First slide

Functions (XII)

Question

Let f be a real valued function with domain R satisfying f (x + k) = 1 + [2 – 5 f (x) + 10 { f (x)}²– 10 { f (x)}³ + 5 { f (x)}⁴ – { f (x)}5]^1/5
for all real x and some positive constant k, then the period of the function f (x) is

Moderate

A

k
B

2k
C

Non periodic
D

None of these

Solution

We have,
f (x + k) = 1 + [1 + {1 – f (x)}⁵]^1/5
⇒ f (x + k) – 1 = [1 – ( f (x) – 1)⁵]^1/5
$\begin{array}{l} \Rightarrow ϕ (x + k) = {[1 - {ϕ (x)}^{5}]}^{1 / 5} \\ where ϕ (x) = f (x) - 1 \\ \Rightarrow ϕ (x + 2 k) = {[1 - {ϕ (x + k)}^{5}]}^{1 / 5} \\ \Rightarrow ϕ (x + 2 k) = {[1 - \{1 - (ϕ (x))^{5}\}]}^{1 / 5} = ϕ (x), \forall x \in R \end{array}$
⇒ f (x + 2k) – 1 = f (x) – 1
⇒ f (x + 2k) = f (x), ∀ x ∈ R
$∴$ f (x) is periodic with period 2k.

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