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EAMCET Engineering 2003 Solved Paper

Section: Mathematics

Question : 136 of 160

Marks: +1, -0

e^{f(x)}=\frac{10+x}{10-x}, x\in(-10,10)

f(x)=k f\left(\frac{200x}{100+x^2}\right)

k

Solution:

We have,

e^{f(x)}=\frac{10+x}{10-x}

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

\text{ Given that } f(x)=k f\left(\frac{200x}{100+x^2}\right)

\Rightarrow \log\frac{10+x}{10-x}=k\cdot\log\left\{\frac{10+\frac{200x}{100+x^2}}{10-\frac{200x}{100+x^2}}\right\}

=k\log\left\{\frac{1000+10x^2+200x}{1000+10x^2-200x}\right\}

=k\log\left\{\frac{x^2+100+20x}{x^2+100-20x}\right\}

=k\log\left\{\frac{(x+10)^2}{(10-x)^2}\right\}

=2k\log\frac{10+x}{10-x}

\Rightarrow \log\frac{10+x}{10-x}=2k\log\frac{10+x}{10-x}

\therefore 2k=1 \Rightarrow k=\frac{1}{2}=0.5

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