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


A
0.5
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B
0.6
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C
0.7
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D
0.8
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Solution

The correct option is D $$0.5$$
Given $$\displaystyle f(x)=k.f\left ( \frac{200x}{100+x^{2}} \right ) $$      .....(1)
Also given $$e^{f(x)}=\displaystyle \frac{10+x}{10-x}$$

$$\Rightarrow \displaystyle f(x)=\log \frac{10+x}{10-x}$$

$$\Rightarrow\displaystyle f\left(\frac{200x}{100+x^{2}}\right)=\log  \left(\frac { 10+\displaystyle \frac { 200x }{ 100+x^{ 2 } }  }{ 10-\displaystyle \frac { 200x }{ 100+x^{ 2 } }  } \right)$$

$$\displaystyle =\log \frac{(10+x)^{2}}{(10-x)^{2}}$$

$$\displaystyle =2 \log \frac{10+x}{10-x}$$

$$=2f(x)$$

$$\therefore\displaystyle  f(x)=\frac{1}{2}f\left(\frac{200x}{100+x^{2}}\right)$$

On comparing with (1), we get

$$ \displaystyle k=\frac{1}{2}=0.5$$

Mathematics

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