A = {1, 2, 3, 4, 5}
R = { (a,b) ; |a – b| is even}
It is clear that for any element a ∈A, we have |a -a| = 0(which is even).
∴R is reflexive.
Let (a, b) ∈ R.
=> |a –b| is even.
=> |- (a –b)| = |b - a| is also even.
=> (b, a) ∈ R is even.
A = {1, 2, 3, 4, 5}
R = { (a, b) : | a – b| is even}
It is clear that for any element a ∈A, we have |a - a | = 0 (which is even).
∴R is reflexive.
Let (a, b) ∈ R.
⇒ |a –b| is even.
⇒ |- (a –b)| = |b - a| is also even.
⇒ (b, a) ∈ R is even.
∴R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ |a –b| is even and |(b –c)| is even.
⇒ (a – b) is even and (b –c ) is even.
⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]
⇒ |a – c | is even.
⇒ (a, c) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
∴R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ |a –b| is even and |(b –c)| is even.
⇒ (a – b) is even and (b –c ) is even.
⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]
⇒ |a – c | is even.
⇒ (a, c) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
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 : N → N be defined by
State whether the function f is bijective. Justify your answer.
The total revenue in Rupees received from the sale of x units of a product is given by
R (x) = 13x2 + 26x + 15
Find the marginal revenue when x = 7.
y = x2 + 2x + C : y' - 2x - 2 = 0
Let f: X → Y be an invertible function. Show that the inverse of f –1 is f, i.e., (f–1)–1 = f.
y = ex +1 : yn -y' = 0
Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto,where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R* ?