It's the 2nd question. Circled up n written as doubt. Can anyone help me get through this?

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Answer 1 · 2018-06-17T13:08:47

Kindly refer to the Explanation.

Given that, $e^(f(x))=((10+x)/(10-x)), x in (-10,10).$

$:. lne^(f(x))=ln((10+x)/(10-x))$ .

$:. f(x)*lne=ln((10+x)/(10-x)),$

$i.e., f(x)=ln((10+x)/(10-x))..........................(ast_1)$ .#,

$or, f(x)=ln(10+x)-ln(10-x)$ .

Plugging in $(200x)/(100+x^2)$ in place of $x$ , we get,

$f((200x)/(100+x^2))$ ,

$=ln{10+(200x)/(100+x^2)}-ln{10-(200x)/(100+x^2)}$ ,

$=ln{(1000+10x^2+200x)/(100+x^2)}-ln{(1000+10x^2-200x)/(100+x^2)}$ ,

$=ln[{10(100+x^2+20x)}/(100+x^2)]-ln[{10(100+x^2-20x)}/(100+x^2)]$ ,

$=ln[{10(100+x^2+20x)}/(100+x^2)-:{10(100+x^2-20x)}/(100+x^2)]$ ,

$=ln{(100+x^2+20x)/(100+x^2-20x)}$ ,

$=ln{((10+x)/(10-x))^2}$ .

Thus, $f((200x)/(100+x^2))=ln{((10+x)/(10-x))^2}...........(ast_2)$ .

Now, utilising $(ast_1) and (ast_2)$ in

$f(x)=k*f((200x)/(100+x^2))......................."[Given]"$ , we get,

$ln((10+x)/(10-x))=k*ln{((10+x)/(10-x))^2}$ ,

$i.e., ln((10+x)/(10-x))=ln((10+x)/(10-x))^(2k)$ .

$:. 1=2k, or, k=1/2=0.5," which is the option "(1).$