If a, b be two fixed positive integers such that f(a + x) = b + {[{b^3} + 1

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1.Relation and Function

hard

If $a, b$ be two fixed positive integers such that $f(a + x) = b + {[{b^3} + 1 - 3{b^2}f(x) + 3b{\{ f(x)\} ^2} - {\{ f(x)\} ^3}]^{\frac{1}{3}}}$ for all real $x$, then $f(x)$ is a periodic function with period

$a$

$2a$

$b$

$2b$

Solution

(b) $f(a + x) = b + {(1 + {\{ b – f(x)\} ^3})^{1/3}}$
==> $f(a + x) – b = {\{ 1 – {\{ f(x) – b\} ^3}\} ^{1/3}}$
==> $\phi (a + x) = {\{ 1 – {\{ \phi (x)\} ^3}\} ^{1/3}}$, [$\phi (x) = f(x) – b$]
==> $\phi (x + 2a) = {\{ 1 – {\{ \phi (x + a)\} ^3}\} ^{1/3}} = \phi (x)$
==> $f(x + 2a) – b = f(x) – b \Rightarrow f(x + 2a) = f(x)$
$\therefore$ $f(x)$ is periodic with period $2a$.

Standard 12

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hard

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medium

(IIT-1995)

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