→ R given by f ( x ) = sin x and g : , → R given by g ( x ) = cos x . Show that f and g are one-one, but f + g is not one-one. Solution Since for any two distinct elements x and x in , , sin x ≠ sin x and cos x ≠ cos x , both f and g must be one-one. But ( f + g ) ( ) = sin + cos = and ( f + g ) = sin cos + = .
Therefore, f + g is not one-one. Miscellaneous Exercise on Chapter . Show that the function f : R → { x ∈ R : – < x < } defined by ( ) | | f x = + , x ∈ R is one one and onto function. .
Show that the function f : R → R given by f ( x ) = x is injective. . Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B.
Is R an equivalence relation on P(X)? Justify your answer. . Find the number of all onto functions from the set { , , ,......, n } to itself.
. Let A = {– , , , }, B = {– , – , , } and f , g : A → B be functions defined by f ( x ) = x – x , x ∈ A and ( ) , g x x ∈ A. Are f and g equal? Justify your answer.
(Hint: One may note that two functions f : A → B and g : A → B