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.
y = ex +1 : yn -y' = 0
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(y). Use one-one ness of f).
Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π (B) 12π (C) 8π (D) 11π
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.