(i) Let A = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.
Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.
=> (5, 6), (6, 5) ∈ R, but (5, 5) ∉ R
∴R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
(ii) Consider a relation R in R defined as:
R = {(a, b): a < b}
For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself. In fact, a = a.
∴ R is not reflexive.
Now,
(1, 2) ∈ R (as 1 < 2)
But, 2 is not less than 1.
∴ (2, 1) ∉ R
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
⇒ a < b and b < c
⇒ a < c
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.
(iii) Let A = {4, 6, 8}.
Define a relation R on A as:
A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}
Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.
Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.
Hence, relation R is reflexive and symmetric but not transitive.
(iv) Define a relation R in R as:
R = {a, b): a3 ≥ b3}
Clearly (a, a) ∈ R as a3 = a3.
∴ R is reflexive.
Now,
(2, 1) ∈ R (as 23 ≥ 13)
But,
(1, 2) ∉ R (as 13 < 23)
∴ R is not symmetric.
Now,
Let (a, b), (b, c) ∈ R.
⇒ a3 ≥ b3 and b3 ≥ c3
⇒ a3 ≥ c3
⇒ (a, c) ∈ R
∴ R is transitive.
Hence, relation R is reflexive and transitive but not symmetric.
(v) Let A = {−5, −6}.
Define a relation R on A as:
R = {(−5, −6), (−6, −5), (−5, −5)}
Relation R is not reflexive as (−6, −6) ∉ R.
Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.
It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.
∴ The relation R is transitive.
Hence, relation R is symmetric and transitive but not reflexive.
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(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto.
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Represent graphically a displacement of 40 km, 30° east of north.
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A balloon, which always remains spherical, has a variable diameter
\begin{align} \frac{3}{2}(2x+1)\end{align}
Find the rate of change of its volume with respect to x.
y = x2 + 2x + C : y' - 2x - 2 = 0
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Classify the following measures as scalars and vectors.
(i) 10 kg (ii) 2 metres north-west (iii) 40°
(iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2
The total revenue in Rupees received from the sale of x units of a product is given by
R (x) = 3x2 + 36x + 5. The marginal revenue, when x = 15 is
(A) 116 (B) 96 (C) 90 (D) 126
in part v set is trans. then (-6,-5) & (-5,-6) both are in relation
Thanks for the help
In v. If -6,-6 belongs to R then it will be reflexive (a,a) belongs to R therefore v answer is correct
Try to improve much more
I think, it is correct because (-6,-6) does not belongs to relation set R. Properties of Relation is A realtion R on set A is reflexive if aRa for all a belongs to A i.e. is (a,a) belongs to R for all a belongs to R => each element a of A is related to itself. Ex: Let A = {a,b} and R = {(a,a),(a,b),(b,a)} then R is reflexive as aRa belongs to R but it is not reflexive for pair (b,b) does not belongs to R.
plz check part v it does not seems correct as -6,-6 doesnot belongs to R