SATHEE: Relations Question 38

Relations Question 38

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

Solution:

Answer: 0.5

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Relations Question 38

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

Solution:

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