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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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Correct option is A

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

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Ready to Test Your Skills?

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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 =