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

Jitender Singh IIT Delhi
6 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