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Differential Calculus> If f (x)+f (y) = f [x+y/1-xy] and f'(0) =...

question mark

If f (x)+f (y) = f [x+y/1-xy] and f'(0) =3
Then f [tan1/3]=

ashu , 10 Years ago

Grade 11

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

$f(x) + f(y) = f(\frac{x+y}{1-xy})$

$y\rightarrow -x$

$f(x) + f(-x) = f(\frac{x+(-x)}{1-x(-x)}) = f(0)$

$x = 0$

$2f(0) = f(0)$

$f(0) = 0$

$\Rightarrow f(x) + f(-x) = 0$

f(x) is a odd function.

$f(x) + f(-y) = f(\frac{x+(-y)}{1-x(-y)})$

$f(-y) = -f(y)$

$f(x) - f(-y) = f(\frac{x-y}{1+xy})$

$\frac{dy}{dx} = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$\frac{dy}{dx} = \lim_{h\rightarrow 0}\frac{f(\frac{x+h-x}{1+x(x+h)})}{h}$

$\frac{dy}{dx} = \lim_{h\rightarrow 0}\frac{f(\frac{h}{1+x(x+h)})}{\frac{h}{1+x(x+h)}}.\frac{1}{1+x(x+h)}$

$f'(0) = 3$

$\lim_{x\rightarrow 0}\frac{f(x)}{x} = f'(0) = 3$

$\frac{dy}{dx} = \frac{3}{1+x^{2}}$

$f(x) = 3tan^{-1}x +c$

$0 = tan^{-1}(0) +c$

$c = 0$

$f(x) = 3tan^{-1}(x)$

$f(tan\frac{1}{3}) = 3tan^{-1}(tan\frac{1}{3}) = 3.\frac{1}{3} = 1$

Thanks & Regards

Jitender Singh

IIT Delhi

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