let R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. Then R is
R is reflexive and transitive but not symmetric
Let A {a, b, c} and let R = {(a, a)(a, b), (b, a)}. Then, R is
R is symmetric and transitive but not reflexive
Let A = {1, 2, 3} then total number of element in A x A is
Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b ⇔ a ⊥ b. then, R is
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.
Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ a – b < 1. Then, R is
(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 ^{1}x is [1,1]
Let R be the relation in the set N given by R = {(a, b): a = b – 2, b > 6}. Choose the correct answer.
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
Let R = {(3, 3), (6, 6), (9, 9), (3,6), (3, 9), (9, 12), (3,12), (6, 12), (12, 12)}, be a relation on the set A = {3, 6, 9, 12} Then the relation is
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
If f(x) = (a – x^{n})^{1/n}. then f(f(x)) =
f(x) = (a – x^{n})^{1/n} = y
The function is injective (onetoone) 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) = x^{2},
Let A and B be two sets of real numbers.
Let x_{1}, x_{2} ∈ A such that f(x_{1}) = f(x_{2}).
⇒1 + (x_{1})^{2} = 1 + (x_{2})^{2}
⇒ (x_{1})^{2} − (x_{2})^{2} = 0
⇒(x_{1 }− x_{2})(x_{1} + x_{2})=0
⇒ x_{1 }= ± x_{2}.
Thus f(x_{1}) = f(x_{2}) does not imply that x_{1} = x_{2}.
For instance, f(1) = f(−1) = 1, i.e. , two elements (1, 1) of A have the same image in B. So, f is manyone function.
⇒ x^{2 }= x^{2}
∴ f is oneone
Let y∈R and let y=x^{3}. then x=y^{⅓} ∈ R
Thus for each y in the codomain R there exists y^{1/3} in R such that
f(y^{13})=(y^{13})^{3}=y
∴ f is onto.
Hence f is one one onto.
f(x) = sinx
f(x) = sinx is a oneone function.
codomain = range
therefore, f(x) = sinx is an onto function.
cos(2π  θ) = cosq ⇒ f is manyone.
Range (f) = [1,1] ⊂ R ⇒ f is in to.
The domain of the function f = {(1, 3), (3, 5), (2, 6)} is
Domain = {1, 3, 2}
Let f(x) = x  1/x +1 , x ≠ 1, then f^{1} (x) is
If f(x) = x/x 1 , x ≠ 1, then f^{1} (x) is
If f(x) = cos (log x), then has the value
= 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 – x^{3})^{1/3}} = f(y) where y = (3 – x^{3})^{1/3}
= (3 – y^{3})^{1/3} = [3 (3 – x^{3})]^{1/3} = (x^{3})^{1/3} = x
If f(x) = x^{2} – 3x + 2, then (fof) (x) = ?
(fof) (x) = f[f(x)] = f(x^{2}  3x + 2)^{2}  3(x^{2} – 3x + 2)
= y^{2} – 3y + 2 = (x^{2} – 3x + 2)^{2} – 3(x^{2} – 3x + 2) + 2 = (x^{4} – 6x^{3} + 10x^{2} – 3x)
Let S = {1, 2, 3}. The function f : S → S defined as below have inverse for
Since f(2) = f(3) = 1, then inverse does not exists
Inverse exist for (C) f^{1} = {(3, 1), (2, 3), (1, 2)}.
The relation R defined on the set N of natural numbers by xRy ⇔2x^{2}  3xy + y^{2} = 0 is
(i) xRx ⇔ 2x^{2}  3x.x + x^{2}
∴ R is reflexive
(ii) For x = 1, y= 2; 2x^{2}  3xy + y^{2} = 0
∴ 1 R 2 but 2.2^{2}  3.2.1 + 1^{2} = 3 ≠ 0.
So, 2 is not Rrelated to 1.
∴ R is not symmetric.
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6) be a relation on the set A = {3, 6, 9, 12}. The relation is
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.
Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y)R(u, v) if and only if xv = yu, then
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.
If R be a relation defined as aRb iff a b> 0, then the relation is
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 of a set A, then R^{1} is
If R is an equivalence relation, then R^{1} is also an equivalence relation.
Let r be a relation from R (set of real numbers) to R defined by r = {(a, b)a, b∈R and a  b + √3 is an irrational number}. The relation r is
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.
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