If f ( x ) f ( x 1 ) = f ( x ) + f ( x 1 ) ∀ x ∈ R − { 0 } , where f(x) be a polynomial function and f(5)=126, then f ( 3 ) =

f ( x ) = 1 ± x n or f ( 5 ) = 1 ± 5 n or, 126 = 1 ± 5 n or ± 5 n = 125 ⇒ ± 5 n = 5 3 n = 3 f ( 3 ) = 1 + 3 3 = 28

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If $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) \forall x \in R - {0}$ , where f(x) be a polynomial function and f(5)=126, then $f (3) =$

Answer:

Solution:

$f (x) = 1 \pm x^{n}$ or $f (5) = 1 \pm 5^{n}$

or, $126 = 1 \pm 5^{n}$ or $\pm 5^{n} = 125 \Rightarrow \pm 5^{n} = 5^{3}$

$n = 3$

$f (3) = 1 + 3^{3} = 28$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Functional Equation

⏱ 2mℹ️ Source: QB

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