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

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

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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
askIITians Faculty

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