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

Jun 17, 2018

Kindly refer to the Explanation.

#### Explanation:

Given that, ${e}^{f \left(x\right)} = \left(\frac{10 + x}{10 - x}\right) , x \in \left(- 10 , 10\right) .$

$\therefore \ln {e}^{f \left(x\right)} = \ln \left(\frac{10 + x}{10 - x}\right)$.

$\therefore f \left(x\right) \cdot \ln e = \ln \left(\frac{10 + x}{10 - x}\right) ,$

$i . e . , f \left(x\right) = \ln \left(\frac{10 + x}{10 - x}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\ast}_{1}\right)$.#,

$\mathmr{and} , f \left(x\right) = \ln \left(10 + x\right) - \ln \left(10 - x\right)$.

Plugging in $\frac{200 x}{100 + {x}^{2}}$ in place of $x$, we get,

$f \left(\frac{200 x}{100 + {x}^{2}}\right)$,

$= \ln \left\{10 + \frac{200 x}{100 + {x}^{2}}\right\} - \ln \left\{10 - \frac{200 x}{100 + {x}^{2}}\right\}$,

$= \ln \left\{\frac{1000 + 10 {x}^{2} + 200 x}{100 + {x}^{2}}\right\} - \ln \left\{\frac{1000 + 10 {x}^{2} - 200 x}{100 + {x}^{2}}\right\}$,

$= \ln \left[\frac{10 \left(100 + {x}^{2} + 20 x\right)}{100 + {x}^{2}}\right] - \ln \left[\frac{10 \left(100 + {x}^{2} - 20 x\right)}{100 + {x}^{2}}\right]$,

$= \ln \left[\frac{10 \left(100 + {x}^{2} + 20 x\right)}{100 + {x}^{2}} \div \frac{10 \left(100 + {x}^{2} - 20 x\right)}{100 + {x}^{2}}\right]$,

$= \ln \left\{\frac{100 + {x}^{2} + 20 x}{100 + {x}^{2} - 20 x}\right\}$,

$= \ln \left\{{\left(\frac{10 + x}{10 - x}\right)}^{2}\right\}$.

Thus, $f \left(\frac{200 x}{100 + {x}^{2}}\right) = \ln \left\{{\left(\frac{10 + x}{10 - x}\right)}^{2}\right\} \ldots \ldots \ldots . . \left({\ast}_{2}\right)$.

Now, utilising $\left({\ast}_{1}\right) \mathmr{and} \left({\ast}_{2}\right)$ in

$f \left(x\right) = k \cdot f \left(\frac{200 x}{100 + {x}^{2}}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \text{[Given]}$, we get,

$\ln \left(\frac{10 + x}{10 - x}\right) = k \cdot \ln \left\{{\left(\frac{10 + x}{10 - x}\right)}^{2}\right\}$,

$i . e . , \ln \left(\frac{10 + x}{10 - x}\right) = \ln {\left(\frac{10 + x}{10 - x}\right)}^{2 k}$.

$\therefore 1 = 2 k , \mathmr{and} , k = \frac{1}{2} = 0.5 , \text{ which is the option } \left(1\right) .$