Consider the function f : 0 , π 2 → ℝ given by f ( x ) = sin ⁡ x and g : 0 , π 2 → ℝ given by g x = cos x . Is f + g is one-one?why?

Given that the function f : 0 , π 2 → ℝ given by f ( x ) = sin ⁡ x and g : 0 , π 2 → ℝ given by g x = cos x . One-One function states that every element of the function's codomain is the image of at most one element of its domain. We know that f + g x = f x + g x ⇒ f + g x = sin x + cos x Now, f + g 0 = sin 0 + cos 0 = 0 + 1 = 1 f + g π 2 = sin π 2 + cos π 2 = 1 + 0 = 1 Since different elements 0 , π 2 have same image 1 Therefore, f + g is not a one-one function.

Consider the function f : 0 , π 2 → ℝ given by f ( x ) = sin ⁡ x and g : 0 , π 2 → ℝ given by g x = cos x . Show that f and g are one-one functions.

Given that the function f : 0 , π 2 → ℝ given by f ( x ) = sin ⁡ x and g : 0 , π 2 → ℝ given by g x = cos x . One-One function states that every element of the function's codomain is the image of at most one element of its domain. Consider f x = sin x Now, f x 1 = f x 2 ⇒ sin x 1 = sin x 2 ⇒ x 1 = x 2 Consider g x = cos x Now, g x 1 = g x 2 ⇒ cos x 1 = cos x 2 ⇒ x 1 = x 2 Therefore, f and g are one-one functions.

The function P is defined as “To each person on the earth is assigned a date of birth”. Is this function one-one ? Given reason.

One-One function states that every element of the function's codomain is the image of at most one element of its domain. The function is not one-one because many people are born on same day.

Let u and v be two functions defined on ℝ as u ( x ) = 2 x − 3 and v ( x ) = 3 + x 2 . Prove that u and v are inverse to each other.

Given u and v be two functions defined on ℝ as u ( x ) = 2 x − 3 and v ( x ) = 3 + x 2 Now, u ∘ v x = u v x = u 3 + x 2 = 2 3 + x 2 - 3 = 3 + x - 3 = x And v ∘ u x = v u x = v 2 x - 3 = 3 + 2 x - 3 2 = 2 x 2 = x Therefore, u and v are inverse to each other.