Question

e^f(x) =10+x/10-x
x€(-10,10)
f(x)=k.f(200x/100+x^2)
Then k is equal to?
1) 0.5
2) 0.6
3) 0.7
4) 0.8

Answer 1

$\underline{\textsf{GIVEN:}}$

$\mathsf{e^{f(x)}=\dfrac{10+x}{10-x}\;\;and\;\;f(x)=k\,f\left(\dfrac{200x}{100+x^2}\right)}$

$\underline{\textsf{TO FIND:}}$

$\textsf{The value of k}$

$\underline{\textsf{SOLUTION:}}$

$\mathsf{Consider,}$

$\mathsf{e^{f(x)}=\dfrac{10+x}{10-x}}$

$\textsf{Take logarithm on bothsides, we ge}$

$\mathsf{\log{e^{f(x)}}=\log\left(\dfrac{10+x}{10-x}\right)}$

$\mathsf{f(x)\,\log\,e=\log\left(\dfrac{10+x}{10-x}\right)}$

$\implies\mathsf{f(x)=\log\left(\dfrac{10+x}{10-x}\right)}$

$\mathsf{Now,}$

$\mathsf{f(x)=k\,f\left(\dfrac{200x}{100+x^2}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{10+\dfrac{200x}{100+x^2}}{10-\dfrac{200x}{100+x^2}}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{\dfrac{1000+10x^2+200x}{100+x^2}}{\dfrac{1000+10x^2-200x}{100+x^2}}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{10x^2+200x+1000}{10x^2-200x+1000}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{10(x^2+20x+100)}{10(x^2-20x+100)}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{x^2+20x+100}{x^2-20x+100}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{(10+x)^2}{(10-x)^2}\right)}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=k\,\log\left(\dfrac{10+x}{10-x}\right)^2}$

$\textsf{Using power rule of logarithm, we get}$

$\mathsf{\log\left(\dfrac{10+x}{10-x}\right)=2k\,\log\left(\dfrac{10+x}{10-x}\right)}$

$\mathsf{1=2k}$

$\implies\mathsf{k=\dfrac{1}{2}}$

$\implies\boxed{\mathsf{k=0.5}}$

$\underline{\textsf{ANSWER:}}$

$\textsf{Option (1) is correct}$

Answer 2

Answer:

[math/100 + x] = [math/100 - x]