Let f: [0, π/2] → R: f(x) = sin x and g: [0, π/2] → R: g(x) = co

1.	Let f: [0, π/2] → R: f(x) = sin x and g: [0, π/2] → R: g(x) = cos x. Show that each one of f and g is one-one but (f + g) is not one-one.
Answer» For any two distinct elements x₁ and x₂in [0, π/2] We know that sin x₁≠ sin x₂ and cos x₁ ≠ cos x₂ So we get f(x₁) ≠ f(x₂) and g(x₁) ≠ g(x₂) Here, both f and g are one-one We know that (f + g) (x) = f( x) + g(x) = sin x + cos x By substituting the values (f + g) (0) = sin 0 + cos 0 = 1 (f + g) (π/2) = sin π/2 + cos π/2 = 1 We know that 0 ≠ π/2 Here, (f + g) (0) = (f + g) (π/2) So, f + g is not one-one.

Let f: [0, π/2] → R: f(x) = sin x and g: [0, π/2] → R: g(x) = cos x. Show that each one of f and g is one-one but (f + g) is not one-one.

Answer»

For any two distinct elements x₁ and x₂in [0, π/2]

We know that

sin x₁≠ sin x₂ and cos x₁ ≠ cos x₂

So we get

f(x₁) ≠ f(x₂) and g(x₁) ≠ g(x₂)

Here, both f and g are one-one

We know that

(f + g) (x) = f( x) + g(x) = sin x + cos x

By substituting the values

(f + g) (0) = sin 0 + cos 0 = 1

(f + g) (π/2) = sin π/2 + cos π/2 = 1

We know that 0 ≠ π/2

Here, (f + g) (0) = (f + g) (π/2)

So, f + g is not one-one.