(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.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(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.
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.
Consider f : R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f–1 of f given by , where R+ is the set of all non-negative real numbers.
Answer the following as true or false.
\begin{align}(i) \overrightarrow{a}\; and\; \overrightarrow{-a}\; are\; collinear.\end{align}
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Show that the Signum Function f : R → R, given by
is neither one-one nor onto.
Determine order and degree(if defined) of differential equation yn + (y')2 + 2y =0
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