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

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1 Answer
Jun 17, 2018

Answer:

Kindly refer to the Explanation.

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).#