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Question: If we have functions \({e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,10} ...

If we have functions ef(x)=10+x10−x,x∈(−10,10){e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}},x \in \left( { - 10,10} \right) and f(x)=k.f(200x100+x2)f\left( x \right) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) then k =
A. 0.5 B. 0.6 C. 0.7 D. 0.8  {\text{A}}{\text{. 0}}{\text{.5}} \\\ {\text{B}}{\text{. 0}}{\text{.6}} \\\ {\text{C}}{\text{. 0}}{\text{.7}} \\\ {\text{D}}{\text{. 0}}{\text{.8}} \\\

Explanation

Solution

Hint: -To solve this question first we have to find f (x ) taking log on both sides and then we have to find f(200x100+x2)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) and thereafter further calculation to get answer.

Complete step-by-step solution -
We have
ef(x)=10+x10−x{e^{f\left( x \right)}} = \dfrac{{10 + x}}{{10 - x}}
Now taking log on both side with base e we get,
log⁡eef(x)=log⁡e10+x10−x{\log _e}{e^{f\left( x \right)}} = {\log _e}\dfrac{{10 + x}}{{10 - x}}
Now as we know the property of log (log⁡aam=mlog⁡aa=m)\left( {{{\log }_a}{a^m} = m{{\log }_a}a = m} \right)
So now,
f(x)=log⁡e10+x10−xf\left( x \right) = {\log _e}\dfrac{{10 + x}}{{10 - x}} ……(i) \ldots \ldots \left( i \right)
Now we have to find f(200x100+x2)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) so we will replace x by (200x100+x2)\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)
On doing this we get,
f(200x100+x2)=log⁡e(10+200x100+x210−200x100+x2)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{10 + \dfrac{{200x}}{{100 + {x^2}}}}}{{10 - \dfrac{{200x}}{{100 + {x^2}}}}}} \right)
Now on further calculation we get,
f(200x100+x2)=log⁡e(1000+10x2+200x100+x21000+10x2−200x100+x2)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{\dfrac{{1000 + 10{x^2} + 200x}}{{100 + {x^2}}}}}{{\dfrac{{1000 + 10{x^2} - 200x}}{{100 + {x^2}}}}}} \right)
On cancel out we get,
f(200x100+x2)=log⁡e(1000+10x2+200x1000+10x2−200x)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{1000 + 10{x^2} + 200x}}{{1000 + 10{x^2} - 200x}}} \right)
Now we can write it as
f(200x100+x2)=log⁡e(100+x2+20x100+x2−20x)=log⁡e((10+x)2(10−x)2)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{100 + {x^2} + 20x}}{{100 + {x^2} - 20x}}} \right) = {\log _e}\left( {\dfrac{{{{\left( {10 + x} \right)}^2}}}{{{{\left( {10 - x} \right)}^2}}}} \right)
f(200x100+x2)=log⁡e((10+x)(10−x))2=2log⁡e(10+x10−x)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = {\log _e}{\left( {\dfrac{{\left( {10 + x} \right)}}{{\left( {10 - x} \right)}}} \right)^2} = 2{\log _e}\left( {\dfrac{{10 + x}}{{10 - x}}} \right)
Now we have following result
f(200x100+x2)=2log⁡e(10+x10−x)f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right) = 2{\log _e}\left( {\dfrac{{10 + x}}{{10 - x}}} \right) ……(ii) \ldots \ldots \left( {ii} \right)
And we have given in the question
f(x)=k.f(200x100+x2)f\left( x \right) = k.f\left( {\dfrac{{200x}}{{100 + {x^2}}}} \right)
Using results obtained from equation (i) and equation ( ii ) we get,
log⁡e10+x10−x=k.2log⁡(10+x10−x){\log _e}\dfrac{{10 + x}}{{10 - x}} = k.2\log \left( {\dfrac{{10 + x}}{{10 - x}}} \right)
On cancel out we get,
1 = 2k
∴k=12=0.5\therefore k = \dfrac{1}{2} = 0.5
Hence option A is the correct option.

Note:- Whenever we get this type of question the key concept of solving is we have knowledge of function as well as good understanding of function and relation. This solution is a standard method of solving this type of question so keep in mind this method for further this type of question.