If e f ( x ) = 10 − x 10 + x , x ∈ ( − 10 , 10 ) and f ( x ) = k f ( 100 + x 2 200 x ) , then k =

f ( x ) = lo g e ( 10 − x 10 + x ) f ( 100 + x 2 200 x ) = lo g e ( 10 − 100 + x 2 200 x 10 + 100 + x 2 200 x ) = 2 lo g ( 10 − x 10 + x ) = 2 f ( x )

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Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsFunctional EquationEasy2 minQB

MathematicsEasysingle choice

If $e^{f (x)} = \frac{10 + x}{10 - x}, x \in (- 10, 10)$ and $f (x) = k f (\frac{200 x}{100 + x ^{2}})$ , then $k =$

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Answer:

Solution:

$f (x) = lo g_{e} (\frac{10 + x}{10 - x})$

$f (\frac{200 x}{100 + x ^{2}}) = lo g_{e} (\frac{10 + \frac{200 x}{100 + x ^{2}}}{10 - \frac{200 x}{100 + x ^{2}}}) = 2 lo g (\frac{10 + x}{10 - x}) = 2 f (x)$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Functional Equation

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