Let f : R → R be defined by f x = x 2 1 + x 2 . Prove that f is neither injective nor surjective.

Given f : R → R and f x = x 2 1 + x 2 For one-one function, Let x 1 , x 2 ∈ R , then f x 1 = f x 2 ⇒ x 1 2 1 + x 1 2 = x 2 2 1 + x 2 2 ⇒ x 1 2 1 + x 2 2 = x 2 2 1 + x 1 2 ⇒ x 1 2 + x 1 2 x 2 2 = x 2 2 + x 1 2 x 2 2 ⇒ x 1 2 = x 2 2 ⇒ x 1 = ± x 2 ⇒ f is not one-one function For onto function, Let, y = x 2 1 + x 2 ⇒ y 1 + x 2 = x 2 ⇒ x 2 = y 1 - y ≥ 0 ⇒ y y - 1 ≤ 0 ⇒ y ∈ [ 0 , 1 ) Range of f ≠ The codomain of f ⇒ f is not onto function. Therefore, f is neither one-one nor onto function.

Let f : R → R be defined by f x = x 2 1 + x 2 . Prove that f is neither injective nor surjective.

Given f : R → R and f x = x 2 1 + x 2 For one-one function, Let x 1 , x 2 ∈ R , then f x 1 = f x 2 ⇒ x 1 2 1 + x 1 2 = x 2 2 1 + x 2 2 ⇒ x 1 2 1 + x 2 2 = x 2 2 1 + x 1 2 ⇒ x 1 2 + x 1 2 x 2 2 = x 2 2 + x 1 2 x 2 2 ⇒ x 1 2 = x 2 2 ⇒ x 1 = ± x 2 ⇒ f is not one-one function For onto function, Let, y = x 2 1 + x 2 ⇒ y 1 + x 2 = x 2 ⇒ x 2 = y 1 - y ≥ 0 ⇒ y y - 1 ≤ 0 ⇒ y ∈ [ 0 , 1 ) Range of f ≠ The codomain of f ⇒ f is not onto function. Therefore, f is neither one-one nor onto function.

If the function f : R → A given by f x = x 2 x 2 + 1 is surjection, then find A .

A function f : R → A , is given by f x = x 2 x 2 + 1 ∵ f is onto ⇒ Codomain of f = Range of f = A For Range, Let y = x 2 x 2 + 1 ⇒ y + x 2 y = x 2 ⇒ x 2 = y 1 - y ≥ 0 ⇒ y y - 1 ≤ 0 ⇒ y ∈ [ 0 , 1 ) ∴ A = [ 0 , 1 )

If the function f : R → A given by f x = e x - e - x e x + e x is a surjection, then find A .

A function f : R → A , is given by f x = e x - e - x e x + e x ∵ x = x , x ≥ 0 - x , x < 0 If x ≥ 0 , f x = e x - e - x e x + e x = e x - e - x 2 e x Let, y = e 2 x - 1 2 e 2 x ⇒ 2 y e 2 x = e 2 x - 1 ⇒ e 2 x 2 y - 1 = - 1 ⇒ e 2 x = 1 1 - 2 y ⇒ 2 x = log e 1 1 - 2 y ⇒ x = 1 2 log e 1 1 - 2 y ≥ 0 ⇒ log e 1 1 - 2 y ≥ 0 ⇒ 1 1 - 2 y ≥ e 0 ⇒ 1 1 - 2 y ≥ 1 ⇒ 1 1 - 2 y - 1 ≥ 0 ⇒ 1 - 1 + 2 y 1 - 2 y ≥ 0 ⇒ 2 y 2 y - 1 ≤ 0 ⇒ y ∈ [ 0 , 1 2 ) , for x ≥ 0 If x < 0 , f x = e x - e x e x + e - x = 0 Thus, Range of f = R f = 0 ∪ [ 0 , 1 2 ) = [ 0 , 1 2 ) ∴ A = [ 0 , 1 2 )

Let f x = a x 3 + b x 2 + c x + d sin x , a ≠ 0 . Find the condition that f x is always one-one function.

f x is always one-one. ⇒ f is either strictly increasing or strictly decreasing. f x = a x 3 + b x 2 + c x + d sin x f ' x = 3 a x 2 + 2 b x + c + d cos x f is strictly increasing function then f ' x > 0 . ⇒ 3 a x 2 + 2 b x + c + d cos x > 0 , cos x ∈ - 1 , 1 ⇒ 3 a x 2 + 2 b x + c - d > 0 , d > 0 ⇒ 3 a > 0 and D < 0 ⇒ a > 0 and 4 b 2 - 12 a c - d < 0 ⇒ a > 0 and b 2 < 3 a c - d . . . i If f is a strictly decreasing function then f ' x < 0 . ⇒ 3 a x 2 + 2 b x + c + d cos x < 0 , cos x ∈ - 1 , 1 ⇒ 3 a x 2 + 2 b x + c + d < 0 , d > 0 ⇒ 3 a < 0 and D < 0 ⇒ a < 0 and 4 b 2 - 12 a c + d < 0 ⇒ a < 0 and b 2 < 3 a c + d . . . ii by equation i and ii , we take intersection of these ∴ b 2 < 3 a c - d

Let f : X → Y be a function defined by f x = a sin x + π 4 + b cos x + c . If f is both one-one and onto, then find the sets X and Y .

f : X → Y , f x = a sin x + π 4 + b cos x + c ∵ f is one-one and onto. f x = a sin x · cos π 4 + cos x · sin π 4 + b cos x + c = a 2 sin x + a 2 cos x + b cos x + c = a 2 sin x + a 2 + b cos x + c Let, a 2 = r cos α , a 2 + b = r sin α ⇒ r 2 = a 2 2 + a 2 + b 2 and tan α = b + a 2 a 2 ⇒ r = a 2 2 + a 2 2 + 2 a b + b 2 and α = tan - 1 2 b + a a ⇒ r = a 2 + 2 a b + b 2 and α = tan - 1 2 b + a a Then, f x = r sin x · cos α + r cos x · sin α + c = r sin x + α + c Hence, sin x strictly increasing in - π 2 , π 2 ⇒ sin x + α is strictly increasing in - π 2 - α , π 2 - α If f is strictly increasing, then it is one-one. So, x ∈ - π 2 - α , π 2 - α For onto function Codomain of f = Range of f ∵ - 1 ≤ sin x + α ≤ 1 ⇒ - r ≤ r sin x + α ≤ r ⇒ c - r ≤ c + r sin x + α ≤ c + r ⇒ f x ∈ c - r , c + r So, y = c - r , c + r Where, r = a 2 + 2 a b + b 2 and α = tan - 1 2 b + a a Therefore, x ∈ - π 2 - α , π 2 - α and y ∈ c - r , c + r

Find the inverse of the function: f x = sin - 1 x 3 , x ∈ - 3 , 3

Given: f x = sin - 1 x 3 , x ∈ - 3 , 3 The given function is one-one and onto, therefore its inverse can be calculated. y = sin - 1 x 3 ⇒ sin y = x 3 ⇒ 3 sin y = x Replace x by y and y by x in 3 sin y = x we get, ⇒ y = 3 sin x Hence, f - 1 x = 3 sin x

Find the inverse of the function f x = log e x 2 + 3 x + 1 , x ∈ 1 , 3 . Assuming it to be an onto function.

Given: f x = log e x 2 + 3 x + 1 , x ∈ 1 , 3 ⇒ f ' x = 2 x + 3 x 2 + 3 x + 1 ⇒ f ' ( x ) > 0 , ∀ x ∈ 1 , 3 i.e. f x is strictly increasing function. Thus, f x is injective, and it is given that f x is onto. Hence the given function is invertible. Now, let y = f x y = ln x 2 + 3 x + 1 ⇒ e y = x 2 + 3 x + 1 ⇒ x 2 + 3 x + 1 - e y = 0 ⇒ x = - 3 ± 9 - 4 1 - e y 2 ⇒ x = - 3 ± 5 + 4 e y 2 But, x ∈ 1 , 3 (given) ∴ x = - 3 + 5 + 4 e y 2 We replace ' x ' by ' y ' and ' y ' by ' x ' in x = - 3 + 5 + 4 e y 2 we get, y = - 3 + 5 + 4 e x 2 ⇒ f - 1 x = - 3 + 5 + 4 e x 2

Find the inverse of the function: f x = log e x + x 2 + 1 .

Given: f x = log e x + x 2 + 1 Given function is bijective, hence its inverse can be calculated. Let, y = log e x + x 2 + 1 ⇒ e y = x + x 2 + 1 ⇒ e y - x = x 2 + 1 ⇒ e 2 y + x 2 - 2 e y x = 1 + x 2 ⇒ e 2 y - 2 e y x - 1 = 0 ⇒ x = e 2 y - 1 2 e y ⇒ x = 1 2 e y - e - y . . . i Replace ' x ' by ' y ' and ' y ' by ' x ' in equation i , we get ⇒ y = 1 2 e x - e - x ⇒ f - 1 x = 1 2 e x - e - x