If ef(x)=10+x10−x,x∈(−10,10) and f(x)=kf200x100+x2then k =

ef(x)=10+x10−x⇒f(x)=log⁡10+x10−x∴f200x100+x2=log⁡10+200x100+x210−200x100+x2=log⁡1000+200x+10x21000−200x2+10x2=log⁡100+20x+x2100−20x+x2=log⁡10+x10−x2=2log⁡10+x10−x=2f(x) By the given condition f(x)=k.2f(x)⇒k=12

If ef(x)=10+x10−x,x∈(−10,10) and f(x)=kf200x100+x2then k =

ef(x)=10+x10−x⇒f(x)=log⁡10+x10−x∴f200x100+x2=log⁡10+200x100+x210−200x100+x2=log⁡1000+200x+10x21000−200x2+10x2=log⁡100+20x+x2100−20x+x2=log⁡10+x10−x2=2log⁡10+x10−x=2f(x) By the given condition f(x)=k.2f(x)⇒k=12

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

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Detailed Solution

$\begin{array}{l} e^{f (x)} = \frac{10 + x}{10 - x} \Rightarrow f (x) = \log (\frac{10 + x}{10 - x}) \\ ∴ f (\frac{200 x}{100 + x^{2}}) = \log (\frac{10 + \frac{200 x}{100 + x^{2}}}{10 - \frac{200 x}{100 + x^{2}}}) \\ = \log (\frac{1000 + 200 x + 10 x^{2}}{1000 - 200 x^{2} + 10 x^{2}}) \\ = \log (\frac{100 + 20 x + x^{2}}{100 - 20 x + x^{2}}) \\ = \log {(\frac{10 + x}{10 - x})}^{2} = 2 \log (\frac{10 + x}{10 - x}) = 2 f (x) \end{array}$