If f is a differentiable function on R and f-0-2 satisfying f x-y -f x -f y -1-f x f y then f - -8 is equal to

The correct option is B 1 Let f(x)=tan(ax) . Where 'a' is a constant.Hence f(x+y)=tan(ax+ay) =tanax+tanay1 tanax.tanay =f(x)+f(y)1 f(x) ...

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Standard XII

Mathematics

Continuity of a Function

If f is a dif...

Question

If f is a differentiable function on R and $f^{'} (0) = 2$ satisfying $f (x + y) = \frac{f (x) + f (y)}{1 - f (x) f (y)}$ then $f (π / 8)$ is equal to

1 / 2

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tan π / 8

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Solution

The correct option is B $1$
Let $f (x) = t a n (a x)$ . Where 'a' is a constant.
Hence
$f (x + y) = t a n (a x + a y)$
$= \frac{t a n a x + t a n a y}{1 - t a n a x . t a n a y}$
$= \frac{f (x) + f (y)}{1 - f (x) . f (y)}$
Now
$f^{'} (0) = 2$
Or
$s e c^{2} (0) . a = 2$
Or
$a = 2$ .
Hence
$f (x) = t a n (2 x)$ .
Then $f (\frac{π}{8}) = t a n (\frac{π}{4}) = 1$ .

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If f is a differentiable function on R and f′(0)=2 satisfying f(x+y)=f(x)+f(y)1−f(x)f(y) then f(π/8) is equal to

The correct option is B 1Let f(x)=tan(ax). Where 'a' is a constant.Hencef(x+y)=tan(ax+ay)=tanax+tanay1−tanax.tanay=f(x)+f(y)1−f(x).f(y)Now f′(0)=2Or sec2(0).a=2Or a=2.Hencef(x)=tan(2x).Then f(π8)=tan(π4)=1.

If f is a differentiable function on R and $f^{'} (0) = 2$ satisfying $f (x + y) = \frac{f (x) + f (y)}{1 - f (x) f (y)}$ then $f (π / 8)$ is equal to