If efx -10-x10-x- x- -10-10 and fx-kf200x100-x2 then k-

F(x)=ln(10+x10 x) Put x=200x100+x^2 ⇒ f (200x100+x^2)=ln(10+200x100+x^210 200x100+x^2) =ln(100+x^2+20x100+x^2 20x) ⇒ f (200x100+x^2)=ln(10+x10 x)^2 =2ln(10+x10 ...

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Byju's Answer

Standard IX

Mathematics

Product Law

If efx =10+x1...

Question

If $e^{f (x)} = \frac{10 + x}{10 - x}, x \in (- 10, 10)$ and
$f (x) = k f (\frac{200 x}{100 + x^{2}})$ then $k =$

0.5

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0.6

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Solution

The correct option is A $0.5$
$f (x) = ln (\frac{10 + x}{10 - x})$
Put $x = \frac{200 x}{100 + x^{2}}$
$\Rightarrow f (\frac{200 x}{100 + x^{2}}) = ln ⎛ ⎜ ⎜ ⎜ ⎝ \frac{10 + \frac{200 x}{100 + x^{2}}}{10 - \frac{200 x}{100 + x^{2}}} ⎞ ⎟ ⎟ ⎟ ⎠$
$= ln (\frac{100 + x^{2} + 20 x}{100 + x^{2} - 20 x})$
$\Rightarrow f (\frac{200 x}{100 + x^{2}}) = ln {(\frac{10 + x}{10 - x})}^{2} = 2 ln (\frac{10 + x}{10 - x}) = 2 f (x)$
$\Rightarrow f (x) = \frac{1}{2} f (\frac{200 x}{100 + x^{2}})$
$∴ k = \frac{1}{2} = 0.5$

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If ef(x)=10+x10−x,x∈(−10,10) and f(x)=kf(200x100+x2) then k=

The correct option is A 0.5f(x)=ln(10+x10−x) Put x=200x100+x2 ⇒f(200x100+x2)=ln⎛⎜ ⎜ ⎜⎝10+200x100+x210−200x100+x2⎞⎟ ⎟ ⎟⎠ =ln(100+x2+20x100+x2−20x) ⇒f(200x100+x2)=ln(10+x10−x)2=2ln(10+x10−x)=2f(x) ⇒f(x)=12f(200x100+x2) ∴k=12=0.5

If $e^{f (x)} = \frac{10 + x}{10 - x}, x \in (- 10, 10)$ and
$f (x) = k f (\frac{200 x}{100 + x^{2}})$ then $k =$