If the function f is defined by f x = x + 2 , x > 1 2 , - 1 ≤ x ≤ 1 x - 1 , - 3 < x < - 1 , then find the values of f 3 , f 0 , f - 1 . 5 , f 2 + f - 2 and f - 5 .

The given function is f x = x + 2 , x > 1 2 , - 1 ≤ x ≤ 1 x - 1 , - 3 < x < - 1 Here, the domain of the function f is - 3 , - 1 ∪ - 1 , 1 ∪ 1 , ∞ . Since, f x = x + 2 , for x > 1 , we have f 3 = 3 + 2 = 5 Since, f x = 2 , for x ∈ - 1 , 1 , we have f 0 = 2 Since, f x = x - 1 , for x ∈ - 3 , - 1 , we have f - 1 . 5 = - 1 . 5 - 1 = - 2 . 5 Since, f x = x + 2 , for x > 1 , we have f 2 = 2 + 2 = 4 and for x ∈ - 3 , - 1 , we have f x = x - 1 ⇒ f - 2 = - 2 - 1 = - 3 ∴ f 2 + f - 2 = 4 - 3 = 1 As we can see that - 5 does not belong to the domain of f , So f - 5 is not defined.

If f : R \ 0 → R is defined by f x = x 3 - 1 x 3 , then show that f x + f 1 x = 0 .

The given function is f x = x 3 - 1 x 3 . . . i Here, f 1 x = 1 x 3 - 1 1 x 3 = 1 x 3 - 1 1 x 3 ∴ f 1 x = 1 x 3 - x 3 . . . ii Now, by adding i and ii , we have f x + f 1 x = x 3 - 1 x 3 + 1 x 3 - x 3 ⇒ x 3 - x 3 + 1 x 3 - 1 x 3 = 0 Hence, f x + f 1 x = 0

If f : R → R is defined by f x = 1 - x 2 1 + x 2 , then show that f tan θ = cos 2 θ .

The given function is f x = 1 - x 2 1 + x 2 . . . i Here, f tan θ = 1 - tan 2 θ 1 + tan 2 θ = 1 - sin 2 θ cos 2 θ 1 + sin 2 θ cos 2 θ = cos 2 θ - sin 2 θ cos 2 θ cos 2 θ + sin 2 θ cos 2 θ = cos 2 θ - sin 2 θ ⇒ cos θ · cos θ - sin θ · sin θ = cos θ + θ ∵ cos A + B = cos A · cos B - sin A · sin B Hence, f tan θ = cos 2 θ

If f : R \ ± 1 → R is defined by f x = log 1 + x 1 - x , then show that f 2 x 1 + x 2 = 2 f x .

The given function is f x = log 1 + x 1 - x Here, f 2 x 1 + x 2 = log 1 + 2 x 1 + x 2 1 - 2 x 1 + x 2 = log 1 + x 2 + 2 x 1 + x 2 1 + x 2 - 2 x 1 + x 2 = log 1 + 2 x + x 2 1 - 2 x + x 2 = log 1 + x 2 1 - x 2 ⇒ log 1 + x 1 - x 2 = 2 log 1 + x 1 - x ∵ log a b = b · log a Hence, f 2 x 1 + x 2 = 2 f x

If f x + y = f x y ∀ x , y ∈ R then prove that f is a constant function.

Given, f x + y = f x y ∀ x , y ∈ R Let, x be any real number and y = 0 Then f x + 0 = f x · 0 ⇒ f x = f 0 for all real x which implies that for any value of x the function f x returns the same value which is f 0 . Hence, the given function is a constant function.